328 free, high-quality past-paper-style questions on Functions for IB Maths AA HL, each with a full mark scheme. Covers: Equations of straight lines, parallel and perpendicular lines; Functions, domain, range and inverse functions; Composite functions and inverse functions; and more. For when you have finished the past papers, or want to practise one topic at a time.Topic 2 at Standard Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.2,686 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1AAHL2-q00012.1 Equations of straight lines, parallel and perpendicular linesPaper 1Easy9 marks
The following diagram shows the triangle ABC with vertices \(\mathrm{A}(-4,\,-2)\), \(\mathrm{B}(0,\,-4)\) and \(\mathrm{C}(3,\,5)\). An altitude of a triangle is the line through a vertex that is perpendicular to the opposite side.
(a)
Find the equation of the altitude through C, giving your answer in the form \(y=mx+c\).
(2)
(b)
Find the equation of the altitude through A, giving your answer in the form \(ax+by+d=0\), where \(a,b,d\in\mathbb{Z}\).
(3)
(c)
The two altitudes in parts (a) and (b) meet at the point H. Find the coordinates of H.
(2)
(d)
Show that H also lies on the altitude through B.
(2)
Show mark scheme
Working
Mark
Notes
(a)
gradient of \([\mathrm{AB}]=\dfrac{-4-(-2)}{0-(-4)}=-\dfrac{1}{2}\), so the altitude has gradient \(2\)
M1
\(y=2x-1\)
A1
From \(y-5=2(x-3)\).
(b)
gradient of \([\mathrm{BC}]=\dfrac{5-(-4)}{3-0}=3\)
(A1)
\(y+2=-\dfrac{1}{3}(x+4)\)
M1
Perpendicular gradient \(-\tfrac{1}{3}\) used with the point A.
\(x+3y+10=0\)
A1
Accept any integer multiple.
(c)
attempt to solve simultaneously, e.g. \(x+3(2x-1)+10=0\)
M1
\(\mathrm{H}(-1,\,-3)\)
A1
(d)
gradient of \([\mathrm{CA}]=\dfrac{5-(-2)}{3-(-4)}=1\), so the altitude through B is \(y=-x-4\)
M1
substituting \(x=-1\): \(y=1-4=-3\), so H lies on it
4AAHL2-q00142.5 Composite functions and inverse functionsPaper 1Easy7 marks
Let \(f(x)=ax+b\), for \(x\in\mathbb{R}\), where \(a,b\in\mathbb{R}\). It is given that \((f\circ f)(x)=9x-8\) for all \(x\in\mathbb{R}\).
(a)
Show that \(a^{2}=9\) and \(b(a+1)=-8\).
(2)
(b)
Find the two possible functions \(f\).
(3)
(c)
For each function \(f\) found in part (b), find \(f^{-1}(x)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\((f\circ f)(x)=a(ax+b)+b=a^{2}x+ab+b\)
M1
comparing coefficients: \(a^{2}=9\) and \(ab+b=-8\)
A1
as required
AG
(b)
\(a=3\) or \(a=-3\)
A1
\(f(x)=3x-2\)
A1
From \(4b=-8\).
\(f(x)=-3x+4\)
A1
From \(-2b=-8\).
(c)
\(f^{-1}(x)=\dfrac{x+2}{3}\)
A1
\(f^{-1}(x)=\dfrac{4-x}{3}\)
A1
FT from (b).
Answers: (b) \(f(x)=3x-2\) or \(f(x)=-3x+4\) · (c) \(\dfrac{x+2}{3}\); \(\dfrac{4-x}{3}\)
5AAHL2-q00172.6 The quadratic function and its formsPaper 1Easy9 marks
A café sells \(q\) cups of a drink each week when the price of a cup is \(p\) dollars, where \(q=1200-40p\), for \(0\le p\le30\). The weekly revenue from the drink is \(R(p)\) dollars.
(a)
Show that \(R(p)=1200p-40p^{2}\).
(1)
(b)
Write \(R(p)\) in the form \(a(p-h)^{2}+k\).
(3)
(c)
Write down the maximum weekly revenue and the price at which it occurs.
(2)
(d)
Find the range of prices for which the weekly revenue is at least $8000.
8AAHL2-q00312.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy10 marks
Let \(P(x)=2x^{3}+ax^{2}+bx+6\), where \(a,b\in\mathbb{R}\). The polynomial \(P(x)\) has a factor \((x+3)\), and when \(P(x)\) is divided by \((x-1)\) the remainder is \(-4\).
(a)
Find the value of \(a\) and the value of \(b\).
(5)
(b)
Hence factorize \(P(x)\) completely.
(3)
(c)
Find the remainder when \(P(x)\) is divided by \((x+1)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(P(-3)=0\): \(2(-27)+9a-3b+6=0\)
M1
Attempt to use the factor theorem.
\(9a-3b=48\) (or \(3a-b=16\))
A1
\(P(1)=-4\): \(2+a+b+6=-4\)
M1
Attempt to use the remainder theorem.
\(a+b=-12\)
A1
\(a=1,\ b=-13\)
A1
Both values required.
(b)
dividing \(2x^{3}+x^{2}-13x+6\) by \((x+3)\), or comparing coefficients
M1
\(P(x)=(x+3)(2x^{2}-5x+2)\)
A1
\(P(x)=(x+3)(2x-1)(x-2)\)
A1
FT from their quadratic factor.
(c)
\(P(-1)=-2+1+13+6\)
M1
Award M1 for substituting \(x=-1\) into their \(P(x)\).
11AAHL2-q00452.13 Rational functions with oblique and horizontal asymptotesPaper 1Easy10 marks
Let \(f(x)=\dfrac{x+3}{(x-1)^{2}}\), for \(x\neq1\). The graph of \(f\) has a local minimum point at \(\left(-7,\,-\tfrac{1}{16}\right)\) and no other turning points.
(a)
Write down the equation of the vertical asymptote and the equation of the horizontal asymptote of the graph of \(f\).
(2)
(b)
Find the coordinates of the points where the graph of \(f\) meets the axes.
(2)
(c)
Sketch the graph of \(y=f(x)\), showing clearly the asymptotes, the axis intercepts and the local minimum point.
(3)
(d)
Write down the range of \(f\).
(1)
(e)
Find the set of values of \(k\) for which the equation \(f(x)=k\) has exactly two solutions.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(x=1\)
A1
\(y=0\)
A1
(b)
\((-3,\,0)\)
A1
\((0,\,3)\)
A1
(c)
both branches tending to \(+\infty\) on either side of \(x=1\)
A1
The numerator is positive near \(x=1\) and the denominator is a square.
right branch above the \(x\)-axis approaching \(y=0\); left branch passing through \((0,3)\) and \((-3,0)\)
A1
left branch with minimum at \(\left(-7,-\tfrac{1}{16}\right)\), approaching \(y=0\) from below as \(x\to-\infty\)
A1
(d)
\(f(x)\ge-\dfrac{1}{16}\)
A1
(e)
\(-\dfrac{1}{16}\lt k\lt 0\)
A1
For \(k=0\) there is one solution, \(x=-3\); for \(k=-\tfrac{1}{16}\) there is one solution, \(x=-7\).
\(k\gt 0\)
A1
Award A1A0 for \(k\gt-\tfrac{1}{16}\) without excluding \(k=0\).
12AAHL2-q00472.13 Rational functions with oblique and horizontal asymptotesPaper 1Easy8 marks
Let \(f(x)=\dfrac{ax+b}{x^{2}+dx+e}\), where \(a,b,d,e\in\mathbb{Z}\). The graph of \(f\) has vertical asymptotes \(x=-1\) and \(x=2\), crosses the \(x\)-axis at \(\left(\tfrac{3}{4},\,0\right)\) and crosses the \(y\)-axis at \(\left(0,\,\tfrac{3}{2}\right)\).
15AAHL2-q00622.15 Solving inequalities g(x) ≥ f(x)Paper 1Easy5 marks
Let \(p(x)=x^{3}-2x^{2}-5x+6\).
(a)
Show that \((x-1)\) is a factor of \(p(x)\).
(1)
(b)
Hence factorize \(p(x)\) fully.
(2)
(c)
Solve the inequality \(p(x)\ge0\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(p(1)=1-2-5+6=0\)
A1
Must see the substitution.
so \((x-1)\) is a factor by the factor theorem
AG
(b)
\(p(x)=(x-1)(x^{2}-x-6)\)
(M1)
By division, synthetic division or comparing coefficients.
\(p(x)=(x-1)(x+2)(x-3)\)
A1
(c)
critical values \(-2,\ 1,\ 3\) and a sign diagram or sketch of the positive cubic
(M1)
\(-2\le x\le1\) or \(x\ge3\)
A1
FT from (b). Award A0 for strict inequalities or for only one interval.
Answers: (b) \((x-1)(x+2)(x-3)\) · (c) \(-2\le x\le1\) or \(x\ge3\)
16AAHL2-q00632.15 Solving inequalities g(x) ≥ f(x)Paper 1Easy6 marks
Let \(f(x)=x^{2}-2x-3\) and \(g(x)=x+1\). The following diagram shows the graphs of \(y=f(x)\) and \(y=g(x)\). The graphs intersect at the points P and Q. The point P lies on the \(x\)-axis.
(a)
Find the coordinates of Q.
(3)
(b)
Hence write down the solution of the inequality \(f(x)\le g(x)\).
(1)
(c)
Find the set of values of \(x\) for which \(g(x)\ge f(x)\ge0\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(x^{2}-2x-3=x+1\)
M1
\(x^{2}-3x-4=0\Rightarrow(x-4)(x+1)=0\)
A1
Q\((4,\,5)\)
A1
Award A0 if \((-1,\,0)\) is also given as Q.
(b)
\(-1\le x\le4\)
A1
FT from (a).
(c)
\(f(x)\ge0\) for \(x\le-1\) or \(x\ge3\), combined with (b)
18AAHL2-q00762.16 Modulus, reciprocal and other graph transformationsPaper 1Easy6 marks
The following diagram shows the graph of \(y=f(x)\), for \(-4\le x\le4\). The graph consists of three line segments joining the points \((-4,\,-2)\), \((-2,\,2)\), \((1,\,-1)\) and \((4,\,2)\).
(a)
Sketch the graph of \(y=|f(x)|\), for \(-4\le x\le4\).
(2)
(b)
Sketch the graph of \(y=f(|x|)\), for \(-4\le x\le4\).
(2)
(c)
Write down the number of solutions of the equation \(|f(x)|=1\).
(1)
(d)
Write down the solutions of the equation \(f(|x|)=1\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
parts below the \(x\)-axis reflected in the \(x\)-axis; parts above unchanged
A1
vertices at \((-4,\,2)\), \((-3,\,0)\), \((-2,\,2)\), \((0,\,0)\), \((1,\,1)\), \((2,\,0)\), \((4,\,2)\)
A1
Award for at least five correct points.
(b)
graph for \(x\ge0\) unchanged and reflected in the \(y\)-axis
A1
vertices at \((\pm4,\,2)\), \((\pm1,\,-1)\), \((0,\,0)\)
A1
(c)
5
A1
\(x=-3.5,\ -2.5,\ -1,\ 1,\ 3\).
(d)
\(x=\pm3\)
A1
Answers: (c) 5 · (d) \(x=\pm3\)
19AAHL2-q00782.16 Modulus, reciprocal and other graph transformationsPaper 1Easy7 marks
(a)
Solve \(|2x-5|=3x\).
(3)
(b)
Solve \(|2x-5|=|x+1|\).
(2)
(c)
Hence solve \(|2x-5|\lt |x+1|\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(2x-5=3x\) or \(2x-5=-3x\)
M1
\(x=-5\) or \(x=1\)
A1
\(x=1\) only, since \(3x\ge0\) (or \(|2(-5)-5|=15\neq-15\))
A1
Award A0 if \(x=-5\) is not rejected.
(b)
\(2x-5=\pm(x+1)\), or squaring: \(3x^{2}-22x+24=0\)
M1
\(x=6\) or \(x=\dfrac{4}{3}\)
A1
(c)
sketch of both graphs, or test value \(x=2\): \(1\lt 3\)
21IBMA2H-047-01-012.10 Solving equations graphically and analyticallyPaper 1Easy5 marks
No calculator
On a winter night, the air temperature, \(T\) °C, inside a greenhouse t hours after midnight is modelled by \(T(t)=t^2-8t+22\), for \(0\le t\le 8\). A heater stays on whenever \(T(t)\le 15\).
(a)
Find the set of values of t for which the heater is on.
(4)
(b)
Hence write down the length of time for which the heater is on.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms the inequality \(t^2-8t+7\le 0\)
✓ 1
A1 factorises to \((t-1)(t-7)\le 0\) or finds the roots \(t=1,\ t=7\)
✓ 1
M1 identifies the region between the roots
✓ 1
A1 \(1\le t\le 7\)
✓ 1
Part (b)
A1 6 hours
✓ 1
Exam tip: For a quadratic with a positive leading coefficient, '≤ 0' always means the interval between the roots. A quick mental sketch of the U-shaped graph confirms this.
22IBMA2H-047-01-022.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy6 marks
No calculator
In a model of a laser alignment system, the cross-section of a reflective surface is the curve \(y=x^2-3x+6\). A laser beam travels along the line \(y=mx+2\), where \(m>0\). The beam touches the surface at exactly one point.
(a)
Find the value of m.
(4)
(b)
Find the coordinates of the point where the beam touches the surface.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates the line and the curve: \(x^2-3x+6=mx+2\)
✓ 1
A1 \(x^2-(3+m)x+4=0\)
✓ 1
M1 sets the discriminant to zero: \((3+m)^2-16=0\)
✓ 1
A1 \(m=1\) (the solution \(m=-7\) is rejected because \(m>0\))
✓ 1
Part (b)
M1 substitutes \(m=1\): \(x^2-4x+4=0\), so \(x=2\)
✓ 1
A1 \((2,\,4)\)
✓ 1
Common error: Both values of m make the discriminant zero. The condition \(m>0\) is given so that you reject \(m=-7\). Always check the solutions against the stated conditions.
23IBMA2H-047-01-052.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy6 marks
No calculator
A designer models the profile of a curved skate ramp by \(y=x^2+kx+9\), where y m is the height above the ground and \(k\) is a constant. For safety, the profile must never touch or cross the ground, so \(y>0\) for all real x.
(a)
Find the set of possible values of k.
(4)
(b)
In a draft design \(k=-6\). Find the value of x at which this profile touches the ground.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the condition that the discriminant is negative: \(k^2-4\times 1\times 9<0\)
✓ 1
A1 \(k^2-36<0\)
✓ 1
M1 solves the quadratic inequality, e.g. \((k-6)(k+6)<0\)
✓ 1
A1 \(-6<k<6\)
✓ 1
Part (b)
M1 \(x^2-6x+9=0\Rightarrow (x-3)^2=0\)
✓ 1
A1 \(x=3\)
✓ 1
Exam tip: 'Positive for all x' with a positive leading coefficient means no real roots, so the strict inequality \(\Delta<0\) applies. Equal roots (\(\Delta=0\)) would mean the curve touches the axis.
24IBMA2H-047-04-022.14 Odd and even functions, inverse and self-inverse functionsPaper 1Easy6 marks
No calculator
The function \(f\) is even and the function \(g\) is odd, both with domain \(\mathbb{R}\).
(a)
Show that \(f\circ g\) is an even function.
(2)
(b)
Show that \(g\circ g\) is an odd function.
(2)
(c)
Given that \(f(3)=5\) and \(g(-2)=3\), find the value of \((f\circ g)(2)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \((f\circ g)(-x)=f(g(-x))=f(-g(x))\) using \(g\) odd
✓ 1
A1 \(=f(g(x))\) using \(f\) even, so \(f\circ g\) is even
✓ 1
Part (b)
M1 \((g\circ g)(-x)=g(g(-x))=g(-g(x))\)
✓ 1
A1 \(=-g(g(x))\), so \(g\circ g\) is odd
✓ 1
Part (c)
M1 \(g(2)=-g(-2)=-3\)
✓ 1
A1 \((f\circ g)(2)=f(-3)=f(3)=5\)
✓ 1
Common error: Replacing \(g(2)\) by \(g(-2)=3\) ignores that \(g\) is odd; the sign changes, then the evenness of \(f\) removes it again.
25IBMA2H-047-04-032.5 Composite functions and inverse functionsPaper 1Easy5 marks
No calculator
A simple number cipher encodes a number \(x\) as \(f(x)=\frac{3x+5}{2x-3}\), \(x\neq\frac{3}{2}\).
(a)
Find \(f^{-1}(x)\).
(3)
(b)
Hence state what your answer to part (a) shows about the function \(f\).
(1)
(c)
A coded message contains the number 11. Find the original number.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets \(y=\frac{3x+5}{2x-3}\) (or swaps \(x\) and \(y\) first) and multiplies out: \(2xy-3y=3x+5\)
✓ 1
M1 collects the \(x\) terms and factorises: \(x(2y-3)=3y+5\), so \(x=\frac{3y+5}{2y-3}\)
R1 \(f^{-1}=f\), so \(f\) is self-inverse: applying the cipher twice returns the original number
✓ 1
Part (c)
A1 \(f^{-1}(11)=\frac{33+5}{22-3}=\frac{38}{19}=2\). Also accept solving \(f(x)=11\): \(3x+5=22x-33\), so \(x=2\)
✓ 1
Exam tip: For a self-inverse function, decoding uses the same rule as encoding. Check your answer by encoding it again: \(f(2)=\frac{11}{1}=11\).
26IBMA2H-047-03-012.5 Composite functions and inverse functionsPaper 1Easy7 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{3x-2}{x+4}\), for \(x\in\mathbb{R}\), \(x\neq -4\).
(a)
Write down the equation of (i) the vertical asymptote and (ii) the horizontal asymptote of the graph of \(y=f(x)\).
(2)
(b)
Find \(f^{-1}(x)\).
(3)
(c)
Write down (i) the domain and (ii) the range of \(f^{-1}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 (i) \(x=-4\)
✓ 1
A1 (ii) \(y=3\)
✓ 1
Part (b)
M1 interchanges \(x\) and \(y\) (at any stage) and multiplies out, e.g. \(xy+4x=3y-2\)
✓ 1
A1 collects the terms in \(y\), e.g. \(y(x-3)=-4x-2\)
✓ 1
A1 \(f^{-1}(x)=\frac{4x+2}{3-x}\) or equivalent
✓ 1
Part (c)
A1 (i) \(x\in\mathbb{R}\), \(x\neq 3\)
✓ 1
A1 (ii) \(y\in\mathbb{R}\), \(y\neq -4\)
✓ 1
Exam tip: The domain and range of an inverse are the range and domain of the original function swapped, so you can read them straight from the asymptotes of \(f\) without any algebra.
27IBMA2H-047-03-032.5 Composite functions and inverse functionsPaper 1Easy6 marks
No calculator
Let \(f(x)=\frac{x^2+3x-1}{x-2}\), for \(x\neq 2\).
(a)
Express \(f(x)\) in the form \(ax+b+\frac{c}{x-2}\), where \(a,b,c\in\mathbb{Z}\).
(3)
(b)
Hence write down the equations of the asymptotes of the graph of \(y=f(x)\).
(2)
(c)
Write down the \(y\)-intercept of the graph.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts polynomial division or compares coefficients in \(x^2+3x-1=(x-2)(ax+b)+c\)
✓ 1
A1 \(a=1\), \(b=5\)
✓ 1
A1 \(c=9\), so \(f(x)=x+5+\frac{9}{x-2}\)
✓ 1
Part (b)
A1 vertical asymptote \(x=2\)
✓ 1
A1 oblique asymptote \(y=x+5\)
✓ 1
Part (c)
A1 \(f(0)=\frac{-1}{-2}=\frac{1}{2}\), i.e. \(\left(0,\frac{1}{2}\right)\)
✓ 1
Common error: The oblique asymptote is the quotient only, \(y=x+5\). Including the remainder term, or forgetting to change sign when subtracting during division, gives the wrong line.
28IBMA2H-047-06-012.5 Composite functions and inverse functionsPaper 1Easy7 marks
No calculator
A fitness app estimates a user's maximum heart rate, in beats per minute, using \(h(a)=220-a\), where \(a\) is the user's age in years. It then sets a training target of 70% of the maximum heart rate using \(t(m)=0.7m\).
(a)
Find \((t\circ h)(a)\), giving your answer in the form \(p-qa\).
(2)
(b)
A user's training target is 126 beats per minute. Find the user's age.
(2)
(c)
Find \((t\circ h)^{-1}(x)\) and hence find the age of a user whose training target is 140 beats per minute.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(h(a)\) into \(t\): \(t(220-a)=0.7(220-a)\)
✓ 1
A1 \((t\circ h)(a)=154-0.7a\)
✓ 1
Part (b)
M1 sets \(154-0.7a=126\)
✓ 1
A1 \(a=40\) years
✓ 1
Part (c)
M1 rearranges \(x=154-0.7a\) to make \(a\) the subject (or interchanges \(x\) and \(y\))
A1 \((t\circ h)^{-1}(140)=\frac{140}{7}=20\), so the user is 20 years old
✓ 1
Exam tip: The order matters in a composite: \((t\circ h)(a)\) means apply \(h\) first, then \(t\). Writing out \(t(h(a))\) before simplifying avoids reversing the order.
29IBMA2H-047-06-022.8 The reciprocal function and rational functionsPaper 1Easy5 marks
No calculator
For a camera lens of focal length 10 cm, the distance \(v\) cm of a sharp image from the lens is related to the distance \(u\) cm of the object from the lens by \(v=g(u)=\frac{10u}{u-10}\), \(u>10\). For \(u>10\), the range of \(g\) is \(v>10\).
(a)
Find \(g(15)\).
(1)
(b)
Find an expression for \(g^{-1}(v)\) and hence show that \(g\) is a self-inverse function.
(3)
(c)
Hence find the object distance that produces an image 25 cm from the lens.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(g(15)=\frac{150}{5}=30\)
✓ 1
Part (b)
M1 writes \(v(u-10)=10u\), so \(uv-10v=10u\)
✓ 1
M1 collects terms in \(u\): \(uv-10u=10v\), so \(u(v-10)=10v\)
✓ 1
A1 \(g^{-1}(v)=\frac{10v}{v-10}\) with domain \(v>10\) (the range of \(g\)). This is the same rule on the same domain as \(g\), so \(g\) is self-inverse
✓ 1
Part (c)
A1 \(u=g^{-1}(25)=g(25)=\frac{250}{15}=\frac{50}{3}\) cm (\(\approx 16.7\) cm)
✓ 1
Exam tip: To show a function is self-inverse, state that \(g^{-1}\) has the same rule and the same domain as \(g\). The domain of \(g^{-1}\) is the range of \(g\), so matching formulas alone is not a complete argument.
30IBMA2H-047-06-092.5 Composite functions and inverse functionsPaper 1Easy6 marks
No calculator
The functions \(f\) and \(g\) are defined by \(f(x)=\sqrt{x-3}\), \(x\ge3\), and \(g(x)=x^2+1\), \(x\in\mathbb{R}\).
(a)
Find \((g\circ f)(x)\), stating its domain.
(2)
(b)
Write down the range of \(g\circ f\).
(1)
(c)
Find \((f\circ g)(x)\) and the largest possible domain for which it is defined.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(g(f(x))=\left(\sqrt{x-3}\right)^2+1\)
✓ 1
A1 \((g\circ f)(x)=x-2\), \(x\ge3\)
✓ 1
Part (b)
A1 \((g\circ f)(x)\ge1\)
✓ 1
Part (c)
M1 \(f(g(x))=\sqrt{x^2+1-3}\)
✓ 1
A1 \((f\circ g)(x)=\sqrt{x^2-2}\)
✓ 1
A1 domain \(x\le-\sqrt2\) or \(x\ge\sqrt2\)
✓ 1
Common error: Simplifying \(g\circ f\) to \(x-2\) does not remove the restriction \(x\ge3\); the domain of a composite is limited by the domain of the inner function.
31IBMA2H-047-00-012.3 Graphs of functions and sketching from a contextPaper 1Easy5 marks
No calculator
A basketball is thrown upwards. Its height above the court, h metres, t seconds after release is modelled by \(h(t) = 2 + 6t - 5t^2\), for \(t \ge 0\).
(a)
Write \(h(t)\) in the form \(a(t-p)^2 + q\), where \(a\), \(p\), \(q\) are exact constants.
(3)
(b)
Hence write down the maximum height of the ball and the time at which it occurs.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 factorises −5 from the t terms: \(-5(t^2 - \frac{6}{5}t) + 2\)
✓ 1
M1 completes the square: \(-5\left[(t-\frac{3}{5})^2 - \frac{9}{25}\right] + 2\)
✓ 1
A1 \(h(t) = -5(t-\frac{3}{5})^2 + \frac{19}{5}\)
✓ 1
Part (b)
A1 maximum height \(\frac{19}{5}\) m (3.8 m)
✓ 1
A1 at \(t = \frac{3}{5}\) s (0.6 s)
✓ 1
Exam tip: When the leading coefficient is negative, factorise it out of the t² and t terms only; the constant stays outside the bracket until the end.
32IBMA2H-047-00-022.2 Functions, domain, range and inverse functionsPaper 1Easy5 marks
No calculator
Consider the equation \(|2x - 3| = x + 1\).
(a)
Solve the equation \(|2x - 3| = x + 1\).
(3)
(b)
Hence solve the inequality \(|2x - 3| < x + 1\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 considers both cases \(2x-3 = x+1\) and \(2x-3 = -(x+1)\)
M1 tests a value between the critical points, e.g. \(x=1\): \(1 < 2\) true, or uses graph reasoning
✓ 1
A1 \(\frac{2}{3} < x < 4\)
✓ 1
Common error: Solving only \(2x-3 = x+1\) loses the second solution. Always check that each solution makes the right-hand side non-negative.
33IBMA2H-047-00-032.16 Modulus, reciprocal and other graph transformationsPaper 1Easy6 marks
No calculator
The underside of a footbridge is a parabolic arch. Its two ends are on the ground 24 m apart and its highest point is 6 m above the ground. Using axes with the origin at one end of the arch and the x-axis along the ground, the arch is modelled by \(y = ax(x - 24)\), where x and y are in metres.
Diagram not to scale
(a)
Show that \(a = -\frac{1}{24}\).
(3)
(b)
A sign is to be hung across the arch at a height of 4.5 m. Find the width of the arch at this height.
Exam tip: In factorised form the vertex lies halfway between the roots, which gives its x-coordinate without any differentiation.
34IBMA2H-047-00-042.13 Rational functions with oblique and horizontal asymptotesPaper 1Easy7 marks
No calculator
Let \(f(x) = x^2 - 4x + 3\) and \(g(x) = \frac{1}{f(x)}\).
(a)
Find the equations of the vertical asymptotes of the graph of \(y = g(x)\).
(2)
(b)
Write down the equation of the horizontal asymptote of the graph of \(y = g(x)\).
(1)
(c)
Find the coordinates of the local maximum point on the graph of \(y = g(x)\).
(2)
(d)
Find the range of g.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 factorises \(f(x) = (x-1)(x-3)\) or solves \(f(x) = 0\)
✓ 1
A1 \(x = 1\) and \(x = 3\)
✓ 1
Part (b)
A1 \(y = 0\)
✓ 1
Part (c)
M1 locates \(x = 2\), either from the vertex of f (minimum \(f(2) = -1\), with \(f < 0\) near \(x = 2\), so 1/f has a local maximum there) or by solving \(g'(x) = \frac{-(2x-4)}{(x^2-4x+3)^2} = 0\)
✓ 1
A1 \((2, -1)\)
✓ 1
Part (d)
M1 uses the range of f, \(f(x) \ge -1\), splitting it into \(-1 \le f(x) < 0\) and \(f(x) > 0\)
✓ 1
A1 \(g(x) \le -1\) or \(g(x) > 0\)
✓ 1
Common error: A minimum of f below the x-axis becomes a local maximum of 1/f, because the reciprocal of a negative number closer to zero is more negative.
35IBMA2H-047-07-012.11 Transformations of graphsPaper 1Easy6 marks
No calculator
An audio engineer models a sound waveform by the graph of \(y=f(x)\). The graph of \(f\) passes through the point \(P(6,\,-2)\). To create a processed signal, the engineer defines \(g(x)=3f(2x-4)+1\).
(a)
Find the coordinates of the image of \(P\) on the graph of \(y=g(x)\).
(2)
(b)
Describe a sequence of transformations that maps the graph of \(y=f(x)\) onto the graph of \(y=g(x)\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets the input of \(f\) equal to 6: \(2x-4=6\), giving \(x=5\)
✓ 1
A1 \(y=3(-2)+1=-5\), so the image is \((5,\,-5)\)
✓ 1
Part (b)
A1 horizontal translation of 4 units to the right, i.e. by vector \(\binom{4}{0}\)
✓ 1
A1 followed by a horizontal stretch, scale factor \(\frac{1}{2}\), parallel to the x-axis
✓ 1
A1 vertical stretch, scale factor 3, parallel to the y-axis
✓ 1
A1 followed by a vertical translation of 1 unit up, by vector \(\binom{0}{1}\)
✓ 1
Accept a horizontal stretch with scale factor \(\frac{1}{2}\) followed by a translation of 2 units right.
—
Accept a vertical translation of \(\frac{1}{3}\) unit up followed by a vertical stretch with scale factor 3.
—
The horizontal and vertical transformations may be given in either order relative to each other.
—
Exam tip: Rewrite \(f(2x-4)\) as \(f(2(x-2))\) before describing it. If you stretch first, the shift is 2; if you translate first, the shift is 4. Always state the order of your transformations clearly.
36IBMA2H-047-07-052.6 The quadratic function and its formsPaper 1Easy5 marks
No calculator
Let \(f(x)=2x^2-8x+5\). The function g is defined by \(g(x)=f(x+3)+4\).
(a)
Write \(f(x)\) in the form \(a(x-h)^2+k\).
(2)
(b)
Hence write \(g(x)\) in the form \(a(x-h)^2+k\) and state the coordinates of the vertex of the graph of g.
(2)
(c)
State the range of g.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to complete the square, e.g. \(2(x^2-4x)+5\)
✓ 1
A1 \(f(x)=2(x-2)^2-3\)
✓ 1
Part (b)
M1 replaces \(x\) by \(x+3\) and adds 4, or translates vertex \((2,-3)\) by \(\binom{-3}{4}\)
✓ 1
A1 \(g(x)=2(x+1)^2+1\), vertex \((-1,\,1)\)
✓ 1
Part (c)
A1 \(g(x)\ge 1\)
✓ 1
Common error: \(f(x+3)\) moves the graph 3 units to the left, so the vertex x-coordinate decreases from 2 to \(-1\).
37IBMA2H-047-02-012.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy8 marks
No calculator
The polynomial \(p(x) = x^3 + ax^2 + bx - 6\), where \(a, b \in \mathbb{Z}\), has \((x + 1)\) as a factor. When \(p(x)\) is divided by \((x - 1)\) the remainder is \(-8\).
(a)
Find the value of \(a\) and the value of \(b\).
(4)
(b)
Hence factorize \(p(x)\) fully.
(2)
(c)
Solve the inequality \(p(x) \le 0\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(x = -1\) and sets the result equal to 0: \(-1 + a - b - 6 = 0\), so \(a - b = 7\)
✓ 1
M1 substitutes \(x = 1\) and sets the result equal to \(-8\): \(1 + a + b - 6 = -8\), so \(a + b = -3\)
✓ 1
A1 \(a = 2\)
✓ 1
A1 \(b = -5\)
✓ 1
Part (b)
M1 divides \(x^3 + 2x^2 - 5x - 6\) by \((x + 1)\) to obtain the quotient \(x^2 + x - 6\)
✓ 1
A1 \(p(x) = (x + 1)(x - 2)(x + 3)\)
✓ 1
Part (c)
M1 identifies the roots \(-3, -1, 2\) and uses the sign of a positive cubic between and beyond them
✓ 1
A1 \(x \le -3\) or \(-1 \le x \le 2\)
✓ 1
Exam tip: A factor gives you an equation equal to zero, and a remainder gives you an equation equal to that remainder. Write both equations down before you start solving.
38IBMA2H-047-02-022.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy7 marks
No calculator
The equation \(2x^4 - 6x^3 + kx^2 + 5x - 8 = 0\), where \(k \in \mathbb{R}\), has roots \(\alpha, \beta, \gamma, \delta\).
(a)
Write down the value of \(\alpha + \beta + \gamma + \delta\) and the value of \(\alpha\beta\gamma\delta\).
(2)
(b)
Find the value of \(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}\).
(2)
(c)
Given that \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 4\), find the value of \(k\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\alpha + \beta + \gamma + \delta = 3\)
✓ 1
A1 \(\alpha\beta\gamma\delta = -4\)
✓ 1
Part (b)
M1 writes the sum as \(\frac{\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta}{\alpha\beta\gamma\delta}\), with numerator \(-\frac{5}{2}\)
Common error: The signs alternate: the sum of roots is \(-b/a\), the sum of pairs is \(+c/a\), the sum of triples is \(-d/a\), and the product is \(+e/a\) for a quartic. Getting one sign wrong ruins every later part.
39IBMA2H-047-02-042.9 Exponential and logarithmic functionsPaper 1Easy6 marks
No calculator
The sound level \(L\) decibels (dB) of a sound with intensity \(I\) W m−2 is given by \(L = 10\log_{10}\left(\frac{I}{I_0}\right)\), where \(I_0 = 10^{-12}\) W m−2.
(a)
Find the sound level of a busy street, which has an intensity of 10−4 W m−2.
(2)
(b)
Two sounds have intensities \(I_1\) and \(I_2\) and sound levels \(L_1\) and \(L_2\). Show that \(L_2 - L_1 = 10\log_{10}\left(\frac{I_2}{I_1}\right)\).
(2)
(c)
A leaf blower has a sound level of 95 dB and a quiet conversation has a sound level of 55 dB. Find the ratio \(\frac{\text{intensity of the leaf blower}}{\text{intensity of the conversation}}\), that is, how many times as intense the leaf blower is as the conversation.
M1 uses the result shown above with \(L_2 - L_1 = 95 - 55 = 40\): \(40 = 10\log_{10}\left(\frac{I_2}{I_1}\right)\), so \(\log_{10}\left(\frac{I_2}{I_1}\right) = 4\)
✓ 1
A1 \(\frac{I_2}{I_1} = 10^4 = 10000\)
✓ 1
(Finding each intensity separately, \(10^{-2.5}\) and \(10^{-6.5}\) W m−2, and dividing them also earns full marks.)
—
Common error: A difference of 40 dB does not mean 40 times the intensity. Each 10 dB increase multiplies the intensity by 10, so the factor is \(10^4\).
40IBMA2H-047-05-012.1 Equations of straight lines, parallel and perpendicular linesPaper 1Easy5 marks
No calculator
On a site plan, with coordinates in metres, a straight drainage pipe passes through the points A(−2, 7) and B(4, −2). A second straight pipe is to be laid along the perpendicular bisector of AB.
Coordinates in metres
(a)
Find the gradient of AB.
(2)
(b)
Find the equation of the line along which the second pipe is laid. Give your answer in the form \(ax+by+d=0\), where \(a, b, d \in \mathbb{Z}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\frac{-2-7}{4-(-2)}\)
✓ 1
A1 gradient \(=-\frac{3}{2}\)
✓ 1
Part (b)
A1 midpoint of AB is \(\left(1, \frac{5}{2}\right)\)
✓ 1
M1 uses perpendicular gradient \(\frac{2}{3}\) with the midpoint, e.g. \(y-\frac{5}{2}=\frac{2}{3}(x-1)\)
✓ 1
A1 \(4x-6y+11=0\) or any integer multiple
✓ 1
Common error: The perpendicular gradient is the negative reciprocal, \(\frac{2}{3}\), not \(\frac{3}{2}\) or \(-\frac{2}{3}\). Check that the product of the two gradients is −1.
41IBMA2H-047-05-022.5 Composite functions and inverse functionsPaper 1Easy7 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{2x+3}{x-4}\), \(x \neq 4\).
(a)
Write down the equation of (i) the vertical asymptote of the graph of \(f\); (ii) the horizontal asymptote of the graph of \(f\).
(2)
(b)
Find the coordinates of the points where the graph of \(f\) meets the coordinate axes.
(2)
(c)
Find \(f^{-1}(x)\), stating its domain.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=4\)
✓ 1
A1 \(y=2\)
✓ 1
Part (b)
A1 \(\left(-\frac{3}{2}, 0\right)\)
✓ 1
A1 \(\left(0, -\frac{3}{4}\right)\)
✓ 1
Part (c)
M1 interchanges \(x\) and \(y\) and attempts to make \(y\) the subject, e.g. \(x(y-4)=2y+3\)
✓ 1
A1 \(f^{-1}(x)=\frac{4x+3}{x-2}\)
✓ 1
A1 domain \(x \neq 2\)
✓ 1
Exam tip: The domain of \(f^{-1}\) is the range of \(f\). Here the horizontal asymptote \(y=2\) of \(f\) becomes the excluded value \(x=2\) for the inverse, which is a quick way to check your answer.
42IBMA2H-047-05-042.11 Transformations of graphsPaper 1Easy6 marks
No calculator
A designer of a phone app uses the graph of \(y=\frac{1}{x}\) to create a curve for a scrolling animation. The graph of \(y=\frac{1}{x}\) is first stretched vertically by scale factor 2 and then translated by the vector \(\begin{pmatrix}3\\-5\end{pmatrix}\) to give the graph of \(y=h(x)\).
(a)
Find an expression for \(h(x)\).
(2)
(b)
Write down the equations of the asymptotes of the graph of \(h\).
(2)
(c)
Find the exact \(x\)-coordinate of the point where the graph of \(h\) crosses the \(x\)-axis.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 applies stretch to give \(\frac{2}{x}\) then horizontal shift to give \(\frac{2}{x-3}\)
✓ 1
A1 \(h(x)=\frac{2}{x-3}-5\)
✓ 1
Part (b)
A1 \(x=3\)
✓ 1
A1 \(y=-5\)
✓ 1
Part (c)
M1 \(\frac{2}{x-3}=5\)
✓ 1
A1 \(x=\frac{17}{5}\)
✓ 1
Common error: A translation of +3 in the \(x\)-direction replaces \(x\) with \(x-3\), not \(x+3\). The vertical asymptote moves to \(x=3\).
43IBMA3H-2-18r7-012.5 Composite functions and inverse functionsPaper 1Easy6 marks
No calculator
The functions f and g are defined for x ∈ ℝ by f(x) = 4x − 3 and g(x) = x2 + 1.
(a)
Find f−1(x).
(2)
(b)
Find (g ∘ f)(x), giving your answer in the form ax2 + bx + c.
(2)
(c)
Solve (f ∘ g)(x) = 17.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to rearrange \(y=4x-3\) for \(x\) (or swap variables)
✓ 1
A1 \(f^{-1}(x)=\frac{x+3}{4}\)
✓ 1
Part (b)
M1 \(g(4x-3)=(4x-3)^2+1\)
✓ 1
A1 \(16x^2-24x+10\)
✓ 1
Part (c)
M1 \(4(x^2+1)-3=17\), so \(4x^2=16\)
✓ 1
A1 \(x=2\) or \(x=-2\)
✓ 1
Common error: (f ∘ g)(x) means apply g first; swapping the order gives a different function.
44IBMA3H-2-18r7-022.2 Functions, domain, range and inverse functionsPaper 1Easy4 marks
No calculator
The function f is defined by f(x) = √(x − 3) + 2, for the largest possible real domain.
(a)
Write down the domain and the range of f.
(2)
(b)
Find an expression for f−1(x) and state its domain.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 domain \(x\ge 3\)
✓ 1
A1 range \(y\ge 2\)
✓ 1
Part (b)
M1 rearranges \(y-2=\sqrt{x-3}\) and squares both sides
✓ 1
A1 \(f^{-1}(x)=(x-2)^2+3\), with domain \(x\ge 2\) (the range of f)
✓ 1
Exam tip: The domain of an inverse function is the range of the original function, not the expression's natural domain.
45IBMA3H-2-18r7-092.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy5 marks
No calculator
The polynomial p(x) = x3 + kx2 − 4x + 12 has (x + 2) as a factor, where k is a constant.
(a)
Find the value of k.
(2)
(b)
Factorise p(x) completely and hence solve p(x) = 0.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 applies the factor theorem with \(x=-2\): \(-8+4k+8+12=0\)
✓ 1
A1 \(k=-3\)
✓ 1
Part (b)
M1 attempt to divide by \((x+2)\) or factorise by grouping
✓ 1
A1 \((x+2)(x-2)(x-3)\)
✓ 1
A1 \(x=-2,\ 2,\ 3\)
✓ 1
Exam tip: After finding k, check your result by testing another root in the polynomial.
46IBMA3H-2-18r7-102.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy5 marks
No calculator
The roots of the equation \(2x^2-9x+4=0\) are α and β.
(a)
Write down the value of α + β and the value of αβ.
(2)
(b)
Find the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\).
Common error: The sum of roots of ax² + bx + c = 0 is −b/a; forgetting the minus sign is a frequent slip.
47IBMA3H-2-19r8-012.14 Odd and even functions, inverse and self-inverse functionsPaper 1Easy6 marks
No calculator
An archaeology team catalogues pottery fragments. The cost per fragment, in £, when \(x\) hundred fragments are catalogued in one batch is modelled by \(f(x)=\frac{3x+2}{x-1}\), for \(x>1\).
(a)
State the equation of the vertical asymptote and the equation of the horizontal asymptote of the graph of \(y=f(x)\).
(2)
(b)
Find \(f^{-1}(x)\).
(3)
(c)
Write down the domain of \(f^{-1}\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertical asymptote \(x=1\)
✓ 1
A1 horizontal asymptote \(y=3\)
✓ 1
Part (b)
M1 swap variables or make \(x\) the subject of \(y=\frac{3x+2}{x-1}\)
✓ 1
M1 correct expansion and collection, e.g. \(xy-y=3x+2\) so \(x(y-3)=y+2\)
✓ 1
A1 \(f^{-1}(x)=\frac{x+2}{x-3}\)
✓ 1
Part (c)
A1 \(x>3\) (the range of \(f\))
✓ 1
Exam tip: The domain of an inverse function is the range of the original function, so read it from the horizontal asymptote and the restricted domain of \(f\).
48IBMA3H-2-19r8-032.5 Composite functions and inverse functionsPaper 1Easy6 marks
No calculator
A data centre models its cooling cost with \(f(x)=3x+4\) and \(g(x)=x^2-2\), where \(x\) is the number of racks in operation (in hundreds) before an upgrade.
(a)
Find \((g\circ f)(x)\), giving your answer in the form \(ax^2+bx+c\).
(2)
(b)
Find the positive value of \(x\) for which \(f(g(x))=25\).
(3)
(c)
Explain why \(g\) has no inverse function when its domain is all real numbers.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(f(x)\) into \(g\): \((3x+4)^2-2\)
✓ 1
A1 \(9x^2+24x+14\)
✓ 1
Part (b)
M1 \(f(g(x))=3(x^2-2)+4\)
✓ 1
A1 \(3x^2-2=25\) so \(x^2=9\)
✓ 1
A1 \(x=3\)
✓ 1
Part (c)
R1 \(g\) is not one-to-one, e.g. \(g(2)=g(-2)\), so it fails the horizontal line test
✓ 1
Common error: \((g\circ f)(x)\) means apply \(f\) first, then \(g\); reversing the order gives a different function.
49IBMA3H-2-19r13-082.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy6 marks
No calculator
Let \(p(x)=x^3-4x^2+kx+6\), where \(k\) is a constant. \((x-3)\) is a factor of \(p(x)\).
(a)
Find the value of \(k\).
(2)
(b)
Hence factorise \(p(x)\) completely.
(3)
(c)
Find the sum of the squares of the roots of \(p(x)=0\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use the factor theorem: \(p(3)=0\), i.e. \(27-36+3k+6=0\)
✓ 1
A1 \(k=1\)
✓ 1
Part (b)
M1 attempt to divide by \((x-3)\) or compare coefficients to obtain the quadratic factor
✓ 1
A1 \(x^2-x-2\)
✓ 1
A1 \(p(x)=(x-3)(x-2)(x+1)\)
✓ 1
Part (c)
A1 \(9+4+1=14\)
✓ 1
Exam tip: In a factor-theorem question, substitute the root into the polynomial first; the equation for the unknown constant is then linear.
50IBMA3H-2-19r13-092.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy5 marks
No calculator
When \(p(x)=2x^3+ax^2-5x+4\) is divided by \((x-2)\), the remainder is 18.
(a)
Find the value of \(a\).
(3)
(b)
Find the remainder when \(p(x)\) is divided by \((x+1)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use the remainder theorem: \(p(2)=18\)
✓ 1
A1 \(16+4a-10+4=18\)
✓ 1
A1 \(a=2\)
✓ 1
Part (b)
M1 evaluate \(p(-1)\) using \(a=2\)
✓ 1
A1 \(-2+2+5+4=9\)
✓ 1
Common error: Evaluating \(p(-2)\) for a divisor of \((x-2)\). Divisor \((x-c)\) means evaluate \(p(c)\).
51IBMA3H-2-19r21-012.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy7 marks
No calculator
A theatre lighting controller converts a dial setting \(x\) into a power level \(f(x)=5x-3\), and a dimmer unit converts a power level \(x\) into a brightness index \(g(x)=x^2+4\), where \(x\in\mathbb{R}\).
(a)
Find \(f^{-1}(x)\).
(2)
(b)
Find \((g\circ f)(x)\), giving your answer in the form \(ax^2+bx+c\).
(2)
(c)
Solve \((g\circ f)(x)=13\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets \(y=5x-3\) and rearranges for \(x\) (or swaps \(x\) and \(y\))
✓ 1
A1 \(f^{-1}(x)=\frac{x+3}{5}\)
✓ 1
Part (b)
M1 \(g(5x-3)=(5x-3)^2+4\)
✓ 1
A1 \(25x^2-30x+13\)
✓ 1
Part (c)
M1 \(25x^2-30x+13=13\) so \(25x^2-30x=0\)
✓ 1
A1 \(5x(5x-6)=0\)
✓ 1
A1 \(x=0\) or \(x=\frac{6}{5}\) (\(=1.2\))
✓ 1
Common error: For \(g\circ f\), \(f\) is applied first: substitute \(f(x)\) into \(g\), not the other way round.
52IBMA3H-2-19r21-032.5 Composite functions and inverse functionsPaper 1Easy6 marks
No calculator
The function \(f\) is defined by \(f(x)=\dfrac{2x+1}{x-3}\), for \(x\neq 3\).
(a)
Find an expression for \(f^{-1}(x)\).
(3)
(b)
State the domain of \(f^{-1}\).
(1)
(c)
Given that \(f^{-1}(k)=4\), find the value of \(k\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(y(x-3)=2x+1\) (or the same with \(x\) and \(y\) swapped)
✓ 1
M1 collects \(x\) terms: \(x(y-2)=3y+1\)
✓ 1
A1 \(f^{-1}(x)=\frac{3x+1}{x-2}\)
✓ 1
Part (b)
A1 \(x\in\mathbb{R}\), \(x\neq 2\)
✓ 1
Part (c)
M1 \(k=f(4)\) (or solves \(\frac{3k+1}{k-2}=4\))
✓ 1
A1 \(k=9\)
✓ 1
Exam tip: If you are told \(f^{-1}(k)=a\), it is usually quicker to evaluate \(f(a)\) than to find the inverse first.
53IBMA3H-2-19r22-012.7 Quadratic equations, inequalities and the discriminantPaper 1Easy4 marks
No calculator
A tailor models the curved edge of a sleeve pattern by \(y = 2x^2 + kx + 8\), where \(k\) is a constant. The curve touches the \(x\)-axis at exactly one point.
(a)
Find the possible values of \(k\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the discriminant \(b^2-4ac\) with \(a=2\), \(c=8\)
✓ 1
A1 \(k^2 - 4(2)(8)\) or \(k^2-64\) seen
✓ 1
M1 set discriminant equal to zero (one repeated root) and solve for \(k\)
✓ 1
A1 \(k=8\) or \(k=-8\)
✓ 1
Exam tip: 'Touches the axis at exactly one point' means one repeated root, so set the discriminant equal to zero rather than greater than zero.
54IBMA3H-2-19r22-022.14 Odd and even functions, inverse and self-inverse functionsPaper 1Easy5 marks
No calculator
A robot's sensor reading \(x\) is encoded by the function \(f(x)=\dfrac{x+6}{x-1}\), \(x\neq 1\).
(a)
Show that \(f\) is a self-inverse function, that is \(f(f(x)) = x\).
(3)
(b)
Hence find the value of \(f^{-1}(3)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to substitute \(f(x)\) into \(f\)
✓ 1
A1 \(\dfrac{\frac{x+6}{x-1}+6}{\frac{x+6}{x-1}-1}\) simplified to \(\dfrac{x+6+6(x-1)}{(x+6)-(x-1)}\)
✓ 1
A1 \(\dfrac{7x}{7}=x\), so \(f(f(x))=x\) (shown)
✓ 1
Part (b)
M1 use \(f^{-1}=f\) so \(f^{-1}(3)=f(3)\)
✓ 1
A1 \(\dfrac{9}{2}\)
✓ 1
Common error: Self-inverse means \(f^{-1}(x)=f(x)\), not \(f(x)=\dfrac{1}{f(x)}\). Do not confuse the inverse function with the reciprocal.
55IBMA3H-2-19r22-032.14 Odd and even functions, inverse and self-inverse functionsPaper 1Easy5 marks
No calculator
The vertical sway of a sonar buoy is modelled by \(f(x)=x^5-3x^3+x\), and a second model is \(g(x)=x\,f(x)\).
(a)
Show that \(f\) is an odd function.
(2)
(b)
Determine whether \(g\) is odd, even or neither. Justify your answer.
(2)
(c)
Given that \(f(2)=10\), write down the value of \(g(-2)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find \(f(-x)\) by replacing \(x\) with \(-x\)
✓ 1
A1 \(f(-x) = -x^5+3x^3-x = -f(x)\), so \(f\) is odd
✓ 1
Part (b)
M1 \(g(-x)=(-x)f(-x)=(-x)(-f(x))\)
✓ 1
A1 \(g(-x)=x\,f(x)=g(x)\), so \(g\) is even
✓ 1
Part (c)
A1 \(g(-2)=g(2)=2\times 10=20\)
✓ 1
Exam tip: An odd function times an odd function is even. Spotting this lets you find a value at \(-x\) from the value at \(x\) without recalculating.
56IBMA3H-2-19r22-122.8 The reciprocal function and rational functionsPaper 1Easy6 marks
No calculator
A robotic cutter set to dial value \(x\) cuts a ribbon of length \(f(x)=2x-3\) cm, where \(x\ge 2\). The cost of wasted fabric, in euro cents, for a ribbon of length \(L\) cm is \(g(L)=L^2+1\).
(a)
Find \((g\circ f)(x)\) in the form \(ax^2+bx+c\).
(2)
(b)
Find \(f^{-1}(x)\).
(2)
(c)
Find the dial value for which the cost of wasted fabric is 10 euro cents.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(f(x)\) into \(g\) to get \((2x-3)^2+1\)
✓ 1
A1 \(4x^2-12x+10\)
✓ 1
Part (b)
M1 swap \(x\) and \(y\) and rearrange
✓ 1
A1 \(f^{-1}(x)=\dfrac{x+3}{2}\)
✓ 1
Part (c)
M1 set \(4x^2-12x+10=10\) and factorise as \(4x(x-3)=0\)
✓ 1
A1 \(x=3\) (reject \(x=0\) since \(x\ge2\))
✓ 1
Common error: \((g\circ f)(x)\) means apply \(f\) first, then \(g\). Writing \(f(g(x))\) gives a different function.
57IBMA3H-2-20r30-032.14 Odd and even functions, inverse and self-inverse functionsPaper 1Easy5 marks
No calculator
The function f is odd and the function g is even. Both are defined for all real x.
(a)
Write down the value of f(0), giving a reason.
(1)
(b)
Show that h(x) = f(x)g(x) is an odd function.
(2)
(c)
Given that f(3) = 4 and g(3) = −1, find the value of f(−3) + g(−3).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 f(0) = 0 because f(−0) = −f(0) means f(0) = −f(0)
✓ 1
Part (b)
M1 h(−x) = f(−x)g(−x) = (−f(x))(g(x))
✓ 1
A1 = −h(x), so h is odd (AG)
✓ 1
Part (c)
M1 use f(−3) = −f(3) = −4 and g(−3) = g(3) = −1
✓ 1
A1 −5
✓ 1
Common error: For an even function g(−x) = g(x) with no change of sign; only odd functions change sign.
58IBMA3H-2-20r30-042.16 Modulus, reciprocal and other graph transformationsPaper 1Easy5 marks
No calculator
The function f is defined by f(x) = 2x − 6 for x ∈ ℝ. Let g(x) = 1 ⁄ f(x).
(a)
Write down the equation of the vertical asymptote and the equation of the horizontal asymptote of the graph of y = g(x).
(2)
(b)
Describe a sketch of y = |f(x)|.
(2)
(c)
Find the y-intercept of the graph of y = g(x).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 x = 3
✓ 1
A1 y = 0
✓ 1
Part (b)
A1 V-shaped graph with its vertex on the x-axis at (3, 0), with the part of the line below the axis reflected upwards
✓ 1
A1 crosses the y-axis at (0, 6)
✓ 1
Part (c)
A1 g(0) = 1⁄(−6) = −1⁄6 (≈ −0.1667)
✓ 1
Model sketch
Exam tip: The graph of y = 1/f(x) has a vertical asymptote wherever f(x) = 0, and it crosses the y-axis at the reciprocal of f(0).
59IBMA3H-2-18r34-012.3 Graphs of functions and sketching from a contextPaper 1Easy5 marks
No calculator
A dolphin leaps from the sea. Its height above the water surface is \(h(t)=7t-5t^2\) metres, where t is the time in seconds after it leaves the water, and \(h\ge0\) during the leap.
(a)
Find the times at which the dolphin is at the water surface.
(2)
(b)
Find the greatest height of the dolphin above the water, and state the range of h during the leap.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(7t-5t^2=0\), so \(t(7-5t)=0\)
✓ 1
A1 \(t=0\) and \(t=1.4\) s
✓ 1
Part (b)
M1 the axis of symmetry is at \(t=\frac{0+1.4}{2}=0.7\) (or \(t=-\frac{b}{2a}\))
✓ 1
A1 \(h(0.7)=4.9-2.45=2.45\) m
✓ 1
A1 range \(0\le h\le2.45\)
✓ 1
Exam tip: The vertex of a quadratic lies halfway between its roots. This is quicker than completing the square when the roots are easy to find.
60IBMA3H-2-18r34-032.5 Composite functions and inverse functionsPaper 1Easy6 marks
No calculator
The function f is defined by \(f(x)=\sqrt{2x-6}\) for the largest possible real domain.
(a)
Write down the domain and the range of f.
(2)
(b)
Find \(f^{-1}(x)\) and state its domain.
(3)
(c)
Find the value of \(f^{-1}(5)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 domain: \(x\ge3\)
✓ 1
A1 range: \(f(x)\ge0\)
✓ 1
Part (b)
M1 write \(x=\sqrt{2y-6}\) (swap variables) and square: \(x^2=2y-6\)
✓ 1
A1 \(f^{-1}(x)=\frac{x^2+6}{2}\)
✓ 1
A1 domain of \(f^{-1}\): \(x\ge0\) (the range of f)
✓ 1
Part (c)
A1 \(f^{-1}(5)=\frac{5^2+6}{2}=15.5\)
✓ 1
Exam tip: The domain of an inverse function is the range of the original function. State it with the inverse, since squaring on its own would allow negative inputs.
61IBMA3H-2-18r34-042.4 Key features of graphs and intersections using technologyPaper 1Easy5 marks
No calculator
The concentration of a nutrient in a marine aquarium is modelled by \(C(t)=\dfrac{6t+4}{t+2}\) mg per litre, where t is the time in hours after the nutrient is added, and \(t\ge0\).
(a)
Find the concentration of the nutrient when it is first added.
(1)
(b)
Write down the equation of the horizontal asymptote of the graph of \(y=C(t)\), and interpret it in this context.
(2)
(c)
Find the time at which the concentration is 5 mg per litre.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(C(0)=\frac42=2\) mg per litre
✓ 1
Part (b)
A1 \(y=6\)
✓ 1
R1 in the long term the concentration approaches 6 mg per litre (and never exceeds it)
✓ 1
Part (c)
M1 \(6t+4=5(t+2)\)
✓ 1
A1 \(t=6\) hours
✓ 1
Exam tip: For \(\frac{ax+b}{cx+d}\) the horizontal asymptote is \(y=\frac{a}{c}\). Dividing the leading coefficients is quicker than rewriting the fraction.
62IBMA3H-2-19r34-022.9 Exponential and logarithmic functionsPaper 1Easy6 marks
No calculator
A lighting cable is hung between two towers 12 m apart. Its height above the stage floor is modelled by \(h(x) = a(x-6)^2 + 2.5\) for \(0 \le x \le 12\), where \(x\) is the horizontal distance in metres from the first tower. The cable is attached to each tower at a height of 7 m.
Diagram not to scale
(a)
Find the value of \(a\).
(2)
(b)
Find the two values of \(x\) at which the cable is 4 m above the floor. Give your answers in exact form.
(3)
(c)
Hence find the horizontal distance between the two points in part (b). Give your answer in exact form.
Exam tip: When solving \((x-p)^2 = k\) by square roots, remember both \(+\sqrt{k}\) and \(-\sqrt{k}\), and simplify surds such as \(\sqrt{12} = 2\sqrt{3}\).
63IBMA3H-2-19r34-042.2 Functions, domain, range and inverse functionsPaper 1Easy5 marks
No calculator
The water level in a crater lake, relative to its mean level, is modelled by \(d(t) = t^2 - 6t + 5\) metres, where \(t\) is the time in hours and \(0 \le t \le 6\). The size of the departure from the mean level is \(|d(t)|\).
(a)
Factorise \(d(t)\) and hence write down the zeros of \(d\).
(2)
(b)
Find the largest value of \(t\) in the interval \(0 \le t \le 6\) for which \(|d(t)| = 3\). Give your answer in exact form.
A1 all four values lie in the interval, and the largest is \(t = 3 + \sqrt{7}\)
✓ 1
Common error: An equation \(|f(x)| = k\) with \(k > 0\) needs both \(f(x) = k\) and \(f(x) = -k\); solving only one case loses solutions.
64IBMA3H-2-19r36-012.1 Equations of straight lines, parallel and perpendicular linesPaper 1Easy6 marks
No calculator
The charge, C mAh, left in the battery of a handheld game console is modelled as a linear function of the time t hours since it was unplugged. When t = 2 the charge is 740 mAh, and when t = 5 the charge is 500 mAh.
(a)
Find the gradient of the line and state what it represents.
(2)
(b)
Find an equation for C in terms of t.
(2)
(c)
Find the time at which the battery is completely empty.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at gradient, e.g. (500 − 740)/(5 − 2)
✓ 1
A1 gradient = −80, meaning the charge falls by 80 mAh each hour
✓ 1
Part (b)
M1 substitutes a point and the gradient, e.g. C − 740 = −80(t − 2)
✓ 1
A1 C = 900 − 80t
✓ 1
Part (c)
M1 sets 900 − 80t = 0 (ECF from (b))
✓ 1
A1 t = 11.25 hours (ECF)
✓ 1
Exam tip: When a question gives two points on a line, find the gradient first and then substitute one point; check your equation by testing the other point.
65IBMA3H-2-19r36-042.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy6 marks
No calculator
The cross-section of a valley in a video game is modelled by a quadratic function y = f(x), where x is the horizontal position and y the height relative to sea level, both in tens of metres. The graph of f has its vertex at (5, −18) and one x-intercept at x = 2.
Diagram not to scale
(a)
Show that the other x-intercept is at x = 8.
(2)
(b)
Given that f(x) = k(x − 2)(x − 8), find the value of k.
(2)
(c)
Write f(x) in the form ax2 + bx + c.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the axis of symmetry x = 5, so the intercepts are equally spaced either side of it
✓ 1
A1 5 + (5 − 2) = 8, so the other intercept is at x = 8
✓ 1
Part (b)
M1 substitutes (5, −18): k(5 − 2)(5 − 8) = −18
✓ 1
A1 k = 2
✓ 1
Part (c)
M1 expands 2(x − 2)(x − 8) (ECF from (b))
✓ 1
A1 f(x) = 2x2 − 20x + 32
✓ 1
Exam tip: The vertex of a quadratic lies halfway between its x-intercepts; use this symmetry to find a missing root without solving anything.
66IBMA3H-2-18r43-012.13 Rational functions with oblique and horizontal asymptotesPaper 1Easy4 marks
No calculator
The function \(f\) is defined by \(f(x)=\dfrac{2x+1}{x-3}\), for \(x\ne3\).
(a)
Write down the equation of (i) the vertical asymptote, (ii) the horizontal asymptote of the graph of \(y=f(x)\).
(2)
(b)
Find the coordinates of the point where the graph of \(y=f(x)\) crosses (i) the \(x\)-axis, (ii) the \(y\)-axis.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=3\)
✓ 1
A1 \(y=2\)
✓ 1
Part (b)
A1 \(\left(-\frac12,0\right)\)
✓ 1
A1 \(\left(0,-\frac13\right)\)
✓ 1
Exam tip: For \(y=\frac{ax+b}{cx+d}\) the vertical asymptote is where the denominator is zero and the horizontal asymptote is \(y=\frac{a}{c}\).
67IBMA3H-2-18r43-022.14 Odd and even functions, inverse and self-inverse functionsPaper 1Easy6 marks
No calculator
Let \(f(x)=x^3-4x\) and \(g(x)=x^2+5\), for \(x\in\mathbb{R}\).
(a)
Show that \(f\) is an odd function.
(2)
(b)
State, with a reason, whether \(h(x)=f(x)\,g(x)\) is odd, even or neither.
(2)
(c)
Solve \(f(x)=0\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(f(-x)=(-x)^3-4(-x)\)
✓ 1
A1 \(=-x^3+4x=-f(x)\), so \(f\) is odd
✓ 1
Part (b)
M1 \(g(-x)=g(x)\) so \(g\) is even, and \(h(-x)=f(-x)g(-x)=-f(x)g(x)\)
✓ 1
A1 \(h(-x)=-h(x)\), so \(h\) is odd
✓ 1
Part (c)
M1 \(x(x^2-4)=0\)
✓ 1
A1 \(x=-2,\ 0,\ 2\)
✓ 1
Exam tip: To test for odd or even, find \(f(-x)\) and compare it with \(f(x)\) and \(-f(x)\) algebraically; checking one value is not enough.
68IBMA3H-2-18r43-142.8 The reciprocal function and rational functionsPaper 1Easy4 marks
No calculator
Let \(f(x)=2x-6\) and \(g(x)=\dfrac{1}{f(x)}\), for \(x\ne3\).
(a)
Write down the equations of the asymptotes of the graph of \(y=g(x)\).
(2)
(b)
State whether \(g(x)\) is positive or negative for \(x>3\).
(1)
(c)
Find the \(y\)-intercept of the graph of \(y=g(x)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=3\)
✓ 1
A1 \(y=0\)
✓ 1
Part (b)
A1 positive
✓ 1
Part (c)
A1 \(-\frac16\)
✓ 1
Exam tip: The graph of \(y=\frac{1}{f(x)}\) has a vertical asymptote wherever \(f(x)=0\), and it keeps the same sign as \(f\).
69IBMA3H-2-20r48-012.1 Equations of straight lines, parallel and perpendicular linesPaper 1Easy7 marks
No calculator
The line \(L_1\) has equation \(2x+3y=12\).
(a)
Find the gradient of \(L_1\).
(2)
(b)
The line \(L_2\) is parallel to \(L_1\) and passes through the point \((6,-3)\). Find the equation of \(L_2\) in the form \(ax+by=c\).
(2)
(c)
The line \(L_3\) is perpendicular to \(L_1\) and passes through \((6,-3)\). Find the \(x\)-coordinate of the point where \(L_3\) crosses the \(x\)-axis.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rearrange to \(y=-\frac{2}{3}x+4\)
✓ 1
A1 gradient \(=-\frac{2}{3}\)
✓ 1
Part (b)
M1 use gradient \(-\frac{2}{3}\) in \(y+3=-\frac{2}{3}(x-6)\)
✓ 1
A1 \(2x+3y=3\)
✓ 1
Part (c)
M1 perpendicular gradient \(\frac{3}{2}\)
✓ 1
M1 \(y+3=\frac{3}{2}(x-6)\) with \(y=0\)
✓ 1
A1 \(x=8\)
✓ 1
Common error: A perpendicular gradient is the negative reciprocal; flipping the fraction without changing the sign gives a line that is neither parallel nor perpendicular.
70IBMA3H-2-20r48-022.8 The reciprocal function and rational functionsPaper 1Easy7 marks
No calculator
The point \(P(6,4)\) lies on the graph of \(y=f(x)\). Write down the coordinates of the image of \(P\) on the graph of each of the following.
(a)
\(y=f(x-2)\)
(1)
(b)
\(y=3f(x)\)
(1)
(c)
\(y=f\left(\frac{x}{2}\right)\)
(1)
(d)
\(y=1-f(x)\)
(2)
(e)
\(y=f(2x-4)\)
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \((8,4)\)
✓ 1
Part (b)
A1 \((6,12)\)
✓ 1
Part (c)
A1 \((12,4)\)
✓ 1
Part (d)
M1 new \(y\)-value \(=1-4\)
✓ 1
A1 \((6,-3)\)
✓ 1
Part (e)
M1 solve \(2x-4=6\)
✓ 1
A1 \((5,4)\)
✓ 1
Common error: Changes inside the brackets act on \(x\) and work 'backwards': \(f(x-2)\) moves the graph 2 units to the right, and \(f(\frac{x}{2})\) stretches it horizontally by factor 2.
71IBMA3H-2-20r48-042.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Easy6 marks
No calculator
The polynomial \(p(x)=2x^3+ax^2-7x+6\) has \((x+2)\) as a factor, where \(a\) is a constant.
(a)
Find the value of \(a\).
(3)
(b)
Hence factorise \(p(x)\) completely.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 factor theorem: \(p(-2)=0\)
✓ 1
A1 \(-16+4a+14+6=0\)
✓ 1
A1 \(a=-1\)
✓ 1
Part (b)
M1 attempt to divide \(p(x)\) by \((x+2)\) (ECF from (a))
✓ 1
A1 quotient \(2x^2-5x+3\)
✓ 1
A1 \(p(x)=(x+2)(2x-3)(x-1)\)
✓ 1
Exam tip: If \((x-c)\) is a factor then \(p(c)=0\), so a factor like \((x+2)\) gives \(p(-2)=0\); mind the sign of \(c\).
72AAHL2-q00032.1 Equations of straight lines, parallel and perpendicular linesPaper 1Medium9 marks
For each value of \(k\in\mathbb{R}\), the line \(L_{k}\) has equation \((1+2k)x+(3-k)y=8-5k\).
(a)
Show that every line \(L_{k}\) passes through the point \((-1,\,3)\).
(2)
(b)
Find the value of \(k\) for which \(L_{k}\) is perpendicular to the line \(2x+y=7\).
(3)
(c)
For this value of \(k\), find the equation of \(L_{k}\), giving your answer in the form \(ax+by+d=0\), where \(a,b,d\in\mathbb{Z}\).
(2)
(d)
Show that no line \(L_{k}\) is parallel to the line \(y=2x\).
77AAHL2-q00222.8 The reciprocal function and rational functionsPaper 1Medium8 marks
Let \(f(x)=\dfrac{2x+5}{x+1}\), for \(x\neq-1\).
(a)
Write down the equations of the asymptotes of the graph of \(y=f(x)\).
(2)
(b)
Show that \(f(x)=2+\dfrac{3}{x+1}\).
(2)
(c)
Show that \(f(-2-x)+f(x)=4\), for \(x\neq-1\).
(2)
(d)
Hence explain why, if the point \((a,\,b)\) lies on the graph of \(y=f(x)\), then the point \((-2-a,\,4-b)\) also lies on the graph, and write down the coordinates of the midpoint of these two points.
(1)
(e)
The point \((2,\,3)\) lies on the graph of \(y=f(x)\). Use part (d) to write down the coordinates of another point on the graph.
79AAHL2-q00292.11 Transformations of graphsPaper 1Medium8 marks
Let \(f(x)=2x^{2}-12x+13\), for \(x\in\mathbb{R}\). The graph of \(y=f(x)\) is the image of the graph of \(y=x^{2}\) under a sequence of transformations.
(a)
Write \(f(x)\) in the form \(a(x-h)^{2}+k\).
(2)
(b)
Describe a sequence of two transformations, the first of which is a vertical stretch, that maps the graph of \(y=x^{2}\) onto the graph of \(y=f(x)\).
(2)
(c)
Describe a sequence of two transformations, the first of which is a translation and the second of which is a vertical stretch, that maps the graph of \(y=x^{2}\) onto the graph of \(y=f(x)\).
(2)
(d)
Show that the vertical stretch in part (b) can be replaced by a horizontal stretch, and state its scale factor.
80AAHL2-q00322.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
Let \(P(x)=x^{3}+px^{2}+qx-4\), where \(p,q\in\mathbb{R}\). When \(P(x)\) is divided by \((x-1)\) the remainder is \(5\), and when \(P(x)\) is divided by \((x+2)\) the remainder is \(-4\).
(a)
Find the value of \(p\) and the value of \(q\).
(4)
(b)
The polynomial \(P(x)\) can be written as \(P(x)=(x-1)(x+2)Q(x)+mx+n\), where \(Q(x)\) is a polynomial and \(m,n\in\mathbb{R}\). Without performing a polynomial division, find the remainder \(mx+n\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(P(1)=1+p+q-4=5\) and \(P(-2)=-8+4p-2q-4=-4\)
M1
Award for either equation.
\(p+q=8\) and \(2p-q=4\)
A1
Both equations, in any equivalent form.
\(p=4\)
A1
\(q=4\)
A1
(b)
substituting \(x=1\) and \(x=-2\) into the identity: \(m+n=5\) and \(-2m+n=-4\)
M1
The remainder theorem values must be used; no marks for long division (not as instructed).
\(m=3\)
A1
\(n=2\), so the remainder is \(3x+2\)
A1
Answers: (a) \(p=4,\ q=4\) · (b) \(3x+2\)
81AAHL2-q00332.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium10 marks
Let \(p(x)=x^{4}-3x^{3}+ax^{2}+bx+6\), where \(a,b\in\mathbb{R}\). It is given that \((x-1)^{2}\) is a factor of \(p(x)\).
(a)
Find the value of \(a\) and the value of \(b\).
(5)
(b)
Find the quadratic polynomial \(q(x)\) such that \(p(x)=(x-1)^{2}q(x)\).
(2)
(c)
Show that \(x=1\) is the only real root of the equation \(p(x)=0\).
(2)
(d)
Write down the two non-real roots of \(p(x)=0\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(p(1)=0\): \(1-3+a+b+6=0\)
M1
\(a+b=-4\)
A1
a second condition, e.g. \(p'(1)=0\): \(4-9+2a+b=0\), or remainder on division by \(x^{2}-2x+1\) equal to zero
M1
\(2a+b=5\)
A1
\(a=9,\ b=-13\)
A1
(b)
dividing \(x^{4}-3x^{3}+9x^{2}-13x+6\) by \(x^{2}-2x+1\), or comparing coefficients
M1
\(q(x)=x^{2}-x+6\)
A1
(c)
\(\Delta=(-1)^{2}-4(1)(6)=-23\)
M1
\(\Delta\lt 0\), so \(q(x)=0\) has no real roots and the only real root of \(p(x)=0\) is \(x=1\)
83AAHL2-q00362.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
The following diagram shows part of the graph of a quartic polynomial \(y=p(x)\). The graph touches the \(x\)-axis at \((-1,\,0)\), crosses the \(x\)-axis at \((2,\,0)\) and \((4,\,0)\), and passes through the point \((1,\,-6)\).
Diagram not to scale
(a)
Find an expression for \(p(x)\), giving your answer in factorized form.
(3)
(b)
Show that \(p(0)=-4\).
(1)
(c)
The equation \(p(x)=-4\) has four real roots, one of which is \(x=0\). By considering the sum of the roots of \(p(x)+4=0\), find the sum of the other three roots.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(p(x)=a(x+1)^{2}(x-2)(x-4)\)
M1
Must see the repeated factor \((x+1)^{2}\).
substituting \((1,-6)\): \(a(4)(-1)(-3)=-6\)
M1
\(p(x)=-\dfrac{1}{2}(x+1)^{2}(x-2)(x-4)\)
A1
(b)
\(p(0)=-\dfrac{1}{2}(1)^{2}(-2)(-4)\)
A1
\(=-4\)
AG
(c)
finding the coefficients of \(x^{4}\) and \(x^{3}\) in \(p(x)+4\): \(-\dfrac{1}{2}x^{4}+2x^{3}+\ldots\)
M1
The \(x^{3}\) coefficient comes from \((x^{2}+2x+1)(x^{2}-6x+8)\), giving \(-4x^{3}\) inside the bracket.
sum of all four roots \(=-\dfrac{2}{-\frac{1}{2}}=4\)
84AAHL2-q00432.13 Rational functions with oblique and horizontal asymptotesPaper 1Medium11 marks
Let \(f(x)=\dfrac{2x-4}{x^{2}-2x-3}\), for \(x\neq-1,\ x\neq3\). The graph of \(f\) has no turning points.
(a)
Write down the equations of the vertical asymptotes of the graph of \(f\).
(2)
(b)
Write down the equation of the horizontal asymptote of the graph of \(f\).
(1)
(c)
Find the coordinates of the points where the graph of \(f\) meets the axes.
(2)
(d)
Solve the equation \(f(x)=1\).
(3)
(e)
Sketch the graph of \(y=f(x)\), showing clearly the asymptotes and the axis intercepts.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(x=-1\)
A1
\(x=3\)
A1
(b)
\(y=0\)
A1
(c)
\((2,\,0)\)
A1
\(\left(0,\ \dfrac{4}{3}\right)\)
A1
(d)
\(2x-4=x^{2}-2x-3\)
M1
\(x^{2}-4x+1=0\)
A1
\(x=2\pm\sqrt{3}\)
A1
(e)
three branches separated by the asymptotes \(x=-1\) and \(x=3\), with the left branch below and the right branch above the \(x\)-axis, both approaching \(y=0\)
A1
middle branch decreasing from \(+\infty\) near \(x=-1\) to \(-\infty\) near \(x=3\)
A1
middle branch passing through \(\left(0,\tfrac{4}{3}\right)\) and \((2,\,0)\)
85AAHL2-q00442.13 Rational functions with oblique and horizontal asymptotesPaper 1Medium8 marks
Let \(f(x)=\dfrac{ax^{2}+bx+c}{x+d}\), where \(a,b,c,d\in\mathbb{R}\). The graph of \(f\) has a vertical asymptote \(x=-1\) and an oblique asymptote \(y=x-3\), and it passes through the point \((0,\,-1)\).
(a)
Write down the value of \(d\).
(1)
(b)
The function can be written as \(f(x)=x-3+\dfrac{r}{x+1}\). Find the value of \(r\).
(2)
(c)
Hence find the values of \(a\), \(b\) and \(c\).
(3)
(d)
Find the exact \(x\)-intercepts of the graph of \(f\).
86AAHL2-q00532.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium9 marks
The following diagram shows the graph of \(y=f(x)\) for \(0\le x\le4\). It consists of straight-line segments joining the points \((0,0)\), \((1,2)\), \((3,-1)\) and \((4,0)\).
(a)
The function \(f\) is extended to the domain \(-4\le x\le4\) so that it is an even function. Write down the coordinates of the points on the graph for \(-4\le x\lt 0\) at which the gradient changes, and the end point.
(2)
(b)
Instead, \(f\) is extended to the domain \(-4\le x\le4\) so that it is an odd function. Write down the value of \(f(-1)\) and of \(f(-3)\).
(2)
(c)
The odd function in part (b) is then extended to \(x\in\mathbb{R}\) so that it is periodic with period 8. Find \(f(13)\) and \(f(-6)\).
(3)
(d)
For the function in part (c), solve \(f(x)=2\) for \(0\le x\le16\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\((-1,\,2)\) and \((-3,\,-1)\)
A1
\((-4,\,0)\)
A1
(b)
\(f(-1)=-2\)
A1
\(f(-3)=1\)
A1
(c)
\(f(13)=f(13-16)=f(-3)\) and \(f(-6)=f(2)\)
M1
Use of the period to move into \(-4\le x\le4\).
\(f(13)=1\)
A1
FT from (b).
\(f(-6)=\dfrac{1}{2}\)
A1
\(f(2)\) is the midpoint of the segment from \((1,2)\) to \((3,-1)\).
(d)
\(x=1\)
A1
In \(-4\le x\le4\), \(f(x)=2\) only at \(x=1\) (the odd extension has \(-2\le f(x)\le1\) for \(-4\le x\le0\)).
87AAHL2-q00542.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium10 marks
The functions \(f\) and \(g\) are defined for all \(x\in\mathbb{R}\). The function \(f\) is odd and the function \(g\) is even.
(a)
Prove that the product \(fg\), defined by \((fg)(x)=f(x)g(x)\), is odd.
(2)
(b)
Prove that \(g\circ f\) is even.
(2)
(c)
Determine whether \(f\circ f\) is odd, even or neither, justifying your answer.
(2)
(d)
Show that \(f(0)=0\).
(2)
(e)
Hence, without further working, state whether each of \(p(x)=x^{3}\cos x\) and \(q(x)=\cos(\sin x)\) is odd or even.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\((fg)(-x)=f(-x)g(-x)\)
M1
\(=-f(x)g(x)=-(fg)(x)\)
A1
so \(fg\) is odd
AG
(b)
\((g\circ f)(-x)=g(f(-x))=g(-f(x))\)
M1
\(=g(f(x))\) since \(g\) is even
A1
so \(g\circ f\) is even
AG
(c)
\(f(f(-x))=f(-f(x))\)
M1
\(=-f(f(x))\), so \(f\circ f\) is odd
A1
(d)
\(f(-0)=-f(0)\), so \(f(0)=-f(0)\)
M1
\(2f(0)=0\), so \(f(0)=0\)
A1
(e)
\(p\) is odd
A1
Product of odd \(x^{3}\) and even \(\cos x\).
\(q\) is even
A1
Even \(\cos\) composed with odd \(\sin\).
Answers: (c) odd · (e) \(p\) odd, \(q\) even
88AAHL2-q00582.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium8 marks
Let \(f(x)=x^{3}+x-8\), for \(x\in\mathbb{R}\).
(a)
Explain why \(f\) has an inverse function.
(2)
(b)
Find \(f^{-1}(22)\).
(2)
(c)
Explain why any solution of \(f(x)=f^{-1}(x)\) must also satisfy \(f(x)=x\).
(2)
(d)
Hence solve \(f(x)=f^{-1}(x)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(f'(x)=3x^{2}+1\)
M1
Or: \(x^{3}\) and \(x\) are both increasing, so their sum is increasing.
\(f'(x)\gt 0\) for all \(x\), so \(f\) is strictly increasing and hence one-to-one
R1
(b)
\(x^{3}+x-8=22\), i.e. \(x^{3}+x-30=0\)
M1
\(f^{-1}(22)=3\)
A1
\(x=3\) is the only real root since \(f\) is one-to-one.
(c)
if \(f(a)=f^{-1}(a)\) then \(f(f(a))=a\)
M1
if \(f(a)\gt a\) then, as \(f\) is increasing, \(f(f(a))\gt f(a)\gt a\), a contradiction; similarly \(f(a)\lt a\) is impossible, so \(f(a)=a\)
R1
Accept: the graphs of \(f\) and \(f^{-1}\) are reflections in \(y=x\), and for an increasing function they can only meet on \(y=x\), with a valid justification.
(d)
\(x^{3}+x-8=x\), so \(x^{3}=8\)
M1
\(x=2\)
A1
Answers: (b) \(3\) · (d) \(x=2\)
89AAHL2-q00592.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium9 marks
Let \(f(x)=\dfrac{x^{2}+4}{x-1}\), for \(x\gt 1\). The graph of \(f\) has a local minimum point at \(A\).
(a)
Show that the \(x\)-coordinate of \(A\) is \(1+\sqrt{5}\).
(3)
(b)
Show that the \(y\)-coordinate of \(A\) is \(2+2\sqrt{5}\).
(2)
(c)
The function \(g\) is defined by \(g(x)=f(x)\), for \(x\ge c\). Write down the least value of \(c\) for which \(g^{-1}\) exists.
(1)
(d)
For this value of \(c\), find \(g^{-1}(x)\) and write down the domain of \(g^{-1}\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
attempt at the quotient rule: \(f'(x)=\dfrac{2x(x-1)-(x^{2}+4)}{(x-1)^{2}}\)
M1
\(f'(x)=\dfrac{x^{2}-2x-4}{(x-1)^{2}}\)
A1
\(f'(x)=0\Rightarrow x=1\pm\sqrt{5}\), and \(1-\sqrt{5}\lt 1\) is not in the domain
Positive square root: the larger root of the quadratic in \(y\) is at least \(\tfrac{x}{2}\ge1+\sqrt{5}\), so lies on the branch \(x\ge1+\sqrt{5}\); the other root corresponds to \(1\lt x\le1+\sqrt{5}\).
91AAHL2-q00652.15 Solving inequalities g(x) ≥ f(x)Paper 1Medium7 marks
An open box is made from a rectangular sheet of card measuring 30 cm by 20 cm by cutting a square of side \(x\) cm from each corner and folding up the sides. The volume of the box is \(V(x)\) cm3, where \(0\lt x\lt 10\).
Corner squares of side x cm are cut out (dashed). Diagram not to scale
(a)
Show that \(V(x)=4x^{3}-100x^{2}+600x\).
(2)
(b)
Show that \(V(5)=1000\).
(1)
(c)
Hence find the exact set of values of \(x\) for which \(V(x)\ge1000\).
reject \(10+5\sqrt{2}\) as \(x\lt 10\), and sign analysis on \(0\lt x\lt 10\)
R1
\(10-5\sqrt{2}\le x\le5\)
A1
Accept \(10-\sqrt{50}\le x\le5\).
Answers: (c) \(10-5\sqrt{2}\le x\le5\)
92AAHL2-q00662.15 Solving inequalities g(x) ≥ f(x)Paper 1Medium6 marks
Let \(f(x)=x^{3}-3x+1\). The following diagram shows the graph of \(y=f(x)\). The graph has a local maximum at \((-1,\,3)\), a local minimum at \((1,\,-1)\), and crosses the \(x\)-axis at \(x=a\), \(x=b\) and \(x=c\), where \(a\lt b\lt c\).
Diagram not to scale
(a)
Solve \(f(x)\ge1\), giving your answer in exact form.
(3)
(b)
Write down, in terms of \(a\), \(b\) and \(c\), the solution of \(f(x)\le0\).
(1)
(c)
Solve \(f(2x)\gt 0\), giving your answer in terms of \(a\), \(b\) and \(c\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(x^{3}-3x\ge0\Rightarrow x(x^{2}-3)\ge0\)
M1
critical values \(-\sqrt{3},\ 0,\ \sqrt{3}\)
A1
\(-\sqrt{3}\le x\le0\) or \(x\ge\sqrt{3}\)
A1
(b)
\(x\le a\) or \(b\le x\le c\)
A1
(c)
\(a\lt 2x\lt b\) or \(2x\gt c\)
M1
FT from the complement of (b).
\(\dfrac{a}{2}\lt x\lt\dfrac{b}{2}\) or \(x\gt\dfrac{c}{2}\)
A1
Answers: (a) \(-\sqrt{3}\le x\le0\) or \(x\ge\sqrt{3}\) · (b) \(x\le a\) or \(b\le x\le c\) · (c) \(\tfrac{a}{2}\lt x\lt\tfrac{b}{2}\) or \(x\gt\tfrac{c}{2}\)
93AAHL2-q00772.16 Modulus, reciprocal and other graph transformationsPaper 1Medium8 marks
Let \(f(x)=x^{2}-2x-3\), and let \(h(x)=[f(x)]^{2}\).
(a)
Write down the zeros of \(f\) and the coordinates of the vertex of the graph of \(f\).
(2)
(b)
By writing \(h'(x)\) in terms of \(f(x)\) and \(f'(x)\), find the coordinates of all the stationary points on the graph of \(y=h(x)\).
(4)
(c)
Find the set of values of \(k\) for which the equation \([f(x)]^{2}=k\) has exactly four solutions.
(2)
Show mark scheme
Working
Mark
Notes
(a)
zeros \(x=-1\) and \(x=3\)
A1
vertex \((1,\,-4)\)
A1
(b)
\(h'(x)=2f(x)f'(x)\)
M1
Chain rule.
\(h'(x)=0\Rightarrow f(x)=0\) or \(f'(x)=0\)
M1
\((-1,\,0)\) and \((3,\,0)\)
A1
\((1,\,16)\)
A1
(c)
\(f(x)=\pm\sqrt{k}\); \(f(x)=-\sqrt{k}\) has two solutions only when \(-4\lt -\sqrt{k}\lt 0\)
99IBMA2H-047-01-042.15 Solving inequalities g(x) ≥ f(x)Paper 1Medium6 marks
No calculator
Solve the inequality \(\dfrac{x+5}{x-1}\le x+1\), giving your answer in set or interval form.
(a)
Solve the inequality.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Uses a valid method that does not assume \(x-1>0\). Either collect terms on one side, \(\frac{x+5-(x+1)(x-1)}{x-1}\le 0\), or multiply both sides by \((x-1)^2>0\) with \(x\ne 1\).
✓ 1
A1 Obtains \(\frac{(x-3)(x+2)}{x-1}\ge 0\), or equivalently \((x-1)(x-3)(x+2)\ge 0\) with \(x\ne 1\).
✓ 1
M1 Identifies the critical values from the factors.
✓ 1
A1 States \(x=-2,\ 1,\ 3\).
✓ 1
M1 Uses a sign table, test values in each interval or a sketch of the cubic to choose the correct intervals.
Common error: Multiplying both sides by \((x-1)\) assumes it is positive and loses the interval where \(x<1\). Collect the terms over a common denominator, or multiply by \((x-1)^2\) instead. Remember that \(x=1\) can never be included.
100IBMA2H-047-01-082.15 Solving inequalities g(x) ≥ f(x)Paper 1Medium7 marks
No calculator
An open box is made from a square sheet of aluminium of side 12 cm by cutting a square of side x cm from each corner and folding up the sides. The volume of the box is V cm3.
Corner squares of side x cm are cut out (dashed). Diagram not to scale
(a)
Show that \(V=4x(6-x)^2\) and state the domain of x for which a box can be made.
(2)
(b)
Show that the condition \(V\ge 128\) leads to \(x^3-12x^2+36x-32\ge 0\), and hence solve \(V\ge 128\) for x in this domain.
(4)
(c)
Interpret your answer to the previous part in context.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses base side \(12-2x\) and height \(x\) to get \(V=x(12-2x)^2=x\cdot 4(6-x)^2=4x(6-x)^2\) AG
✓ 1
A1 \(0<x<6\)
✓ 1
Part (b)
M1 divides by 4 to get \(x(6-x)^2\ge 32\), then expands to \(x^3-12x^2+36x-32\ge 0\) AG
✓ 1
A1 uses the factor theorem (\(x=2\) gives \(8-48+72-32=0\)) to obtain \((x-2)^2(x-8)\ge 0\)
✓ 1
M1 notes that \((x-2)^2\ge 0\), so the sign is that of \((x-8)\), which is negative for \(0<x<6\), except where \(x=2\)
✓ 1
A1 the only solution in \(0<x<6\) is \(x=2\)
✓ 1
Part (c)
R1 the volume can never exceed 128 cm3; its maximum of 128 cm3 occurs only when \(x=2\)
✓ 1
Exam tip: A repeated factor such as \((x-2)^2\) does not change sign at its root. The cubic only touches the axis there, which is why a single point can satisfy the inequality.
101IBMA2H-047-01-102.7 Quadratic equations, inequalities and the discriminantPaper 1Medium7 marks
No calculator
Consider the quadratic equation \(kx^2+(k+3)x+1=0\), where \(k\ne 0\).
(a)
Show that the equation has two distinct real roots for every non-zero value of k.
(3)
(b)
The roots are \(\alpha\) and \(\beta\). Given that \(\alpha^2+\beta^2=6\), find the possible values of k.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 finds the discriminant: \((k+3)^2-4k\)
✓ 1
A1 \(k^2+2k+9=(k+1)^2+8\)
✓ 1
R1 \((k+1)^2+8\ge 8>0\) for all k, so there are two distinct real roots
✓ 1
Part (b)
M1 uses \(\alpha+\beta=-\frac{k+3}{k}\), \(\alpha\beta=\frac{1}{k}\) and \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\)
✓ 1
A1 \(\frac{(k+3)^2}{k^2}-\frac{2}{k}=6\)
✓ 1
A1 \(5k^2-4k-9=0\)
✓ 1
A1 \(k=-1\) or \(k=\frac{9}{5}\)
✓ 1
Common error: Do not expand \(\alpha^2+\beta^2\) as \((\alpha+\beta)^2\). You must subtract \(2\alpha\beta\). Here part (a) also guarantees that both values of k give real roots, so neither is rejected.
102IBMA2H-047-04-012.2 Functions, domain, range and inverse functionsPaper 1Medium7 marks
No calculator
An audio amplifier uses a 'soft-clipping' function to stop loud signals from distorting. An input voltage \(x\) is converted into an output \(f(x)=\frac{x}{\sqrt{1+x^2}}\), \(x\in\mathbb{R}\).
(a)
Show that \(f\) is an odd function.
(2)
(b)
Verify that \(f(0.75)=0.6\) and hence write down the value of \(f(-0.75)\).
(2)
(c)
Find an expression for \(f^{-1}(x)\), stating its domain.
A1 \(=-\frac{x}{\sqrt{1+x^2}}=-f(x)\), so \(f\) is odd
✓ 1
Part (b)
A1 \(\sqrt{1+0.5625}=\sqrt{1.5625}=1.25\), so \(f(0.75)=\frac{0.75}{1.25}=0.6\)
✓ 1
A1 \(f(-0.75)=-f(0.75)=-0.6\)
✓ 1
Part (c)
M1 sets \(y=\frac{x}{\sqrt{1+x^2}}\), squares and rearranges: \(y^2(1+x^2)=x^2\) so \(x^2=\frac{y^2}{1-y^2}\)
✓ 1
A1 \(x\) has the same sign as \(y\), so \(f^{-1}(x)=\frac{x}{\sqrt{1-x^2}}\)
✓ 1
A1 domain \(-1<x<1\) (the range of \(f\))
✓ 1
Exam tip: When you square to find an inverse, decide which sign of the root to keep: here \(f(x)\) and \(x\) always share a sign, so the positive root is correct.
103IBMA2H-047-04-062.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
The restoring force in a bistable micro-switch is modelled by \(F(x)=ax^3+bx^2+cx+d\), where \(x\) is the displacement from the central position. The model is an odd function, \(F(2)=4\), and the graph of \(F\) has a local maximum where \(x=-1\).
(a)
Show that \(b=d=0\).
(2)
(b)
Find the values of \(a\) and \(c\).
(3)
(c)
Without further differentiation, write down the coordinates of the local minimum of \(F\), justifying your answer.
R1 so \(2bx^2+2d=0\) for all \(x\), giving \(b=0\) and \(d=0\)
✓ 1
Part (b)
M1 \(F'(x)=3ax^2+c\) and \(F'(-1)=0\) gives \(3a+c=0\)
✓ 1
M1 \(F(2)=4\) gives \(8a+2c=4\)
✓ 1
A1 \(a=2\), \(c=-6\)
✓ 1
Part (c)
R1 \(F(-1)=-2+6=4\), so the local maximum is at \((-1,4)\). The graph of an odd function has rotational symmetry of order 2 about the origin, so this maximum maps to a local minimum at \(x=1\)
✓ 1
A1 \((1,-4)\)
✓ 1
Exam tip: To prove coefficients are zero, compare both sides for all \(x\); checking one value of \(x\) is not enough.
104IBMA2H-047-04-072.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium7 marks
No calculator
Let \(f(x)=\frac{2x+5}{x+p}\), \(x\neq-p\), where \(p\neq-\frac{5}{2}\) is a constant.
(a)
Write down the equations of the asymptotes in terms of \(p\).
(2)
(b)
Given that \(f\) is self-inverse, use a property of its graph to find \(p\).
(2)
(c)
Find the coordinates of the points where the graph meets the line \(y=x\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertical asymptote \(x=-p\)
✓ 1
A1 horizontal asymptote \(y=2\)
✓ 1
Part (b)
R1 the graph of a self-inverse function is symmetric in \(y=x\), so the asymptotes are reflections of each other in \(y=x\): the reflection of \(y=2\) is \(x=2\), so \(-p=2\) (equivalently, the reflection of \(x=-p\) is \(y=-p\), which must be \(y=2\))
R1 \(f\) is increasing, so any solution lies on \(y=x\), giving \(2x^2-13x+23=0\)
✓ 1
R1 \(\Delta=169-184=-15<0\), so there are no solutions
✓ 1
Common error: Writing \(\pm\sqrt{\ }\) in an inverse is wrong: the restricted domain \(x\geq 3\) forces the range of \(f^{-1}\) to be \(y\geq 3\), so only the positive root is valid.
107IBMA2H-047-03-112.5 Composite functions and inverse functionsPaper 1Medium10 marks
No calculator
The function \(h\) is defined by \(h(x)=\ln\left(\frac{x-3}{x+1}\right)\) on its largest possible domain.
R1 \(h^{-1}\) is undefined only at \(x=0\) (or: \(\frac{x-3}{x+1}=1-\frac{4}{x+1}\) takes every positive value except 1)
✓ 1
A1 range: \(y\in\mathbb{R}\), \(y\neq 0\)
✓ 1
Part (d)
M1 \(\frac{x-3}{x+1}=3\), giving \(x-3=3x+3\)
✓ 1
A1 \(x=-3\) (which lies in the domain since \(-3<-1\))
✓ 1
Common error: Requiring both \(x-3>0\) and \(x+1>0\) gives only \(x>3\) and misses the interval \(x<-1\), where both factors are negative and the quotient is still positive.
108IBMA2H-047-06-042.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium7 marks
No calculator
The function \(f\) is defined by \(f(x)=x^2-4x+6\), for \(x\ge 2\).
(a)
Write \(f(x)\) in the form \((x-h)^2+k\).
(2)
(b)
Find \(f^{-1}(x)\), stating its domain.
(2)
(c)
Find the coordinates of the points where the graphs of \(y=f(x)\) and \(y=f^{-1}(x)\) intersect.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 completes the square: \((x-2)^2-4+6\)
✓ 1
A1 \(f(x)=(x-2)^2+2\)
✓ 1
Part (b)
M1 interchanges \(x\) and \(y\) and rearranges: \(y=2\pm\sqrt{x-2}\)
✓ 1
A1 \(f^{-1}(x)=2+\sqrt{x-2}\), \(x\ge 2\) (positive root as the range of \(f^{-1}\) is \(y\ge 2\))
✓ 1
Part (c)
R1 \(f\) is increasing on its domain, so any intersection of \(f\) and \(f^{-1}\) lies on \(y=x\)
✓ 1
M1 solves \(x^2-4x+6=x\), i.e. \(x^2-5x+6=0\), \((x-2)(x-3)=0\)
✓ 1
A1 \((2,2)\) and \((3,3)\), both in the domain
✓ 1
Exam tip: Replacing \(f^{-1}(x)\) by \(x\) is only valid with a reason: say that \(f\) is increasing, so the graphs can only meet on the line \(y=x\).
109IBMA2H-047-06-062.15 Solving inequalities g(x) ≥ f(x)Paper 1Medium6 marks
No calculator
A control system for a robotic arm accepts a setting \(x\) only if the error function \(E(x)=|x^2-5x|\) satisfies \(E(x)\le 6\). Solve the inequality \(|x^2-5x|\le 6\) to find all acceptable settings.
(a)
Solve \(|x^2-5x|\le 6\).
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes the inequality as \(-6\le x^2-5x\le 6\)
✓ 1
M1 solves \(x^2-5x-6\le0\): \((x-6)(x+1)\le 0\)
✓ 1
A1 \(-1\le x\le 6\)
✓ 1
M1 solves \(x^2-5x+6\ge0\): \((x-2)(x-3)\ge 0\)
✓ 1
A1 \(x\le 2\) or \(x\ge 3\)
✓ 1
A1 \(-1\le x\le 2\) or \(3\le x\le 6\)
✓ 1
Common error: Solving only \(x^2-5x\le 6\) gives \(-1\le x\le 6\), which wrongly includes \(26\).
110IBMA2H-047-06-112.5 Composite functions and inverse functionsPaper 1Medium12 marks
No calculator
Consider the function \(f(x)=\frac{2x+k}{x-2}\), \(x\ne2\), where \(k\) is a constant and \(k\ne-4\).
M1 multiplies numerator and denominator by \((x-2)\): numerator \(4x+2k+kx-2k=x(k+4)\), denominator \(2x+k-2x+4=k+4\)
✓ 1
A1 \(f(f(x))=\frac{x(k+4)}{k+4}=x\) since \(k\ne-4\)
✓ 1
Part (b)
A1 \(f^{-1}(x)=\frac{2x+k}{x-2}\), \(x\ne2\) (\(f\) is self-inverse)
✓ 1
Part (c)
R1 if \(k=-4\), \(f(x)=\frac{2(x-2)}{x-2}=2\) for all \(x\ne2\), which is constant and not one-to-one, so no inverse exists
✓ 1
Part (d)
M1 \(2x+5=x(x-2)\)
✓ 1
M1 \(x^2-4x-5=0\), \((x-5)(x+1)=0\)
✓ 1
A1 \(x=5\) or \(x=-1\)
✓ 1
Part (e)
M1 forms a single fraction: \(\frac{2x+5-x(x-2)}{x-2}>0\)
✓ 1
A1 \(\frac{-(x-5)(x+1)}{x-2}>0\), i.e. \(\frac{(x-5)(x+1)}{x-2}<0\)
✓ 1
M1 sign analysis using critical values \(-1\), \(2\), \(5\)
✓ 1
A1 \(x<-1\) or \(2<x<5\)
✓ 1
Common error: Multiplying both sides of \(\frac{2x+5}{x-2}>x\) by \((x-2)\) ignores that \(x-2\) can be negative. Rearrange to a single fraction compared with zero instead.
111IBMA2H-047-00-062.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
In a laser alignment rig, a reflector has a curved edge modelled by \(y = x^2 - 2x + 6\). A laser beam starts at the point \((0, 2)\) and travels along the line \(y = mx + 2\). The beam just grazes the reflector, so the line is a tangent to the curve.
(a)
Show that \(m^2 + 4m - 12 = 0\).
(3)
(b)
Find the possible values of m.
(2)
(c)
For the positive value of m, find the coordinates of the point where the beam touches the reflector.
A1 \(3 - \sqrt{7} < x < 2\) or \(4 < x < 3 + \sqrt{7}\)
✓ 1
Part (d)
A1 \(k = 4\) (the reflected vertex height)
✓ 1
Part (e)
A1 \(k = 0\) (solutions \(x = 1, 5\))
✓ 1
A1 \(k > 4\)
✓ 1
Exam tip: The minimum of \(f\) at \((3, -4)\) becomes a local maximum \((3, 4)\) of \(|f|\). Counting intersections with horizontal lines \(y = k\) above and below 4 settles the solution counts.
113IBMA2H-047-00-102.11 Transformations of graphsPaper 1Medium7 marks
No calculator
Let \(f(x) = x^2 + (k - 3)x + k\), where \(k \in \mathbb{R}\).
(a)
Show that the discriminant of \(f(x)\) is \(k^2 - 10k + 9\).
(2)
(b)
Find the set of values of k for which \(f(x) > 0\) for all real x.
(3)
(c)
For \(k = 9\), describe fully a single transformation that maps the graph of \(y = x^2\) onto the graph of \(y = f(x)\), and state the root of \(f(x) = 0\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\Delta = (k-3)^2 - 4k\)
✓ 1
A1 \(k^2 - 6k + 9 - 4k = k^2 - 10k + 9\)
✓ 1
Part (b)
R1 requires \(\Delta < 0\) (since the leading coefficient is positive)
✓ 1
M1 \((k - 1)(k - 9) < 0\), or critical values \(k = 1\) and \(k = 9\)
Exam tip: "Positive for all x" needs both a positive leading coefficient and a negative discriminant; state both conditions in your reasoning.
114IBMA2H-047-07-122.11 Transformations of graphsPaper 1Medium7 marks
No calculator
Let \(f(x)=\dfrac{x+2}{x-1}\), for \(x\ne1\).
(a)
Show that \(f(x)=1+\dfrac{3}{x-1}\).
(1)
(b)
Describe a sequence of transformations that maps the graph of \(y=\frac{1}{x}\) onto the graph of \(y=f(x)\), stating clearly the order in which they are applied.
A1 first a vertical stretch (parallel to the \(y\)-axis), scale factor 3
✓ 1
A1 then a translation by \(\binom{1}{1}\)
✓ 1
Accept other correct sequences, e.g. translation by \(\binom{1}{0}\), then vertical stretch with scale factor 3, then translation by \(\binom{0}{1}\). Do not award the second A1 if a translation by \(\binom{1}{1}\) is applied before the stretch.
—
Part (c)
M1 rearranges to a single fraction without multiplying by \(x-1\): \(\frac{x^2-2x-2}{x-1}\le0\)
✓ 1
A1 critical values \(x=1\pm\sqrt{3}\) and \(x=1\)
✓ 1
M1 sign analysis over the four intervals
✓ 1
A1 \(x\le1-\sqrt{3}\) or \(1<x\le1+\sqrt{3}\)
✓ 1
Common error: Multiplying both sides by \(x-1\) ignores that it can be negative and loses the interval \(x\le1-\sqrt{3}\); collect terms on one side instead.
115IBMA2H-047-02-052.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium9 marks
No calculator
The cubic polynomial \(p(x) = x^3 + ax^2 + bx - 15\) has real coefficients. One of the roots of \(p(x) = 0\) is \(1 + 2i\).
(a)
Write down a second root of \(p(x) = 0\).
(1)
(b)
Use the product of the roots to find the real root of \(p(x) = 0\).
(2)
(c)
Find the value of \(a\) and the value of \(b\).
(3)
(d)
Solve the inequality \(p(x) > 0\) for \(x \in \mathbb{R}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(1 - 2i\)
✓ 1
Part (b)
M1 product of roots \(= 15\), so \((1 + 2i)(1 - 2i)r = 15\)
✓ 1
A1 \(5r = 15\), so \(r = 3\)
✓ 1
Part (c)
M1 uses the sum of roots \(= -a\), or expands \((x^2 - 2x + 5)(x - 3)\)
✓ 1
A1 \(a = -5\)
✓ 1
A1 \(b = 11\)
✓ 1
Part (d)
M1 writes \(p(x) = (x - 3)(x^2 - 2x + 5)\)
✓ 1
R1 \(x^2 - 2x + 5 = (x - 1)^2 + 4 > 0\) for all real \(x\), so the sign of \(p(x)\) is the sign of \(x - 3\)
✓ 1
A1 \(x > 3\)
✓ 1
Exam tip: Complex roots of a real polynomial come in conjugate pairs, and their product is a positive real number. Use this to reach the real root quickly without long division.
116IBMA2H-047-02-072.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium10 marks
No calculator
A polynomial \(p(x)\) leaves a remainder of 3 when divided by \((x - 1)\) and a remainder of \(-6\) when divided by \((x + 2)\).
(a)
Find the remainder when \(p(x)\) is divided by \((x - 1)(x + 2)\).
(4)
(b)
\(p(x)\) is a cubic with leading coefficient 2, and \(p(0) = 4\). Show that \(p(x) = 2x^3 - 3x + 4\).
(3)
(c)
By considering the stationary points of \(y = p(x)\), show that \(p(x) = 0\) has exactly one real root.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the remainder is linear, so write \(p(x) = (x - 1)(x + 2)q(x) + ax + b\)
✓ 1
M1 substitutes \(x = 1\) and \(x = -2\) to give \(a + b = 3\) and \(-2a + b = -6\)
A1 the local minimum value is \(p\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} - \frac{3}{\sqrt{2}} + 4 = 4 - \sqrt{2}\)
✓ 1
R1 \(4 - \sqrt{2} > 0\), so the curve is above the axis for \(x > -\frac{1}{\sqrt{2}}\) and crosses the axis only once, to the left of the local maximum, since \(p(x) \to -\infty\) as \(x \to -\infty\)
✓ 1
Exam tip: When the divisor is a quadratic, the remainder has degree at most one. Always write it as \(ax + b\) and use one substitution for each linear factor.
117IBMA2H-047-02-102.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium11 marks
No calculator
The length, width and height, in metres, of a rectangular shipping crate are the three roots of the equation \(2x^3 - 19x^2 + px - 50 = 0\), where \(p\) is a constant.
(a)
Find the volume of the crate.
(2)
(b)
Find the total length of all twelve edges of the crate.
(2)
(c)
The length of the space diagonal of the crate is \(\frac{\sqrt{141}}{2}\) m. Find the value of \(p\) and hence write down the total surface area of the crate.
(4)
(d)
Given that the height of the crate is 2 m, find its other two dimensions.
A1 the quotient is \(2x^2 - 15x + 25 = (2x - 5)(x - 5)\)
✓ 1
A1 the other dimensions are 2.5 m and 5 m
✓ 1
Exam tip: Each symmetric function of the roots has a physical meaning here: the sum gives the edges, the sum of pairs gives the surface area, the product gives the volume, and the sum of squares gives the diagonal.
118IBMA2H-047-05-052.1 Equations of straight lines, parallel and perpendicular linesPaper 1Medium6 marks
No calculator
In a computer-aided design program, two straight laser-cut edges lie along the lines \(L_1: (k-1)x+4y=7\) and \(L_2: 3x-(k+1)y=2\), where \(k \in \mathbb{R}\) and \(k \neq -1\).
(a)
Find the value of \(k\) for which \(L_1\) and \(L_2\) are perpendicular.
(3)
(b)
Show that there is no real value of \(k\) for which \(L_1\) and \(L_2\) are parallel.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 finds both gradients: \(m_1=\frac{1-k}{4}\), \(m_2=\frac{3}{k+1}\)
✓ 1
M1 sets \(m_1 m_2=-1\), e.g. \(3(1-k)=-4(k+1)\)
✓ 1
A1 \(k=-7\)
✓ 1
Part (b)
M1 sets \(\frac{1-k}{4}=\frac{3}{k+1}\)
✓ 1
A1 obtains \(1-k^2=12\), so \(k^2=-11\)
✓ 1
R1 \(k^2 \geq 0\) for all real \(k\), so no real solution exists
✓ 1
Exam tip: For a 'show that' with no solution, you must state the reason explicitly (here, a square cannot be negative) to earn the reasoning mark.
119IBMA2H-047-05-072.1 Equations of straight lines, parallel and perpendicular linesPaper 1Medium7 marks
No calculator
On a map with a coordinate grid in kilometres, a straight cycle path lies along the line \(2x-y+3=0\). A farmhouse is at P(7, 0). A straight access lane is to be built from P to the point F on the cycle path, so that the lane is perpendicular to the path.
Coordinates in kilometres
(a)
Find the equation of the line containing the access lane.
(2)
(b)
Find the coordinates of F.
(3)
(c)
Find the exact length of the access lane.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 perpendicular gradient \(-\frac{1}{2}\)
✓ 1
A1 \(y=-\frac{1}{2}x+\frac{7}{2}\) or \(x+2y-7=0\)
✓ 1
Part (b)
M1 solves simultaneously, e.g. \(2x+3=-\frac{1}{2}x+\frac{7}{2}\)
Common error: Rearrange \(2x-y+3=0\) to \(y=2x+3\) before reading off the gradient; the coefficient of \(x\) in the general form is not the gradient unless \(y\) has coefficient 1.
120IBMA2H-047-05-082.15 Solving inequalities g(x) ≥ f(x)Paper 1Medium7 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{x+5}{x-1}\), \(x \neq 1\).
(a)
Solve the inequality \(f(x) \leq 3\).
(4)
(b)
Hence solve \(|f(x)| \leq 3\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rearranges to a single fraction, e.g. \(\frac{8-2x}{x-1} \leq 0\)
A1 combines with part (a): \(x \leq -\frac{1}{2}\) or \(x \geq 4\)
✓ 1
Common error: Multiplying both sides by \(x-1\) ignores that \(x-1\) may be negative, which reverses the inequality. Move everything to one side and analyse signs instead.
121IBMA2H-047-05-102.13 Rational functions with oblique and horizontal asymptotesPaper 1Medium11 marks
No calculator
Consider the function \(f(x)=\frac{2x+4}{x^2-2x-3}\), defined for all real \(x\) except where the denominator is zero.
(a)
Find the equations of all asymptotes of the graph of \(f\).
(3)
(b)
Find the coordinates of the intercepts of the graph of \(f\) with the axes.
(2)
(c)
Solve \(f(x)>0\).
(3)
(d)
Solve \(f(x)=1\), giving your answers in exact form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 factorises denominator as \((x-3)(x+1)\)
✓ 1
A1 vertical asymptotes \(x=3\) and \(x=-1\)
✓ 1
A1 horizontal asymptote \(y=0\)
✓ 1
Part (b)
A1 \((-2, 0)\)
✓ 1
A1 \(\left(0, -\frac{4}{3}\right)\)
✓ 1
Part (c)
M1 sign analysis using critical values \(-2\), \(-1\) and \(3\)
✓ 1
A1 \(-2<x<-1\)
✓ 1
A1 \(x>3\)
✓ 1
Part (d)
M1 \(2x+4=x^2-2x-3\)
✓ 1
A1 \(x^2-4x-7=0\)
✓ 1
A1 \(x=2\pm\sqrt{11}\)
✓ 1
Exam tip: When the degree of the denominator exceeds the degree of the numerator, the horizontal asymptote is \(y=0\). The graph may still cross this asymptote, here at \(x=-2\).
122IBMA3H-2-18r7-052.5 Composite functions and inverse functionsPaper 1Medium7 marks
No calculator
The function f is defined by \(f(x)=\frac{3x+1}{x-2}\), for x ∈ ℝ, x ≠ 2.
(a)
Find an expression for f−1(x), and state its domain.
(3)
(b)
Write down the range of f.
(1)
(c)
Find the values of x for which f(x) = f−1(x). Give your answers to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(y(x-2)=3x+1\) and collects x terms: \(x(y-3)=2y+1\)
✓ 1
A1 \(f^{-1}(x)=\frac{2x+1}{x-3}\)
✓ 1
A1 domain \(x\ne 3\)
✓ 1
Part (b)
A1 \(y\ne 3\)
✓ 1
Part (c)
M1 cross-multiplies: \((3x+1)(x-3)=(2x+1)(x-2)\)
✓ 1
A1 \(x^2-5x-1=0\)
✓ 1
A1 \(x=\frac{5\pm\sqrt{29}}{2}\), i.e. \(x=5.19\) or \(x=-0.193\)
✓ 1
Exam tip: The domain of f⁻¹ excludes the value that is missing from the range of f.
123IBMA3H-2-18r7-082.16 Modulus, reciprocal and other graph transformationsPaper 1Medium6 marks
No calculator
Consider the graphs of \(y=\frac{2}{x-1}\) and \(y=x-2\).
(a)
Find the x-coordinates of the points of intersection of the two graphs.
(3)
(b)
Hence solve the inequality \(\frac{2}{x-1}\ge x-2\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets \(\frac{2}{x-1}=x-2\)
✓ 1
A1 \(x^2-3x=0\)
✓ 1
A1 \(x=0\) and \(x=3\)
✓ 1
Part (b)
R1 the vertical asymptote \(x=1\) is a critical value and is excluded
✓ 1
A1 \(x\le 0\)
✓ 1
A1 \(1<x\le 3\)
✓ 1
Common error: Do not multiply both sides by (x − 1): its sign is unknown, so the inequality direction may flip. Compare the graphs or use a sign diagram.
124IBMA3H-2-19r8-062.13 Rational functions with oblique and horizontal asymptotesPaper 1Medium7 marks
No calculator
Let \(f(x)=x^2-2x-3\) and \(h(x)=\frac{1}{f(x)}\).
(a)
Factorise \(f(x)\) and hence write down the equations of the vertical asymptotes of the graph of \(y=h(x)\).
(2)
(b)
Find the coordinates of the vertex of \(y=f(x)\) and hence the coordinates of the local maximum point of \(y=h(x)\).
(2)
(c)
Sketch the graph of \(y=h(x)\), showing the asymptotes, the local maximum and the \(y\)-intercept.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(f(x)=(x-3)(x+1)\)
✓ 1
A1 vertical asymptotes \(x=3\) and \(x=-1\)
✓ 1
Part (b)
A1 vertex of \(f\) at \((1,-4)\)
✓ 1
A1 local maximum of \(h\) at \(\left(1,-\frac{1}{4}\right)\)
✓ 1
Part (c)
A1 outer branches (\(x<-1\) and \(x>3\)) above the \(x\)-axis, tending to \(+\infty\) at the vertical asymptotes and to \(0\) as \(|x|\to\infty\) (horizontal asymptote \(y=0\))
✓ 1
A1 middle branch below the \(x\)-axis, tending to \(-\infty\) at both vertical asymptotes, with local maximum \(\left(1,-\frac{1}{4}\right)\)
Exam tip: Where \(f(x)\) has a minimum, \(\frac{1}{f(x)}\) has a maximum (if \(f\) is negative there), and zeros of \(f\) become vertical asymptotes.
125IBMA3H-2-19r8-072.5 Composite functions and inverse functionsPaper 1Medium6 marks
No calculator
Let \(g(x)=\frac{x+1}{x-2}\), for \(x\neq 2\), and \(f(x)=2\). Find all values of \(x\) for which \(g(x)\geq f(x)\).
(a)
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to form \(\frac{x+1}{x-2}-2\geq 0\)
✓ 1
M1 combine over a common denominator
✓ 1
A1 \(\frac{5-x}{x-2}\geq 0\)
✓ 1
M1 critical values \(x=2\) and \(x=5\) with a sign analysis or sketch
✓ 1
R1 \(x=2\) excluded because \(g\) is undefined there
✓ 1
A1 \(2<x\leq 5\)
✓ 1
Common error: Do not multiply both sides by \((x-2)\) without considering its sign; moving everything to one side and analysing the sign of the fraction avoids the problem.
126IBMA3H-2-19r8-102.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
A cycle path is modelled by the curve \(C\) with equation \(y=x^2-6x+11\). A straight fence is modelled by the line \(L\) with equation \(y=2x+k\), where \(k\) is a constant.
(a)
Write \(y=x^2-6x+11\) in the form \(y=(x-p)^2+q\).
(2)
(b)
Find the value of \(k\) for which \(L\) is a tangent to \(C\).
(3)
(c)
Find the \(y\)-coordinate of the point where \(L\) touches \(C\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 completing the square, e.g. \((x-3)^2\)
✓ 1
A1 \(y=(x-3)^2+2\)
✓ 1
Part (b)
M1 equate to obtain \(x^2-8x+11-k=0\)
✓ 1
M1 set discriminant equal to zero: \(64-4(11-k)=0\)
✓ 1
A1 \(k=-5\)
✓ 1
Part (c)
M1 repeated root \(x=\frac{6+2}{2}=4\) and substitute into \(C\) or \(L\)
✓ 1
A1 \(y=3\)
✓ 1
Exam tip: A line is a tangent to a parabola exactly when the quadratic formed by equating them has a repeated root, that is, discriminant zero.
127IBMA3H-2-19r8-142.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium6 marks
No calculator
A data centre compares two cost models, in thousands of euros (€), for \(x\) hundred racks, where \(x>0\). Plan A costs \(g(x)=x^3\) and plan B costs \(f(x)=7x-6\).
(a)
Show that \((x-1)\) is a factor of \(x^3-7x+6\) and factorise the expression completely.
(3)
(b)
Hence find the values of \(x\) for which plan A costs at least as much as plan B.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 evaluate \(1-7+6=0\), so by the factor theorem \((x-1)\) is a factor
✓ 1
M1 attempt to divide or compare coefficients to obtain a quadratic factor \(x^2+x-6\)
✓ 1
A1 \((x-1)(x-2)(x+3)\)
✓ 1
Part (b)
M1 \(g(x)-f(x)\geq 0\) with a sign diagram or sketch using roots \(-3\), \(1\), \(2\)
✓ 1
A1 \(-3\leq x\leq 1\) or \(x\geq 2\)
✓ 1
A1 apply the domain \(x>0\): \(0<x\leq 1\) or \(x\geq 2\)
✓ 1
Exam tip: A positive cubic with three distinct roots is positive on the outer-right interval and on alternate intervals after it; check one test value in each interval.
128IBMA3H-2-19r13-062.5 Composite functions and inverse functionsPaper 1Medium7 marks
No calculator
Let \(f(x)=\frac{2x+1}{x-3}\) for \(x\in\mathbb{R},\ x\ne 3\).
(a)
Find \(f^{-1}(x)\), stating clearly any value of \(x\) that must be excluded from its domain.
(3)
(b)
State the range of \(f\).
(1)
(c)
Solve \(f(x)=f^{-1}(x)\), giving your answers in exact form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 swap \(x\) and \(y\) (or rearrange) and clear the fraction: \(x(y-3)=2y+1\)
✓ 1
M1 collect \(y\) terms and factorise: \(y(x-2)=3x+1\)
✓ 1
A1 \(f^{-1}(x)=\frac{3x+1}{x-2}\), \(x\ne 2\)
✓ 1
Part (b)
A1 \(f(x)\in\mathbb{R},\ f(x)\ne 2\)
✓ 1
Part (c)
M1 equate and cross-multiply: \((2x+1)(x-2)=(3x+1)(x-3)\)
✓ 1
M1 rearrange to \(x^2-5x-1=0\)
✓ 1
A1 \(x=\frac{5\pm\sqrt{29}}{2}\)
✓ 1
Exam tip: The range of \(f\) equals the domain of \(f^{-1}\), which is a quick check that your inverse has the right excluded value.
129IBMA3H-2-19r13-072.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium7 marks
No calculator
Let \(f(x)=x^2-6x+11\) for \(x\ge k\), where \(k\) is a constant.
(a)
Find the smallest value of \(k\) for which \(f\) has an inverse function.
(2)
(b)
For this value of \(k\), find \(f^{-1}(x)\) and state its domain.
(4)
(c)
Find \(f^{-1}(18)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find the vertex, e.g. complete the square \(f(x)=(x-3)^2+2\)
✓ 1
A1 \(k=3\)
✓ 1
Part (b)
M1 set \(y=(x-3)^2+2\) and swap or rearrange
✓ 1
M1 \(x-3=\sqrt{y-2}\) (positive root since \(x\ge 3\))
✓ 1
A1 \(f^{-1}(x)=3+\sqrt{x-2}\)
✓ 1
A1 domain \(x\ge 2\)
✓ 1
Part (c)
A1 \(3+\sqrt{16}=7\)
✓ 1
Common error: Writing \(\pm\sqrt{x-2}\) in the inverse. The restriction \(x\ge 3\) chooses the positive root only.
130IBMA3H-2-19r13-102.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
The roots of the equation \(2x^3-9x^2+4x+15=0\) are \(\alpha,\beta,\gamma\).
(a)
Write down the values of \(\alpha+\beta+\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\) and \(\alpha\beta\gamma\).
(3)
(b)
Find the value of \(\alpha^2+\beta^2+\gamma^2\).
(2)
(c)
Find the value of \(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\alpha+\beta+\gamma=\frac{9}{2}\)
✓ 1
A1 \(\alpha\beta+\beta\gamma+\gamma\alpha=2\)
✓ 1
A1 \(\alpha\beta\gamma=-\frac{15}{2}\)
✓ 1
Part (b)
M1 use \((\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)\)
✓ 1
A1 \(\frac{81}{4}-4=\frac{65}{4}\)
✓ 1
Part (c)
M1 write as \(\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}\)
✓ 1
A1 \(-\frac{4}{15}\)
✓ 1
Common error: Sign slips. For \(ax^3+bx^2+cx+d=0\) the sum of roots is \(-\frac{b}{a}\), the sum of pairs is \(\frac{c}{a}\) and the product is \(-\frac{d}{a}\).
131IBMA3H-2-19r13-112.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
The roots of the equation \(x^2-5x+k=0\) are \(\alpha\) and \(\beta\), where \(\alpha>\beta\) and \(\alpha-\beta=3\).
(a)
Use \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\) to find the value of \(k\).
(2)
(b)
Find a quadratic equation, with integer coefficients, whose roots are \(\alpha^2\) and \(\beta^2\).
(3)
(c)
Find the value of \(\alpha^3+\beta^3\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(9=25-4k\)
✓ 1
A1 \(k=4\)
✓ 1
Part (b)
M1 attempt to find the sum \(\alpha^2+\beta^2=25-2\times 4\)
✓ 1
A1 sum \(=17\) and product \(=16\)
✓ 1
A1 \(x^2-17x+16=0\)
✓ 1
Part (c)
M1 use \((\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)\)
✓ 1
A1 \(125-60=65\)
✓ 1
Exam tip: Express every symmetric quantity in terms of \(\alpha+\beta\) and \(\alpha\beta\); you never need to find the roots themselves.
132IBMA3H-2-19r21-072.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
The polynomial \(p(x)=2x^3+ax^2-5x+b\), where \(a,b\in\mathbb{Z}\), has \((x+1)\) as a factor. When \(p(x)\) is divided by \((x-2)\) the remainder is 12.
(a)
Find the value of \(a\) and the value of \(b\).
(4)
(b)
Hence factorise \(p(x)\) completely.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 factor theorem: \(p(-1)=0\) so \(-2+a+5+b=0\), giving \(a+b=-3\)
✓ 1
M1 remainder theorem: \(p(2)=12\) so \(16+4a-10+b=12\), giving \(4a+b=6\)
✓ 1
A1 \(a=3\)
✓ 1
A1 \(b=-6\)
✓ 1
Part (b)
M1 divides by \((x+1)\) or equivalent to obtain a quadratic factor
✓ 1
A1 \(2x^2+x-6\)
✓ 1
A1 \(p(x)=(x+1)(2x-3)(x+2)\)
✓ 1
Common error: The remainder on dividing by \((x-k)\) is the value of the polynomial at \(x=k\), not at \(x=-k\).
133IBMA3H-2-19r21-082.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
The roots of the equation \(2x^3-6x^2+3x+4=0\) are \(\alpha\), \(\beta\) and \(\gamma\).
(a)
Write down the values of \(\alpha+\beta+\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\) and \(\alpha\beta\gamma\).
(3)
(b)
Find the value of \(\alpha^2+\beta^2+\gamma^2\).
(2)
(c)
Find the value of \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}\).
Common error: For \(ax^3+bx^2+cx+d=0\) the sum of the roots is \(-\frac{b}{a}\) and the product is \(-\frac{d}{a}\); the sign of the product differs from that of the quadratic case.
134IBMA3H-2-19r21-112.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
A cubic polynomial \(p(x)\) has leading coefficient 1 and \(p(0)=-3\). When \(p(x)\) is divided by \((x-1)\) the remainder is 5, and when it is divided by \((x+2)\) the remainder is \(-7\).
(a)
Find the remainder when \(p(x)\) is divided by \((x-1)(x+2)\).
(4)
(b)
Find \(p(x)\), giving your answer in the form \(x^3+bx^2+cx+d\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(p(x)=(x-1)(x+2)q(x)+rx+s\)
✓ 1
M1 substituting \(x=1\) and \(x=-2\): \(r+s=5\) and \(-2r+s=-7\)
✓ 1
A1 \(r=4\)
✓ 1
A1 \(s=1\), remainder \(4x+1\)
✓ 1
Part (b)
M1 \(p(x)=(x-1)(x+2)(x-k)+4x+1\)
✓ 1
A1 using \(p(0)=-3\): \(2k+1=-3\) so \(k=-2\)
✓ 1
A1 \(p(x)=x^3+3x^2+4x-3\)
✓ 1
Exam tip: When dividing by a quadratic the remainder has the form \(rx+s\); use the values of the polynomial at the two roots of the divisor to find \(r\) and \(s\).
135IBMA3H-2-19r22-052.7 Quadratic equations, inequalities and the discriminantPaper 1Medium6 marks
No calculator
A robot arm follows the path given by the equation \(kx^2+4x+(k-3)=0\), where \(k\) is a real constant. The equation is to have two distinct real roots.
(a)
Find the set of possible values of \(k\).
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use discriminant greater than zero
✓ 1
A1 \(16-4k(k-3)>0\)
✓ 1
M1 rearrange to \(k^2-3k-4<0\) and attempt to find the critical values
✓ 1
A1 critical values \(k=-1\) and \(k=4\)
✓ 1
A1 \(-1<k<4\)
✓ 1
R1 exclude \(k=0\) because the equation is then linear with one root, so \(-1<k<0\) or \(0<k<4\)
✓ 1
Common error: The word 'quadratic' needs \(k\neq 0\). After solving the inequality, check whether any value in your interval makes the leading coefficient zero.
136IBMA3H-2-19r22-072.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
A robot's positioning error is modelled by the polynomial \(p(x)=x^3+ax^2+bx+6\), where \(a\) and \(b\) are constants. When \(p(x)\) is divided by \((x-2)\) the remainder is \(-4\), and when \(p(x)\) is divided by \((x+1)\) the remainder is \(8\).
(a)
Find the values of \(a\) and \(b\).
(4)
(b)
Hence fully factorise \(p(x)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use the remainder theorem, \(p\) evaluated at \(x=2\) equals \(-4\)
✓ 1
A1 \(4a+2b=-18\) (or \(2a+b=-9\))
✓ 1
M1 use \(p\) evaluated at \(x=-1\) equals \(8\) to obtain \(a-b=3\)
✓ 1
A1 \(a=-2\), \(b=-5\)
✓ 1
Part (b)
M1 show that \(x=1\) gives \(p=0\), or attempt polynomial division
✓ 1
M1 obtain a quadratic factor \(x^2-x-6\) (or equivalent)
✓ 1
A1 \(p(x)=(x-1)(x+2)(x-3)\)
✓ 1
Exam tip: The remainder on dividing by \((x-c)\) is \(p\) evaluated at \(x=c\). Dividing by \((x+1)\) means substituting \(x=-1\), not \(x=1\).
137IBMA3H-2-19r22-082.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium7 marks
No calculator
Three calibration constants of a robotic gripper, \(\alpha\), \(\beta\) and \(\gamma\), are the roots of the equation \(2x^3-8x^2+2x+12=0\).
(a)
Write down the values of \(\alpha+\beta+\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\) and \(\alpha\beta\gamma\).
(3)
(b)
Show that \(\alpha^2+\beta^2+\gamma^2=14\).
(2)
(c)
Find the value of \(\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\alpha+\beta+\gamma=4\)
✓ 1
A1 \(\alpha\beta+\beta\gamma+\gamma\alpha=1\)
✓ 1
A1 \(\alpha\beta\gamma=-6\)
✓ 1
Part (b)
M1 use \(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)\)
✓ 1
A1 \(16-2=14\) (shown)
✓ 1
Part (c)
M1 write as \(\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}\)
✓ 1
A1 \(-\dfrac16\)
✓ 1
Exam tip: Sums of reciprocals of roots come from combining fractions: the numerator is the sum of products in pairs and the denominator is the product of all the roots.
138IBMA3H-2-19r22-132.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium7 marks
No calculator
The response of a robot joint is modelled by \(f(x)=x^3+x\), for \(x\in\mathbb{R}\). It is known that \(f\) has an inverse function \(f^{-1}\).
(a)
Show that \(f\) is an odd function.
(2)
(b)
Given that \(f\) at \(x=1\) equals \(2\), find the value of \(f^{-1}(-2)\).
(2)
(c)
Hence prove that \(f^{-1}\) is also an odd function.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 find \(f(-x)=(-x)^3+(-x)\)
✓ 1
A1 \(f(-x)=-x^3-x=-f(x)\), so \(f\) is odd
✓ 1
Part (b)
M1 use oddness to write \(f(-1)=-2\)
✓ 1
A1 \(f^{-1}(-2)=-1\)
✓ 1
Part (c)
M1 let \(y=f(x)\) so that \(x=f^{-1}(y)\)
✓ 1
M1 use \(f(-x)=-y\) to obtain \(f^{-1}(-y)=-x\)
✓ 1
R1 \(f^{-1}(-y)=-x=-f^{-1}(y)\) for all \(y\), so \(f^{-1}\) is odd
✓ 1
Exam tip: In a proof, let \(y=f(x)\) and translate every statement into the language of \(f^{-1}\) using \(f^{-1}(y)=x\).
139IBMA3H-2-20r30-052.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium7 marks
No calculator
The function f is defined by f(x) = (3x + 1) ⁄ (x − 3) for x ∈ ℝ, x ≠ 3.
(a)
Find the range of f.
(2)
(b)
Show that f is a self-inverse function, that is, f(f(x)) = x.
(3)
(c)
The function f is applied 101 times in succession to the number 5. Find the result.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rearrange y(x − 3) = 3x + 1 to make x the subject, or identify the horizontal asymptote
M1 sign analysis (sketch or table) of the expression
✓ 1
A1 −1 ≤ x < 1
✓ 1
A1 x ≥ 3
✓ 1
Common error: Do not multiply both sides by (x − 1): its sign changes with x, so the inequality direction would be unknown. Move all terms to one side instead.
142IBMA3H-2-20r30-112.5 Composite functions and inverse functionsPaper 1Medium7 marks
No calculator
(a)
Solve log3x + log3(x − 6) = 3.
(4)
(b)
Solve 4x − 5 × 2x + 4 = 0.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 combine: log3[x(x − 6)] = 3
✓ 1
M1 x² − 6x = 27
✓ 1
M1 (x − 9)(x + 3) = 0
✓ 1
A1 x = 9, rejecting x = −3 because log3x is undefined for negative x
✓ 1
Part (b)
M1 let u = 2x, giving u² − 5u + 4 = 0
✓ 1
A1 u = 1 or u = 4
✓ 1
A1 x = 0 or x = 2
✓ 1
Common error: After solving a logarithmic equation, check every root against the original domain; an invalid root earns no credit.
143IBMA3H-2-18r34-052.7 Quadratic equations, inequalities and the discriminantPaper 1Medium8 marks
No calculator
The curved roof of a pavilion is modelled by the curve \(y=x^2-4x+7\). A straight steel support beam has equation \(y=2x+c\), where c is a constant.
(a)
Show that the beam is a tangent to the roof when \(c=-2\).
(3)
(b)
Find the coordinates of the point where the beam touches the roof when \(c=-2\).
(2)
(c)
The beam is replaced by the line \(y=2x-5\). Find the discriminant of the equation formed when the line and the curve are equated, and state what it shows.
R1 the discriminant is negative, so the beam does not meet the roof
✓ 1
Common error: Setting the discriminant to zero before the equation is in the form \(ax^2+bx+c=0\) leads to errors in b and c. Collect all terms on one side first.
144IBMA3H-2-18r34-082.8 The reciprocal function and rational functionsPaper 1Medium8 marks
No calculator
A rectangular atrium floor has perimeter 60 m. Its width is w m and its length is l m, where \(0<w<30\).
Diagram not to scale
(a)
Show that the area of the floor is \(A=w(30-w)\) m2.
(2)
(b)
The floor area must be at least 200 m2. Solve the inequality \(w(30-w)\ge200\).
(3)
(c)
The length must also be at least 1.5 times the width. Find all possible values of w.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2w+2l=60\), so \(l=30-w\)
✓ 1
A1 \(A=wl=w(30-w)\)
✓ 1
Part (b)
M1 \(w^2-30w+200\le0\)
✓ 1
A1 \((w-10)(w-20)=0\), so the roots are 10 and 20
✓ 1
A1 \(10\le w\le20\)
✓ 1
Part (c)
M1 \(30-w\ge1.5w\)
✓ 1
A1 \(w\le12\)
✓ 1
A1 \(10\le w\le12\)
✓ 1
Exam tip: For a quadratic inequality, sketch the parabola or use a sign diagram. A positive \(w^2\) coefficient with \(\le0\) means the solution lies between the roots.
145IBMA3H-2-18r34-112.7 Quadratic equations, inequalities and the discriminantPaper 1Medium8 marks
No calculator
The concentration of plankton, C mg per litre, at a monitoring buoy is modelled by \(C(t)=\dfrac{12t}{t^2+9}\), where t is the time in hours after sunrise, and \(t\ge0\).
(a)
Write down the equation of the horizontal asymptote of the graph of \(y=C(t)\), and interpret it in this context.
(2)
(b)
Let \(C(t)=k\). Show that \(kt^2-12t+9k=0\), and use the discriminant to show that the greatest possible value of the concentration is 2 mg per litre.
(4)
(c)
Find the time at which the concentration is greatest.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(y=0\)
✓ 1
R1 in the long term the concentration of plankton tends to zero
✓ 1
Part (b)
A1 \(k(t^2+9)=12t\), so \(kt^2-12t+9k=0\)
✓ 1
M1 there is a real value of t only if the discriminant \(\ge0\)
✓ 1
A1 \(144-36k^2\ge0\)
✓ 1
A1 \(k^2\le4\); since \(C>0\) for \(t>0\), the greatest value is \(k=2\) mg per litre
✓ 1
Part (c)
M1 for \(k=2\) the equation has a repeated root: \(2t^2-12t+18=0\), so \(t^2-6t+9=0\)
✓ 1
A1 \(t=3\) hours
✓ 1
Exam tip: A maximum or minimum value of a rational function can be found without calculus. A repeated root of the rearranged equation occurs exactly at the extreme value.
146IBMA3H-2-19r34-072.7 Quadratic equations, inequalities and the discriminantPaper 1Medium7 marks
No calculator
A shipping firm models its profit on a voyage as \(P(x) = -2x^2 + kx - 8\) thousand pounds, where \(x\) is the cargo carried in hundreds of tonnes and \(k > 0\) is a constant that depends on the freight rate.
(a)
Find the set of values of \(k\) for which the firm can break even at two different cargo sizes, that is, \(P(x) = 0\) has two distinct real solutions.
(3)
(b)
For \(k = 10\), find the values of \(x\) for which the voyage makes a profit.
(2)
(c)
For \(k = 10\), find the maximum profit, in pounds.
Common error: After multiplying by \((x-1)\), do not forget that the equation \(f(x) = 2\) may have no solution at all, because the line \(y = 2\) is the horizontal asymptote.
148IBMA3H-2-19r36-052.15 Solving inequalities g(x) ≥ f(x)Paper 1Medium8 marks
No calculator
The concentration of a drug in the blood, C mg per litre, t hours after it is taken is modelled by C(t) = at(t − 8) for 0 ≤ t ≤ 8, where a is a constant. The greatest concentration is 12 mg per litre.
(a)
Find the value of a.
(3)
(b)
The drug is effective while C > 9. Find the values of t for which the drug is effective.
(4)
(c)
Hence find the length of time for which the drug is effective.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 states the maximum occurs at t = 4, halfway between the zeros 0 and 8
✓ 1
M1 substitutes: a × 4 × (−4) = 12
✓ 1
A1 a = −0.75
✓ 1
Part (b)
M1 writes −0.75t2 + 6t > 9 (ECF from (a))
✓ 1
M1 rearranges to t2 − 8t + 12 < 0
✓ 1
A1 critical values t = 2 and t = 6
✓ 1
A1 2 < t < 6
✓ 1
Part (c)
A1 4 hours (ECF from (b))
✓ 1
Common error: When dividing or multiplying an inequality by a negative number, the inequality sign reverses; rearranging so the squared term is positive avoids this.
149IBMA3H-2-19r36-062.1 Equations of straight lines, parallel and perpendicular linesPaper 1Medium9 marks
No calculator
In a strategy game, a straight road L1 has equation 3x + 4y = 24, where the units are kilometres. A town T is at the point (6, 2). A new straight road L2 is built through T perpendicular to L1.
Units in kilometres
(a)
Find an equation of L2 in the form ax + by + c = 0, where a, b and c are integers.
(4)
(b)
Find the coordinates of the point R where L2 meets L1.
(3)
(c)
Find the shortest distance from T to the road L1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 gradient of L1 is −3/4
✓ 1
A1 gradient of L2 is 4/3 (perpendicular gradients multiply to −1)
✓ 1
M1 uses y − 2 = (4/3)(x − 6)
✓ 1
A1 4x − 3y − 18 = 0 (or any integer multiple)
✓ 1
Part (b)
M1 attempt to solve 3x + 4y = 24 and 4x − 3y = 18 simultaneously
Exam tip: The shortest distance from a point to a line is along the perpendicular, so the length TR from the perpendicular's foot is the answer.
150IBMA3H-2-19r36-072.1 Equations of straight lines, parallel and perpendicular linesPaper 1Medium7 marks
No calculator
In a video game, a hillside is modelled by the curve y = x2 − 4x + 10 and a zip-line by the straight line y = 2x + c, where c is a constant. The zip-line just touches the hillside at a single point P.
(a)
Show that the x-coordinate of P satisfies x2 − 6x + (10 − c) = 0.
(1)
(b)
Find the value of c.
(3)
(c)
Find the coordinates of P.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 equates x2 − 4x + 10 = 2x + c and rearranges to the given form
✓ 1
Part (b)
M1 states the discriminant must equal zero for a single point of contact
✓ 1
M1 36 − 4(10 − c) = 0
✓ 1
A1 c = 1
✓ 1
Part (c)
M1 solves the repeated equation x2 − 6x + 9 = 0 (ECF from (b))
✓ 1
A1 x = 3
✓ 1
A1 y = 7
✓ 1
Exam tip: A line that touches a curve at exactly one point gives a quadratic with a repeated root, so set the discriminant equal to zero.
151IBMA3H-2-19r36-112.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Medium8 marks
No calculator
In a water treatment plant, the times t1 and t2 in hours, with t1 < t2, at which the chlorine level in a tank equals its target value are the roots of the equation 3t2 − 12t + k = 0, where k is a constant. It is given that t2 = 3t1.
(a)
Find t1, t2 and the value of k.
(5)
(b)
Find a quadratic equation, with integer coefficients, whose roots are 1/t1 and 1/t2.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses sum of roots t1 + t2 = 4
✓ 1
M1 substitutes t2 = 3t1 to get 4t1 = 4
✓ 1
A1 t1 = 1 and t2 = 3
✓ 1
M1 uses product of roots k/3 = t1t2
✓ 1
A1 k = 9
✓ 1
Part (b)
M1 sum of new roots = 1 + 1/3 = 4/3
✓ 1
M1 product of new roots = 1/3
✓ 1
A1 3x2 − 4x + 1 = 0 (or any integer multiple)
✓ 1
Exam tip: For ax² + bx + c = 0, the sum of the roots is −b/a and the product is c/a; these let you build new equations without finding any roots.
152IBMA3H-2-18r43-062.16 Modulus, reciprocal and other graph transformationsPaper 1Medium5 marks
No calculator
Consider the equation \(|3x-5|=x+7\).
(a)
Solve the equation.
(3)
(b)
Hence solve the inequality \(|3x-5|\le x+7\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 consider \(3x-5=x+7\) and \(3x-5=-(x+7)\)
✓ 1
A1 \(x=6\)
✓ 1
A1 \(x=-\frac12\)
✓ 1
Part (b)
M1 the V-shaped graph of \(y=|3x-5|\) lies below the line \(y=x+7\) between the two intersection points
✓ 1
A1 \(-\frac12\le x\le6\)
✓ 1
Common error: For \(|f(x)|=g(x)\), solve \(f(x)=g(x)\) and \(f(x)=-g(x)\), then check each answer in the original equation.
153IBMA3H-2-18r43-082.8 The reciprocal function and rational functionsPaper 1Medium7 marks
No calculator
Let \(f(x)=x^2-6x+5\) and \(g(x)=\dfrac{1}{f(x)}\), for \(x\in\mathbb{R}\) where \(g\) is defined.
(a)
Write \(f(x)\) in the form \((x-h)^2+k\) and state the coordinates of the vertex of the graph of \(y=f(x)\).
(2)
(b)
Write down the equations of all asymptotes of the graph of \(y=g(x)\).
(2)
(c)
State the coordinates of the local maximum point of the graph of \(y=g(x)\), giving a reason.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \((x-3)^2-9+5\)
✓ 1
A1 \((x-3)^2-4\), vertex \((3,-4)\)
✓ 1
Part (b)
A1 \(x=1\) and \(x=5\)
✓ 1
A1 \(y=0\)
✓ 1
Part (c)
A1 \(x=3\)
✓ 1
A1 \(y=-\frac14\)
✓ 1
R1 on the interval between the asymptotes \(f(x)<0\) and \(f\) has its minimum value \(-4\) at \(x=3\), so \(|g|=\frac{1}{|f|}\) is smallest there, making \(g\) (which is negative) greatest at \(x=3\)
✓ 1
Exam tip: Turning points of \(y=\frac{1}{f(x)}\) occur at the same \(x\)-values as those of \(f\), and a minimum of \(f\) becomes a maximum of the reciprocal when \(f\) is negative.
154IBMA3H-2-18r43-092.11 Transformations of graphsPaper 1Medium7 marks
No calculator
The graph of \(y=\dfrac{1}{x}\) is transformed to give the graph of \(y=\dfrac{2x+5}{x+1}\), for \(x\ne-1\).
(a)
Show that \(\dfrac{2x+5}{x+1}=2+\dfrac{3}{x+1}\).
(2)
(b)
Describe a sequence of three transformations that maps \(y=\frac{1}{x}\) onto \(y=\frac{2x+5}{x+1}\).
(3)
(c)
Hence write down the equations of the asymptotes of the graph of \(y=\frac{2x+5}{x+1}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2(x+1)+3=2x+5\) (or algebraic division)
✓ 1
A1 \(\frac{2(x+1)+3}{x+1}=2+\frac{3}{x+1}\), so shown
✓ 1
Part (b)
A1 translation of 1 unit to the left, i.e. by \(\begin{pmatrix}-1\\0\end{pmatrix}\)
✓ 1
A1 vertical stretch with scale factor 3
✓ 1
A1 translation of 2 units up, i.e. by \(\begin{pmatrix}0\\2\end{pmatrix}\)
✓ 1
(accept any equivalent sequence in which the stretch comes before the vertical translation)
—
Part (c)
A1 \(x=-1\)
✓ 1
A1 \(y=2\)
✓ 1
Common error: Order matters: stretching after translating up would scale the shift as well, so stretch first, then translate vertically.
155IBMA3H-2-20r48-052.5 Composite functions and inverse functionsPaper 1Medium8 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{3x+2}{x-4}\), for \(x\neq 4\).
(a)
Write down the equation of the vertical asymptote and the equation of the horizontal asymptote of the graph of \(f\).
(2)
(b)
Find \(f^{-1}(x)\).
(3)
(c)
Write down the range of \(f\).
(1)
(d)
Hence, or otherwise, solve \(f(x)=5\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=4\)
✓ 1
A1 \(y=3\)
✓ 1
Part (b)
M1 swap \(x\) and \(y\): \(x=\frac{3y+2}{y-4}\) and multiply out
✓ 1
A1 collect \(y\) terms: \(y(x-3)=4x+2\)
✓ 1
A1 \(f^{-1}(x)=\frac{4x+2}{x-3}\)
✓ 1
Part (c)
A1 \(f(x)\in\mathbb{R}\), \(f(x)\neq 3\)
✓ 1
Part (d)
M1 \(x=f^{-1}(5)\) (ECF from (b)) or \(3x+2=5(x-4)\)
✓ 1
A1 \(x=11\)
✓ 1
Exam tip: The range of \(f\) equals the domain of \(f^{-1}\), so the horizontal asymptote of \(f\) tells you the excluded value in \(f^{-1}\).
156IBMA3H-2-20r48-062.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium7 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{px+10}{x-3}\), for \(x\neq 3\), where \(p\) is a constant.
(a)
Find an expression for \(f^{-1}(x)\) in terms of \(p\).
(3)
(b)
Given that \(f\) is a self-inverse function, find the value of \(p\).
(2)
(c)
Using \(p=3\), find the value of \(f(f(7))\) without evaluating \(f(7)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 swap \(x\) and \(y\): \(x=\frac{py+10}{y-3}\)
✓ 1
M1 multiply out and collect \(y\): \(y(x-p)=3x+10\)
✓ 1
A1 \(f^{-1}(x)=\frac{3x+10}{x-p}\)
✓ 1
Part (b)
M1 self-inverse means \(f^{-1}(x)=f(x)\), so compare denominators \(x-p\) and \(x-3\)
✓ 1
A1 \(p=3\) (numerators then agree)
✓ 1
Part (c)
R1 \(f\) is self-inverse, so \(f(f(x))=x\) for all \(x\) in the domain
✓ 1
A1 \(7\)
✓ 1
Exam tip: A self-inverse function satisfies \(f(f(x))=x\), so applying it twice returns the starting value; use this to skip long arithmetic.
157IBMA3H-2-20r48-082.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium7 marks
No calculator
Let \(f(x)=2^{2x+1}-9\times 2^x+4\).
(a)
Show that \(f(x)=0\) can be written as \(2u^2-9u+4=0\), where \(u=2^x\).
(2)
(b)
Hence solve \(f(x)=0\).
(3)
(c)
Hence solve \(f(x)\ge 0\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2^{2x+1}=2\times(2^x)^2\)
✓ 1
A1 \(2u^2-9u+4=0\) (AG)
✓ 1
Part (b)
M1 factorise \((2u-1)(u-4)=0\)
✓ 1
A1 \(u=\frac{1}{2}\) and \(u=4\)
✓ 1
A1 \(x=-1\) and \(x=2\)
✓ 1
Part (c)
M1 quadratic with positive leading coefficient is non-negative outside its roots: \(u\le\frac{1}{2}\) or \(u\ge 4\)
✓ 1
A1 \(x\le -1\) or \(x\ge 2\)
✓ 1
Common error: Always convert back from \(u\) to \(x\); the inequality direction is kept because \(2^x\) is an increasing function.
158IBMA3H-2-20r48-092.14 Odd and even functions, inverse and self-inverse functionsPaper 1Medium9 marks
No calculator
Let \(f(x)=x^2-2x-3\).
(a)
Write down the range of \(y=|f(x)|\).
(1)
(b)
Solve \(|f(x)|=3\).
(5)
(c)
Find all values of \(k\) for which \(|f(x)|=k\) has exactly two solutions.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(y\ge 0\)
✓ 1
Part (b)
M1 consider both \(f(x)=3\) and \(f(x)=-3\)
✓ 1
A1 \(x^2-2x-6=0\), giving \(x=1\pm\sqrt{7}\)
✓ 1
A1 \(x^2-2x=0\)
✓ 1
A1 \(x=0\) or \(x=2\)
✓ 1
A1 all four solutions stated: \(1-\sqrt{7}\), \(0\), \(2\), \(1+\sqrt{7}\)
✓ 1
Part (c)
A1 minimum of \(f\) is \(-4\) at \(x=1\), so \(|f|\) has a local maximum value of \(4\) there
✓ 1
A1 \(k=0\)
✓ 1
A1 \(k>4\)
✓ 1
Exam tip: For \(|f(x)|=k\) with \(k>0\), solve \(f(x)=k\) and \(f(x)=-k\) separately and check each root against the original equation.
159AAHL2-q00022.1 Equations of straight lines, parallel and perpendicular linesPaper 1Hard11 marks
The line \(L\) has equation \(x-2y-7=0\). The points \(\mathrm{A}(5,\,4)\) and \(\mathrm{B}(1,\,2)\) lie on the same side of \(L\).
Diagram not to scale
(a)
Find the equation of the line through A that is perpendicular to \(L\).
(2)
(b)
This line meets \(L\) at the point M. Find the coordinates of M.
(2)
(c)
The point \(\mathrm{A}'\) is the reflection of A in the line \(L\). Find the coordinates of \(\mathrm{A}'\).
(2)
(d)
The point P lies on \(L\). Explain why \(\mathrm{AP}+\mathrm{PB}=\mathrm{A}'\mathrm{P}+\mathrm{PB}\).
(1)
(e)
Hence find the coordinates of the point P on \(L\) for which \(\mathrm{AP}+\mathrm{PB}\) is least, and write down this least value.
(4)
Show mark scheme
Working
Mark
Notes
(a)
gradient of \(L\) is \(\dfrac{1}{2}\), so the perpendicular gradient is \(-2\)
M1
\(y=-2x+14\)
A1
Or \(2x+y-14=0\).
(b)
\(x-2(14-2x)-7=0\)
M1
\(\mathrm{M}(7,\,0)\)
A1
(c)
M is the midpoint of \([\mathrm{AA}']\), e.g. \(\mathrm{A}'=2\mathrm{M}-\mathrm{A}\)
M1
\(\mathrm{A}'(9,\,-4)\)
A1
(d)
\(L\) is the perpendicular bisector of \([\mathrm{AA}']\), so every point P on \(L\) is equidistant from A and \(\mathrm{A}'\): \(\mathrm{AP}=\mathrm{A}'\mathrm{P}\)
R1
(e)
\(\mathrm{A}'\mathrm{P}+\mathrm{PB}\) is least when \(\mathrm{A}'\), P and B are collinear; gradient of \(\mathrm{A}'\mathrm{B}=\dfrac{2-(-4)}{1-9}=-\dfrac{3}{4}\)
M1
line \(\mathrm{A}'\mathrm{B}\): \(3x+4y=11\)
A1
\(\mathrm{P}(5,\,-1)\)
A1
From \(x-2y=7\) and \(3x+4y=11\).
least value \(=\mathrm{A}'\mathrm{B}=\sqrt{8^{2}+6^{2}}=10\)
A1
Answers: (a) \(y=-2x+14\) · (b) \(\mathrm{M}(7,\,0)\) · (c) \(\mathrm{A}'(9,\,-4)\) · (e) \(\mathrm{P}(5,\,-1)\); least value 10
160AAHL2-q00082.3 Graphs of functions and sketching from a contextPaper 1Hard14 marks
A vessel consists of an inverted right circular cone of radius 6 cm and height 9 cm, with a cylinder of radius 6 cm and height 10 cm on top of it, as shown in the following diagram. The vertex of the cone is at the bottom and the axis of the vessel is vertical.
Diagram not to scale
The vessel is initially empty. Water is poured in at a constant rate of \(12\pi\) cm3 s−1. The depth of the water, measured from the vertex of the cone, is \(h\) cm after \(t\) seconds.
(a)
Show that, while \(0\le h\le9\), the volume of water in the vessel is \(V=\dfrac{4\pi h^{3}}{27}\) cm3.
(2)
(b)
Hence show that \(h=3\sqrt[3]{3t}\) while the water is in the cone.
(2)
(c)
Show that the cone is full after exactly 9 seconds.
(1)
(d)
Find an expression for \(h\) in terms of \(t\) after the cone is full, until the vessel is full.
(3)
(e)
Find the time at which the vessel becomes full.
(1)
(f)
Find the time at which the depth of the water is 6 cm.
(2)
(g)
Sketch the graph of \(h\) against \(t\), from the moment the water is first poured in until the vessel is full.
(3)
Show mark scheme
Working
Mark
Notes
(a)
by similar triangles, the radius of the water surface is \(r=\dfrac{6}{9}h=\dfrac{2h}{3}\)
Or solving \(\mathrm{e}^{2x}-\tfrac{3}{2}\mathrm{e}^{x}-1=0\).
\(x=\ln2\)
A1
Answers: (c) \(x\in\mathbb{R}\) · (d) \(x=\ln2\)
162AAHL2-q00182.6 The quadratic function and its formsPaper 1Hard15 marks
The curve \(C\) has equation \(y=\dfrac{x^{2}}{4}\). The point \(\mathrm{P}\left(2p,\,p^{2}\right)\), where \(p\gt 0\), lies on \(C\), and O is the origin.
The line \(L\) passes through P and is perpendicular to [OP]. The line \(L\) meets \(C\) again at the point Q.
Diagram not to scale
(a)
Write down the gradient of [OP].
(1)
(b)
Show that the equation of \(L\) is \(2x+py=p^{3}+4p\).
(3)
(c)
Show that the \(x\)-coordinates of the points where \(L\) meets \(C\) satisfy \(px^{2}+8x-4p^{3}-16p=0\).
(2)
(d)
Hence show that the \(x\)-coordinate of Q is \(-2p-\dfrac{8}{p}\).
(3)
(e)
Find the coordinates of Q when \(p=1\).
(2)
(f)
Show that \(2p+\dfrac{8}{p}=2\left(\sqrt{p}-\dfrac{2}{\sqrt{p}}\right)^{2}+8\).
(2)
(g)
Hence find the least possible distance of Q from the \(y\)-axis, and the value of \(p\) for which it occurs.
For \(k=-1\) the repeated root is \(t=-1\), which gives no solution.
one positive and one negative root: product \(k+2\lt 0\)
M1
When \(k=-2\) the roots are 0 and −4, giving no solution.
\(k\lt -2\)
A1
Final answer: \(k\lt -2\) or \(k=2\).
Answers: (c) \(k\gt 2\) · (d) \(k\lt -2\) or \(k=2\)
165AAHL2-q00252.9 Exponential and logarithmic functionsPaper 1Hard12 marks
Let \(f(x)=\mathrm{e}^{2x}-4\mathrm{e}^{x}+3\), for \(x\in\mathbb{R}\).
(a)
Solve \(f(x)=0\).
(3)
(b)
By writing \(f(x)\) in the form \(\left(\mathrm{e}^{x}-a\right)^{2}+b\), find the minimum value of \(f\) and the exact value of \(x\) at which it occurs.
(3)
(c)
Write down the equation of the horizontal asymptote of the graph of \(y=f(x)\).
(1)
(d)
Find the exact coordinates of the point where the graph of \(y=f(x)\) meets its horizontal asymptote.
(2)
(e)
Sketch the graph of \(y=f(x)\), showing clearly the exact coordinates of the points found in parts (a), (b) and (d), and the horizontal asymptote.
167AAHL2-q00382.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Hard9 marks
Let \(p(x)=3x^{4}-x^{3}+2x-6\). The equation \(p(x)=0\) has roots \(\alpha,\ \beta,\ \gamma\) and \(\delta\).
(a)
Write down the value of \(\alpha+\beta+\gamma+\delta\) and the value of \(\alpha\beta\gamma\delta\).
(2)
(b)
Find the sum of the roots of the equation \(p(x-1)=0\).
(2)
(c)
Find the product of the roots of the equation \(p(2x)=0\).
(2)
(d)
By considering the equation \(x^{4}p\!\left(\dfrac{1}{x}\right)=0\), find the value of \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}+\dfrac{1}{\delta}\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\alpha+\beta+\gamma+\delta=\dfrac{1}{3}\)
A1
\(\alpha\beta\gamma\delta=-2\)
A1
(b)
the roots of \(p(x-1)=0\) are \(\alpha+1,\ \beta+1,\ \gamma+1,\ \delta+1\)
M1
Or expanding: \(p(x-1)=3x^{4}-13x^{3}+\ldots\)
\(\dfrac{1}{3}+4=\dfrac{13}{3}\)
A1
(c)
the roots are \(\dfrac{\alpha}{2},\ldots\), so the product is \(\dfrac{\alpha\beta\gamma\delta}{16}\); or \(p(2x)=48x^{4}-8x^{3}+4x-6\)
M1
\(-\dfrac{1}{8}\)
A1
(d)
\(x^{4}p\!\left(\dfrac{1}{x}\right)=-6x^{4}+2x^{3}-x+3\), whose roots are \(\dfrac{1}{\alpha},\ldots,\dfrac{1}{\delta}\)
168AAHL2-q00412.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Hard15 marks
Let \(f(x)=x^{3}-3x^{2}-9x+5\), for \(x\in\mathbb{R}\).
(a)
Show that \(f(x)-10=(x-5)(x+1)^{2}\).
(2)
(b)
Hence write down the solutions of \(f(x)=10\).
(1)
(c)
Show that \(f(x)+22=(x-3)^{2}(x+3)\).
(2)
(d)
Using parts (a) and (c), and without using calculus, explain why the graph of \(f\) has a local maximum point at \((-1,\,10)\) and a local minimum point at \((3,\,-22)\).
(2)
(e)
Sketch the graph of \(y=f(x)\), showing the coordinates of the turning points and of the \(y\)-intercept.
(3)
(f)
Find the set of values of \(k\) for which the equation \(f(x)=k\) has three distinct real roots.
(2)
(g)(i)
For a value of \(k\) found in part (f), the roots of \(f(x)=k\) are \(\alpha\), \(\beta\) and \(\gamma\).
Write down the value of \(\alpha+\beta+\gamma\).
(1)
(g)(ii)
Find an expression for \(\alpha\beta\gamma\) in terms of \(k\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\((x-5)(x^{2}+2x+1)\)
M1
Attempt to expand the right-hand side.
\(=x^{3}-3x^{2}-9x-5\), which equals \(f(x)-10\)
A1
(b)
\(x=-1,\ x=5\)
A1
(c)
\((x^{2}-6x+9)(x+3)\)
M1
\(=x^{3}-3x^{2}-9x+27\), which equals \(f(x)+22\)
A1
(d)
near \(x=-1\), \(x-5\lt 0\) and \((x+1)^{2}\ge0\), so \(f(x)-10\le0\): \(f(x)\le10=f(-1)\), a local maximum
R1
near \(x=3\), \(x+3\gt 0\) and \((x-3)^{2}\ge0\), so \(f(x)+22\ge0\): \(f(x)\ge-22=f(3)\), a local minimum
R1
(e)
positive cubic shape (rising to the right, falling to the left)
A1
local maximum \((-1,\,10)\) and local minimum \((3,\,-22)\) in approximately correct positions
A1
\(y\)-intercept \((0,\,5)\)
A1
The curve crosses the \(x\)-axis three times: once in \(x\lt -1\), once between \(0\) and \(3\), and once in \(x\gt 3\).
(f)
horizontal line \(y=k\) between the turning values
(M1)
\(-22\lt k\lt 10\)
A1
Award A0 for non-strict inequalities.
(g)(i)
\(\alpha+\beta+\gamma=3\)
A1
The \(x^{2}\) coefficient of \(f(x)-k\) does not depend on \(k\).
169AAHL2-q00462.13 Rational functions with oblique and horizontal asymptotesPaper 1Hard11 marks
Let \(f(x)=\dfrac{x^{2}-x+2}{x+1}\), for \(x\neq-1\).
(a)
Find the equations of the asymptotes of the graph of \(f\).
(3)
(b)
Show that the line \(y=k\) meets the graph of \(f\) where \(x^{2}-(k+1)x+(2-k)=0\).
(2)
(c)
Hence find the range of \(f\).
(4)
(d)
Hence find the coordinates of the local maximum point and of the local minimum point of the graph of \(f\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(x=-1\)
A1
\(x^{2}-x+2=(x+1)(x-2)+4\)
M1
\(y=x-2\)
A1
(b)
\(k(x+1)=x^{2}-x+2\)
M1
\(x^{2}-x-kx+2-k=0\)
A1
\(x^{2}-(k+1)x+(2-k)=0\)
AG
(c)
\(k\) is in the range when the quadratic has real roots: \(\Delta\ge 0\), \((k+1)^{2}-4(2-k)\ge 0\)
M1
Note \(x=-1\) is never a root of the quadratic, since it would give \(1+k+1+2-k=4\neq0\).
\(k^{2}+6k-7\ge 0\)
A1
\((k+7)(k-1)\ge 0\), critical values \(-7\) and \(1\)
A1
\(f(x)\le-7\) or \(f(x)\ge1\)
A1
Accept \(]-\infty,-7]\cup[1,\infty[\).
(d)
\(k=-7\): \(x^{2}+6x+9=0\), local maximum \((-3,\,-7)\)
A1
\(k=1\): \(x^{2}-2x+1=0\), local minimum \((1,\,1)\)
A1
Answers: (a) \(x=-1,\ y=x-2\) · (c) \(f(x)\le-7\) or \(f(x)\ge1\) · (d) max \((-3,-7)\), min \((1,1)\)
170AAHL2-q00562.14 Odd and even functions, inverse and self-inverse functionsPaper 1Hard9 marks
Let \(f(x)=\ln\!\left(\dfrac{\mathrm{e}^{x}+1}{\mathrm{e}^{x}-1}\right)\), for \(x\gt 0\).
(a)
Show that \(f(x)\gt 0\) for all \(x\gt 0\).
(2)
(b)
Find \(f^{-1}(x)\), and hence show that \(f\) is self-inverse.
(4)
(c)
Solve the equation \(f(x)=x\), giving your answer in the form \(\ln(a+\sqrt{b})\), where \(a,b\in\mathbb{Z}^{+}\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
for \(x\gt 0\), \(\mathrm{e}^{x}+1\gt\mathrm{e}^{x}-1\gt 0\)
M1
so \(\dfrac{\mathrm{e}^{x}+1}{\mathrm{e}^{x}-1}\gt 1\) and its natural logarithm is positive
R1
(b)
interchanging \(x\) and \(y\): \(x=\ln\!\left(\dfrac{\mathrm{e}^{y}+1}{\mathrm{e}^{y}-1}\right)\Rightarrow\mathrm{e}^{x}=\dfrac{\mathrm{e}^{y}+1}{\mathrm{e}^{y}-1}\)
the range of \(f\) is \(x\gt 0\) (from (a), and \(f\) takes all positive values), so \(f^{-1}\) has the same domain and rule as \(f\): \(f\) is self-inverse
R1
(c)
\(\mathrm{e}^{x}=\dfrac{\mathrm{e}^{x}+1}{\mathrm{e}^{x}-1}\Rightarrow u^{2}-2u-1=0\), where \(u=\mathrm{e}^{x}\)
171AAHL2-q00572.14 Odd and even functions, inverse and self-inverse functionsPaper 1Hard12 marks
For a function \(f\) defined for all \(x\) in a domain that is symmetric about \(0\), let \(E(x)=\dfrac{1}{2}\big(f(x)+f(-x)\big)\) and \(O(x)=\dfrac{1}{2}\big(f(x)-f(-x)\big)\).
(a)
Show that \(E\) is even and that \(O\) is odd.
(4)
(b)
Write down \(E(x)\) and \(O(x)\) for \(f(x)=\mathrm{e}^{2x}\), \(x\in\mathbb{R}\).
(2)
(c)
For \(f(x)=\dfrac{1}{1-x}\), \(x\neq\pm1\), find \(E(x)\) and \(O(x)\), giving each as a single fraction.
(4)
(d)
Suppose that \(f(x)=A(x)+B(x)\), where \(A\) is even and \(B\) is odd. Show that \(A(x)=E(x)\).
172AAHL2-q00612.14 Odd and even functions, inverse and self-inverse functionsPaper 1Hard15 marks
Let \(f(x)=\dfrac{x}{x^{2}+1}\), for \(x\in\mathbb{R}\).
(a)
Show that \(f\) is an odd function.
(2)
(b)
Write down the equation of the horizontal asymptote of the graph of \(f\).
(1)
(c)
By considering the equation \(f(x)=k\), show that the range of \(f\) is \(-\dfrac{1}{2}\le y\le\dfrac{1}{2}\).
(4)
(d)
Hence write down the coordinates of the maximum point of the graph of \(f\).
(1)
(e)
The function \(g\) is defined by \(g(x)=f(x)\), for \(-1\le x\le1\). By considering the product of the roots of the equation in part (c), explain why \(g\) has an inverse function.
(2)
(f)
Show that \(g^{-1}(x)=\dfrac{1-\sqrt{1-4x^{2}}}{2x}\) for \(x\neq0\).
(3)
(g)
State the domain and the range of \(g^{-1}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(f(-x)=\dfrac{-x}{(-x)^{2}+1}\)
M1
\(=-\dfrac{x}{x^{2}+1}=-f(x)\)
A1
(b)
\(y=0\)
A1
(c)
\(kx^{2}-x+k=0\)
M1
for \(k\neq0\), real solutions require \(\Delta=1-4k^{2}\ge0\)
M1
\(-\dfrac{1}{2}\le k\le\dfrac{1}{2}\), \(k\neq0\)
A1
\(k=0\) is attained at \(x=0\), so the range is \(-\tfrac{1}{2}\le y\le\tfrac{1}{2}\)
R1
(d)
\(\left(1,\ \dfrac{1}{2}\right)\)
A1
From \(k=\tfrac{1}{2}\): \(x^{2}-2x+1=0\).
(e)
for \(k\neq0\), the product of the roots of \(kx^{2}-x+k=0\) is \(\dfrac{k}{k}=1\)
M1
so when there are two distinct roots, exactly one lies in \(-1\lt x\lt 1\) (and \(k=\pm\tfrac{1}{2},\ 0\) give one root each); each value in the range is taken once, so \(g\) is one-to-one
R1
(f)
interchanging \(x\) and \(y\): \(x=\dfrac{y}{y^{2}+1}\Rightarrow xy^{2}-y+x=0\)
M1
\(y=\dfrac{1\pm\sqrt{1-4x^{2}}}{2x}\)
A1
reject \(+\): for \(0\lt x\le\tfrac{1}{2}\), \(\dfrac{1+\sqrt{1-4x^{2}}}{2x}\ge\dfrac{1}{2x}\ge1\), with equality only at \(x=\tfrac{1}{2}\), so it lies outside the range of \(g^{-1}\) (similarly for \(x\lt 0\))
R1
Accept: the two roots multiply to 1, and \(g^{-1}\) must take values in \([-1,1]\).
173AAHL2-q00682.15 Solving inequalities g(x) ≥ f(x)Paper 1Hard7 marks
Let \(p(x)=x^{3}-12x+k\), where \(k\in\mathbb{R}\).
(a)
Find the least value of \(k\) for which \(p(x)\ge0\) for all \(x\ge0\).
(4)
(b)
For this value of \(k\), solve the inequality \(p(x)\le0\), for \(x\in\mathbb{R}\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(p'(x)=3x^{2}-12=0\)
M1
\(x=2\) (for \(x\ge0\)), a minimum since \(p''(2)=12\gt 0\)
A1
Or sign of \(p'\): negative on \(0\lt x\lt 2\), positive for \(x\gt 2\).
least value of \(p\) on \(x\ge0\) is \(p(2)=k-16\), require \(k-16\ge0\)
M1
\(k=16\)
A1
(b)
\(x^{3}-12x+16=(x-2)^{2}(x+4)\)
M1
Using \(x=2\) as a repeated root.
\((x-2)^{2}(x+4)\le0\)
A1
\(x\le-4\) or \(x=2\)
A1
Award A0 if \(x=2\) is omitted.
Answers: (a) \(k=16\) · (b) \(x\le-4\) or \(x=2\)
174AAHL2-q00792.16 Modulus, reciprocal and other graph transformationsPaper 1Hard10 marks
Let \(f(x)=x^{3}-3x+1\). The graph of \(y=f(x)\) is shown in the following diagram. It has a local maximum at \((-1,\,3)\) and a local minimum at \((1,\,-1)\), and it crosses the \(x\)-axis at \(x=a\), \(x=b\) and \(x=c\), where \(a\lt b\lt c\).
Diagram not to scale
(a)
Sketch the graph of \(y=f(|x|)\).
(3)
(b)
Find the set of values of \(k\) for which the equation \(f(|x|)=k\) has exactly four solutions.
(2)
(c)
Sketch the graph of \(y=|f(x)|\).
(2)
(d)
Find the set of values of \(k\) for which the equation \(|f(x)|=k\) has exactly four solutions.
(3)
Show mark scheme
Working
Mark
Notes
(a)
graph for \(x\ge0\) unchanged and reflected in the \(y\)-axis (symmetric about the \(y\)-axis)
A1
local minima at \((\pm1,\,-1)\)
A1
local maximum \((0,\,1)\) (a corner point), \(x\)-intercepts \(\pm b,\ \pm c\)
A1
(b)
horizontal line \(y=k\) meets the graph of (a) four times when it lies between the minima and the maximum
(M1)
\(-1\lt k\lt 1\)
A1
(c)
parts below the \(x\)-axis (\(x\lt a\) and \(b\lt x\lt c\)) reflected in the \(x\)-axis
A1
local maxima \((-1,\,3)\) and \((1,\,1)\), zeros at \(a,\ b,\ c\)
A1
(d)
considering \(y=k\) against the graph of (c): 6 solutions for \(0\lt k\lt 1\), 5 for \(k=1\)
M1
Any correct analysis of how the number of solutions changes at \(k=1\) and \(k=3\).
4 solutions for \(1\lt k\lt 3\); 3 for \(k=3\); 2 for \(k\gt 3\)
175AAHL2-q00892.16 Modulus, reciprocal and other graph transformationsPaper 1Hard13 marks
Let \(f(x)=x^{3}-3x^{2}+4\). The following diagram shows the graph of \(y=f(x)\). The graph meets the \(x\)-axis at \((-1,\,0)\) and \((2,\,0)\), and has a local maximum point at \((0,\,4)\).
Diagram not to scale
(a)
Use the factor theorem to show that \((x+1)\) is a factor of \(f(x)\), and hence show that \(f(x)=(x+1)(x-2)^{2}\).
(3)
(b)(i)
Consider the graph of \(y=\dfrac{1}{f(x)}\).
Write down the equations of the vertical asymptotes.
(1)
(b)(ii)
Write down the equation of the horizontal asymptote.
(1)
(b)(iii)
Write down the coordinates of the local minimum point.
(1)
(c)
Sketch the graph of \(y=\dfrac{1}{f(x)}\).
(3)
(d)
Solve the inequality \(\dfrac{1}{f(x)}\ge\dfrac{1}{4}\).
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(f(-1)=-1-3+4=0\)
A1
\(f(x)=(x+1)(x^{2}-4x+4)\)
M1
Division or comparing coefficients.
\(x^{2}-4x+4=(x-2)^{2}\)
A1
\(f(x)=(x+1)(x-2)^{2}\)
AG
(b)(i)
\(x=-1\) and \(x=2\)
A1
(b)(ii)
\(y=0\)
A1
(b)(iii)
\(\left(0,\,\dfrac{1}{4}\right)\)
A1
(c)
for \(x\lt -1\): negative branch, tending to 0 from below as \(x\to-\infty\) and to \(-\infty\) as \(x\to-1^{-}\)
A1
for \(-1\lt x\lt 2\): positive branch with local minimum \(\left(0,\tfrac{1}{4}\right)\), tending to \(+\infty\) at both asymptotes
A1
for \(x\gt 2\): positive branch, tending to \(+\infty\) as \(x\to2^{+}\) and to 0 from above as \(x\to\infty\)
176AAHL2-q00902.16 Modulus, reciprocal and other graph transformationsPaper 1Hard15 marks
Let \(f(x)=|x^{2}-4|\), for \(x\in\mathbb{R}\).
(a)
Sketch the graph of \(y=f(x)\), indicating the coordinates of any axis intercepts and of any local maximum point.
(3)
(b)
On the same set of axes, sketch the line \(y=x+2\).
(1)
(c)
Solve the equation \(|x^{2}-4|=x+2\).
(4)
(d)
Hence solve the inequality \(|x^{2}-4|\gt x+2\).
(3)
(e)
Find the set of values of \(k\) for which the equation \(|x^{2}-4|=x+k\) has exactly four solutions.
(4)
Show mark scheme
Working
Mark
Notes
(a)
W shape: \(y=x^{2}-4\) for \(|x|\ge2\), reflected arch between \(x=-2\) and \(x=2\)
A1
\(x\)-intercepts \((\pm2,\,0)\)
A1
local maximum (and \(y\)-intercept) \((0,\,4)\)
A1
(b)
line through \((-2,\,0)\) and \((0,\,2)\), meeting the graph three times
A1
(c)
\(x^{2}-4=x+2\Rightarrow x^{2}-x-6=0\)
M1
\(x=3\) or \(x=-2\)
A1
\(-(x^{2}-4)=x+2\Rightarrow x^{2}+x-2=0\)
M1
\(x=1\) or \(x=-2\); solutions \(x=-2,\ 1,\ 3\)
A1
All three and no others.
(d)
using the graph: the curve is above the line outside the arch section between \(x=1\) and \(x=3\), touching at \(x=-2\)
(M1)
\(x\lt -2\) or \(-2\lt x\lt 1\)
A1
\(x\gt 3\)
A1
Accept \(x\lt 1,\ x\neq-2\) or \(x\gt 3\).
(e)
four solutions require the line to cut the arch \(y=4-x^{2}\) twice as well as each outer branch once
R1
tangency to the arch: \(4-x^{2}=x+k\Rightarrow x^{2}+x+k-4=0\), \(\Delta=1-4(k-4)=0\)
M1
\(k=\dfrac{17}{4}\) (point of contact \(x=-\tfrac12\)), and the line through \((-2,\,0)\) gives \(k=2\)
A1
\(2\lt k\lt\dfrac{17}{4}\)
A1
Answers: (c) \(x=-2,\ 1,\ 3\) · (d) \(x\lt -2\) or \(-2\lt x\lt 1\) or \(x\gt 3\) · (e) \(2\lt k\lt\tfrac{17}{4}\)
Model sketch
177IBMA2H-047-01-072.16 Modulus, reciprocal and other graph transformationsPaper 1Hard10 marks
No calculator
This question is about solving the inequality \(|x^2-4|<3x\).
(a)
Explain why any solution must satisfy \(x>0\).
(1)
(b)
Solve the inequality for \(x\ge 2\).
(3)
(c)
Solve the inequality for \(0<x<2\).
(3)
(d)
Hence state the complete solution set of \(|x^2-4|<3x\).
(1)
(e)
Hence solve \(|x^2-4|<3|x|\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 \(|x^2-4|\ge 0\), so \(3x>0\) is needed, which gives \(x>0\)
✓ 1
Part (b)
M1 for \(x\ge 2\) the modulus is removed as \(x^2-4<3x\)
✓ 1
A1 \((x-4)(x+1)<0\)
✓ 1
A1 combines with \(x\ge 2\): \(2\le x<4\)
✓ 1
Part (c)
M1 for \(0<x<2\): \(4-x^2<3x\)
✓ 1
A1 \(x^2+3x-4>0 \Rightarrow (x+4)(x-1)>0\)
✓ 1
A1 combines with \(0<x<2\): \(1<x<2\)
✓ 1
Part (d)
A1 \(1<x<4\)
✓ 1
Part (e)
R1 both sides are even functions of x, so the solution set is symmetric about \(x=0\) (and \(x=0\) fails, since \(4<0\) is false)
✓ 1
A1 \(-4<x<-1\) or \(1<x<4\)
✓ 1
Common error: After solving each case, intersect its result with the case condition. For example, \(-1<x<4\) from the first case is valid only where \(x\ge 2\).
178IBMA2H-047-04-082.5 Composite functions and inverse functionsPaper 1Hard15 marks
No calculator
The shape of a flexible solar-panel support is described using the function \(f(x)=\ln\left(x+\sqrt{x^2+1}\right)\), \(x\in\mathbb{R}\).
(a)
Show that \(f\) is an odd function.
(3)
(b)
Find \(f^{-1}(x)\), giving your answer in terms of exponentials.
(4)
(c)
Show that \(f^{-1}\) is also an odd function.
(2)
(d)
Show that \(f'(x)=\frac{1}{\sqrt{x^2+1}}\).
(3)
(e)
Hence find the exact value of \(\int_{-3/4}^{3/4}\frac{1}{\sqrt{x^2+1}}\,dx\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(f(-x)=\ln\left(-x+\sqrt{x^2+1}\right)\)
✓ 1
M1 uses \(\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)=1\), so \(\sqrt{x^2+1}-x=\frac{1}{x+\sqrt{x^2+1}}\)
✓ 1
A1 \(f(-x)=\ln\frac{1}{x+\sqrt{x^2+1}}=-f(x)\), so \(f\) is odd
✓ 1
Part (b)
M1 lets \(y=f(x)\) (or interchanges \(x\) and \(y\)) and writes \(e^y=x+\sqrt{x^2+1}\)
✓ 1
M1 isolates the root and squares: \(e^{2y}-2xe^y+x^2=x^2+1\)
M1 writes the numerator as \(\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\)
✓ 1
A1 cancels to \(\frac{1}{\sqrt{x^2+1}}\)
✓ 1
Part (e)
M1 uses (d): integral \(=\left[f(x)\right]_{-3/4}^{3/4}\), or \(2\left[f(x)\right]_0^{3/4}\) using evenness
✓ 1
M1 substitutes correctly, e.g. \(\ln\left(\tfrac34+\tfrac54\right)-\ln\left(-\tfrac34+\tfrac54\right)=\ln2-\ln\tfrac12\), or \(2\left(\ln2-\ln1\right)\)
✓ 1
A1 \(2\ln2\) (or \(\ln4\))
✓ 1
Exam tip: The conjugate identity \((\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)=1\) is the key step in part (a). Show it explicitly rather than just asserting the result.
179IBMA2H-047-04-112.14 Odd and even functions, inverse and self-inverse functionsPaper 1Hard13 marks
No calculator
For any function \(f\) with domain \(\mathbb{R}\), define \(E(x)=\frac{f(x)+f(-x)}{2}\) and \(O(x)=\frac{f(x)-f(-x)}{2}\).
(a)
Show that \(E\) is even, \(O\) is odd, and \(f=E+O\).
(3)
(b)
Show that the product of an odd function and an even function is odd.
(2)
(c)
Find the value of \(\int_{-1}^{1}\left(x^3\cos x+3x^2\right)dx\).
(3)
(d)
Let \(f(x)=\frac{1}{1+e^x}\). Show that \(E(x)=\frac12\).
(3)
(e)
Hence find the exact value of \(\int_{-3}^{3}\frac{1}{1+e^x}dx\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(E(-x)=\frac{f(-x)+f(x)}{2}=E(x)\), so \(E\) is even
✓ 1
M1 \(O(-x)=\frac{f(-x)-f(x)}{2}=-O(x)\), so \(O\) is odd
✓ 1
A1 \(E(x)+O(x)=\frac{2f(x)}{2}=f(x)\)
✓ 1
Part (b)
M1 for odd \(p\) and even \(q\): \((pq)(-x)=p(-x)q(-x)=(-p(x))\,q(x)\)
✓ 1
A1 \(=-(pq)(x)\), so \(pq\) is odd
✓ 1
Part (c)
R1 \(x^3\) is odd and \(\cos x\) is even, so \(x^3\cos x\) is odd (by (b)) and its integral over \([-1,1]\) is 0. Accept a correct direct evaluation showing this integral is 0
M1 multiplies the numerator and denominator by \(e^x\): \(f(-x)=\frac{e^x}{e^x+1}\)
✓ 1
A1 \(f(x)+f(-x)=\frac{1+e^x}{1+e^x}=1\), so \(E(x)=\frac12\)
✓ 1
Part (e)
M1 uses \(f=E+O=\tfrac12+O(x)\), with \(\int_{-3}^{3}O(x)\,dx=0\) because \(O\) is odd
✓ 1
A1 \(=\tfrac12\times6=3\)
✓ 1
Exam tip: The integral of an odd function over an interval symmetric about 0 is zero. State why the function is odd before you use this fact.
180IBMA2H-047-03-072.13 Rational functions with oblique and horizontal asymptotesPaper 1Hard11 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{2x^2+px+q}{x-r}\), \(x\neq r\), where \(p,q,r\in\mathbb{R}\). The graph of \(y=f(x)\) has a vertical asymptote \(x=1\), an oblique asymptote \(y=2x+5\) and a \(y\)-intercept of \(-4\).
(a)
Find the values of \(r\), \(p\) and \(q\).
(4)
(b)
Show that the graph of \(y=f(x)\) does not meet the \(x\)-axis.
(2)
(c)
Solve the inequality \(f(x)\geq x+14\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(r=1\)
✓ 1
M1 divides \(2x^2+px+q\) by \(x-1\) to obtain quotient \(2x+(p+2)\)
✓ 1
A1 \(p+2=5\), so \(p=3\)
✓ 1
A1 \(f(0)=-q=-4\), so \(q=4\)
✓ 1
Part (b)
M1 considers the discriminant of \(2x^2+3x+4\): \(9-32\)
✓ 1
R1 \(\Delta=-23<0\), so the numerator is never zero and there are no \(x\)-intercepts
✓ 1
Part (c)
M1 rearranges to a single fraction compared with zero (not multiplying by \(x-1\))
✓ 1
A1 \(\frac{x^2-10x+18}{x-1}\geq 0\)
✓ 1
A1 critical values \(x=1\) and \(x=5\pm\sqrt{7}\)
✓ 1
M1 sign analysis across the critical values
✓ 1
A1 \(1<x\leq 5-\sqrt{7}\) or \(x\geq 5+\sqrt{7}\)
✓ 1
Common error: Multiplying both sides of an inequality by \(x-1\) assumes it is positive and loses half the solution. Collect everything on one side as a single fraction and use a sign diagram instead.
181IBMA2H-047-03-122.5 Composite functions and inverse functionsPaper 1Hard11 marks
No calculator
Let \(f(x)=\frac{x^2+kx+4}{x+1}\), for \(x\neq -1\), where \(k\in\mathbb{R}\).
(a)
Show that \(f'(x)=\frac{x^2+2x+k-4}{(x+1)^2}\).
(3)
(b)
Find the set of values of \(k\) for which the graph of \(y=f(x)\) has no stationary points.
(3)
(c)
Let \(k=3\). Find the equation of the oblique asymptote of the graph of \(y=f(x)\).
(2)
(d)
For \(k=3\), find the range of \(f\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 applies the quotient rule: \(\frac{(2x+k)(x+1)-(x^2+kx+4)}{(x+1)^2}\)
✓ 1
A1 expands numerator: \(2x^2+2x+kx+k-x^2-kx-4\)
✓ 1
A1 simplifies to \(x^2+2x+k-4\), as required
✓ 1
Part (b)
M1 requires the discriminant of \(x^2+2x+k-4\) to be negative: \(4-4(k-4)<0\)
✓ 1
A1 \(k>5\)
✓ 1
R1 when \(k=5\), \(f(x)=\frac{(x+1)(x+4)}{x+1}=x+4\), \(x\neq -1\), which has no stationary points; hence \(k\geq 5\)
A1 stationary values \(1+2\sqrt{2}\) (minimum) and \(1-2\sqrt{2}\) (maximum)
✓ 1
A1 \(f(x)\leq 1-2\sqrt{2}\) or \(f(x)\geq 1+2\sqrt{2}\)
✓ 1
Exam tip: Always check the boundary case where the numerator and denominator share a factor: the graph becomes a line with a hole, which can change whether a boundary value of a parameter is included.
182IBMA2H-047-06-072.5 Composite functions and inverse functionsPaper 1Hard15 marks
No calculator
The functions \(f\) and \(g\) are defined by \(f(x)=2+\ln(x-1)\), \(x>1\), \(g(x)=e^{x}+1\), \(x\in\mathbb{R}\).
(a)
Find \(f^{-1}(x)\) and state its domain.
(3)
(b)
Find \((f\circ g)(x)\), simplifying your answer.
(2)
(c)
Show that \((g\circ f)(x)=e^2(x-1)+1\), \(x>1\).
(3)
(d)
Solve \((g\circ f)(x)=(f\circ g)(x)\), giving your answer in exact form.
(3)
(e)
Show that the graphs of \(y=f(x)\) and \(y=f^{-1}(x)\) meet at exactly one point, and find its coordinates.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 interchanges \(x\) and \(y\): \(x=2+\ln(y-1)\)
✓ 1
A1 \(f^{-1}(x)=1+e^{x-2}\)
✓ 1
A1 domain \(x\in\mathbb{R}\) (the range of \(f\))
✓ 1
Part (b)
M1 \(f(e^x+1)=2+\ln(e^x+1-1)\)
✓ 1
A1 \((f\circ g)(x)=x+2\), \(x\in\mathbb{R}\)
✓ 1
Part (c)
M1 \(g(2+\ln(x-1))=e^{2+\ln(x-1)}+1\)
✓ 1
M1 uses index laws: \(e^{2}\cdot e^{\ln(x-1)}\)
✓ 1
A1 \(e^{\ln(x-1)}=x-1\), giving \(e^2(x-1)+1\) for \(x>1\)
✓ 1
Part (d)
M1 sets \(e^2(x-1)+1=x+2\)
✓ 1
M1 rearranges to \(x(e^2-1)=e^2+1\)
✓ 1
A1 \(x=\frac{e^2+1}{e^2-1}\), which is greater than 1 so lies in the domain
✓ 1
Part (e)
R1 \(f\) is increasing, so intersections of \(f\) and \(f^{-1}\) lie on \(y=x\)
✓ 1
M1 forms \(2+\ln(x-1)=x\), i.e. \(\ln u=u-1\) where \(u=x-1>0\)
✓ 1
A1 \(u=1\) satisfies this, giving the point \((2,2)\)
✓ 1
R1 for \(h(u)=u-1-\ln u\), \(h'(u)=1-\frac1u\), so \(h\) has its minimum at \(u=1\) where \(h=0\); hence \(h(u)>0\) for \(u\ne1\) and there is exactly one solution
✓ 2
Exam tip: Finding one solution by inspection is not the same as showing it is the only one. A minimum-value argument on \(u-1-\ln u\) shows the root is unique.
183IBMA2H-047-00-052.14 Odd and even functions, inverse and self-inverse functionsPaper 1Hard15 marks
No calculator
Let \(f(x) = x^3 - 3x^2 - x + 3\), \(x \in \mathbb{R}\).
(a)
Show that \((x-1)\) is a factor of \(f(x)\) and hence factorise \(f(x)\) completely.
(3)
(b)
Show that \(f(x) = (x-1)^3 - 4(x-1)\).
(2)
(c)
Find the exact minimum value of \(f(x)\) for \(x \ge 0\).
(4)
(d)
Find the set of values of k for which the equation \(f(|x|) = k\) has exactly four solutions.
(3)
(e)
Find the exact solutions of the equation \(f(|x|) = 3\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substituting \(x = 1\) gives \(1 - 3 - 1 + 3 = 0\), so \((x-1)\) is a factor
R1 \(f(|x|)\) is even, so each positive solution of \(f(x) = k\) gives two solutions
✓ 1
R1 for \(x \ge 0\), f decreases from \(f(0) = 3\) to the minimum and then increases without bound. Two positive solutions therefore need k strictly between these values; at \(k = 3\) the root \(x = 0\) gives only three solutions
✓ 1
A1 \(-\frac{16\sqrt{3}}{9} < k < 3\)
✓ 1
Part (e)
M1 for \(x \ge 0\): \(x^3 - 3x^2 - x = 0 \Rightarrow x(x^2 - 3x - 1) = 0\)
✓ 1
A1 \(x = 0\) or \(x = \frac{3 + \sqrt{13}}{2}\), rejecting the negative root \(\frac{3-\sqrt{13}}{2}\)
✓ 1
A1 \(x = 0,\ x = \pm\frac{3 + \sqrt{13}}{2}\)
✓ 1
Exam tip: For \(f(|x|)\), work only with \(x \ge 0\) and then reflect in the y-axis. A turning point of f at negative x is irrelevant to \(f(|x|)\).
184IBMA2H-047-00-122.7 Quadratic equations, inequalities and the discriminantPaper 1Hard13 marks
No calculator
This question concerns the inequality \(|x^2 - 4| > 3|x|\) and related equations.
(a)
Explain why it is sufficient to solve the inequality for \(x \ge 0\) and then use symmetry.
(2)
(b)
For \(x \ge 0\), solve the equations \(x^2 - 4 = 3x\) and \(x^2 - 4 = -3x\).
(4)
(c)
Hence solve \(|x^2 - 4| > 3|x|\).
(3)
(d)
Show that, for every \(k > 0\), the equation \(|x^2 - 4| = k|x|\) has exactly four real solutions.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 \(x^2 = |x|^2\), so both sides are functions of \(|x|\) only
✓ 1
R1 both sides are even functions, so the solution set is symmetric about \(x = 0\)
M1 tests regions for \(x \ge 0\), e.g. \(x = 0\): \(4 > 0\) true; \(x = 2\): \(0 > 6\) false; \(x = 5\): \(21 > 15\) true
✓ 1
A1 for \(x \ge 0\): \(0 \le x < 1\) or \(x > 4\)
✓ 1
A1 \(-1 < x < 1\) or \(x < -4\) or \(x > 4\)
✓ 1
Part (d)
R1 with \(u = |x|\), \(u > 0\) (since \(x = 0\) gives \(4 = 0\), impossible), the equation becomes \(u^2 - ku - 4 = 0\) or \(u^2 + ku - 4 = 0\)
✓ 1
R1 each quadratic has product of roots −4, so exactly one positive root each
✓ 1
R1 the two positive roots are distinct, since a common root would need \(ku = -ku\), i.e. \(u = 0\)
✓ 1
A1 two distinct positive values of \(u\) give \(x = \pm u\), so exactly four solutions
✓ 1
Exam tip: Substituting \(u = |x|\) turns an even modulus problem into a one-sided quadratic problem. The sign of the product of roots, \(\frac{c}{a}\), tells you how many roots are positive without solving.
185IBMA2H-047-07-112.16 Modulus, reciprocal and other graph transformationsPaper 1Hard13 marks
No calculator
A continuous function f, defined for all real x, has the following features only: • zeros at \(x=-4\), \(x=1\) and \(x=5\); • y-intercept \((0,\,2)\); • a local maximum at \((-2,\,5)\) and a local minimum at \((3,\,-4)\); • f is increasing for \(x<-2\) and \(x>3\), decreasing for \(-2<x<3\), with \(f(x)\to-\infty\) as \(x\to-\infty\) and \(f(x)\to\infty\) as \(x\to\infty\).
(a)
Find the coordinates of the local extrema of \(y=-2f(x-1)+3\), stating the nature of each.
(4)
(b)
For the graph of \(y=f(|x|)\), write down the zeros and the coordinates of all local extrema.
(3)
(c)
For the graph of \(y=\dfrac{1}{f(x)}\), write down the equations of the vertical asymptotes and find the coordinates of the local extrema, stating their nature.
(4)
(d)
Find the number of solutions of the equation \(|f(x)|=4\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 applies \(x\mapsto x+1\) and \(y\mapsto -2y+3\) to the turning points
✓ 1
A1 \((-1,\,-7)\)
✓ 1
A1 \((4,\,11)\)
✓ 1
R1 reflection in the x-axis reverses the nature: \((-1,-7)\) is a local minimum and \((4,11)\) a local maximum
✓ 1
Part (b)
A1 zeros \(x=\pm1,\ \pm5\)
✓ 1
A1 local minima \((\pm3,\,-4)\)
✓ 1
A1 local maximum \((0,\,2)\), since \(f\) is decreasing for \(0<x<3\) and the graph is symmetric in the y-axis
✓ 1
Part (c)
A1 \(x=-4\), \(x=1\), \(x=5\)
✓ 1
A1 \((-2,\,\frac{1}{5})\)
✓ 1
A1 \((3,\,-\frac{1}{4})\)
✓ 1
R1 reciprocal reverses order of same-sign values: \((-2,\frac{1}{5})\) is a local minimum, \((3,-\frac{1}{4})\) a local maximum
✓ 1
Part (d)
M1 considers \(f(x)=4\) (one solution on each of \(x<-2\), \(-2<x<3\), \(x>3\)) and \(f(x)=-4\) (one solution for \(x<-4\) and \(x=3\))
✓ 1
A1 5 solutions
✓ 1
Exam tip: A horizontal line at exactly the height of a turning point touches rather than crosses, contributing one solution instead of two.
186IBMA2H-047-02-122.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Hard11 marks
No calculator
The equation \(x^3 + 2x^2 - x + 3 = 0\) has roots \(\alpha, \beta, \gamma\).
(a)
Write down the values of \(\alpha + \beta + \gamma\), \(\alpha\beta + \beta\gamma + \gamma\alpha\) and \(\alpha\beta\gamma\).
(3)
(b)
Find the value of \(\alpha^2 + \beta^2 + \gamma^2\).
(2)
(c)
Find a cubic equation with integer coefficients whose roots are \(\alpha^2, \beta^2, \gamma^2\).
(4)
(d)
Hence deduce that \(x^3 + 2x^2 - x + 3 = 0\) does not have three real roots.
R1 if \(\alpha, \beta, \gamma\) were all real, each of \(\alpha^2\beta^2, \beta^2\gamma^2, \gamma^2\alpha^2\) would be non-negative, so their sum would be non-negative
✓ 1
R1 but \(\sum \alpha^2\beta^2 = -11 < 0\), which is a contradiction, so not all the roots are real
✓ 1
Exam tip: You can check the transformed cubic by substituting \(y = x^2\). Rearrange to \(x(x^2 - 1) = -(2x^2 + 3)\), square both sides, then replace \(x^2\) with \(y\).
187IBMA2H-047-05-112.13 Rational functions with oblique and horizontal asymptotesPaper 1Hard10 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{x^2+3x-1}{x-1}\), \(x \neq 1\).
(a)
Express \(f(x)\) in the form \(x+a+\frac{b}{x-1}\), where \(a, b \in \mathbb{Z}\).
(3)
(b)
Write down the equations of the asymptotes of the graph of \(f\).
(2)
(c)
By considering the intersection of the line \(y=k\) with the graph of \(f\), find the range of \(f\). Give the boundary values in exact form.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts polynomial division or equivalent, e.g. \(x^2+3x-1=(x-1)(x+4)+3\)
✓ 1
A1 \(a=4\)
✓ 1
A1 \(b=3\)
✓ 1
Part (b)
A1 vertical asymptote \(x=1\)
✓ 1
A1 oblique asymptote \(y=x+4\)
✓ 1
Part (c)
M1 sets \(f(x)=k\) and multiplies by \(x-1\)
✓ 1
A1 \(x^2+(3-k)x+(k-1)=0\)
✓ 1
M1 requires discriminant \(\geq 0\) for real intersections
✓ 1
A1 \(k^2-10k+13 \geq 0\)
✓ 1
A1 \(f(x) \leq 5-2\sqrt{3}\) or \(f(x) \geq 5+2\sqrt{3}\)
✓ 1
Exam tip: Check that \(x=1\) is never a root of the quadratic (substituting gives 3 = 0), so every real root corresponds to a genuine point on the graph.
188IBMA2H-047-05-122.13 Rational functions with oblique and horizontal asymptotesPaper 1Hard15 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{ax+b}{x+c}\), where \(a, b, c \in \mathbb{R}\). The graph of \(f\) has a vertical asymptote \(x=-2\) and a horizontal asymptote \(y=3\), and passes through the point (1, 4). For \(m \in \mathbb{R}\), the line \(L_m\) has equation \(y=mx+3\).
(a)
Find the values of \(a\), \(b\) and \(c\).
(4)
(b)
Express \(f(x)\) in the form \(3+\frac{p}{x+2}\), stating the value of \(p\).
(2)
(c)
Show that the \(x\)-coordinates of any points of intersection of \(L_m\) with the graph of \(f\) satisfy \(mx^2+2mx-3=0\).
(2)
(d)
Find the value of \(m\) for which \(L_m\) is a tangent to the graph of \(f\), and the coordinates of the point of contact.
(4)
(e)
Find the set of values of \(m\) for which \(L_m\) does not intersect the graph of \(f\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(c=2\)
✓ 1
A1 \(a=3\)
✓ 1
M1 substitutes (1, 4): \(\frac{3+b}{3}=4\)
✓ 1
A1 \(b=9\)
✓ 1
Part (b)
M1 writes \(3x+9=3(x+2)+3\)
✓ 1
A1 \(p=3\)
✓ 1
Part (c)
M1 equates \(3+\frac{3}{x+2}=mx+3\)
✓ 1
A1 obtains \(3=mx(x+2)\) and hence \(mx^2+2mx-3=0\)
✓ 1
Part (d)
M1 sets discriminant to zero: \(4m^2+12m=0\)
✓ 1
A1 \(m=-3\) (rejecting \(m=0\), which does not give a quadratic)
✓ 1
M1 substitutes: \(-3x^2-6x-3=0\), so \(x=-1\)
✓ 1
A1 point of contact \((-1, 6)\)
✓ 1
Part (e)
M1 requires \(4m(m+3)<0\)
✓ 1
A1 \(-3<m<0\)
✓ 1
R1 when \(m=0\), \(L_0\) is the asymptote \(y=3\) (equation becomes \(-3=0\)), so no intersection; set is \(-3<m \leq 0\)
✓ 1
Common error: The discriminant method assumes the coefficient of \(x^2\) is non-zero. Treat \(m=0\) separately: the horizontal line \(y=3\) is the asymptote and never meets the graph.
189IBMA3H-2-18r7-132.7 Quadratic equations, inequalities and the discriminantPaper 1Hard7 marks
No calculator
The cubic equation x3 − 6x2 + kx + m = 0, where k and m are constants, has three real roots that form an arithmetic sequence with common difference d > 0. The product of the three roots is −10.
(a)
Show that the middle root is 2.
(2)
(b)
Find the value of d.
(3)
(c)
Find the value of k and the value of m.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes the roots as \(a-d,\ a,\ a+d\) and uses sum of roots \(=6\)
✓ 1
A1 \(3a=6\), so \(a=2\) (AG)
✓ 1
Part (b)
M1 product of roots: \(2(2-d)(2+d)=-10\)
✓ 1
M1 \(4-d^2=-5\)
✓ 1
A1 \(d=3\)
✓ 1
Part (c)
A1 roots \(-1,\ 2,\ 5\), so \(k=(-1)(2)+(-1)(5)+(2)(5)=3\)
✓ 1
A1 \(m=-(\text{product})=10\)
✓ 1
Common error: For a monic cubic x³ + bx² + cx + d, the product of roots is −d, so the constant term is the negative of the product.
190IBMA3H-2-18r7-142.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Hard13 marks
No calculator
The polynomial p(x) is defined by p(x) = 2x3 + x2 − 13x + 6.
(a)
Show that (x − 2) is a factor of p(x).
(2)
(b)
Factorise p(x) completely.
(2)
(c)
Solve the inequality p(x) ≥ 0.
(2)
(d)
Solve the inequality p(x) ≥ −3(x − 2).
(4)
(e)
The roots of p(x) = 0 are α, β and γ. Using the coefficients of p(x), find the value of α2 + β2 + γ2.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 evaluates p at \(x=2\): \(16+4-26+6\)
✓ 1
A1 \(=0\), so \((x-2)\) is a factor (AG)
✓ 1
Part (b)
M1 attempt to divide by \((x-2)\) to obtain \(2x^2+5x-3\)
Exam tip: In a polynomial inequality, factorise first and test the sign in each interval: positive leading coefficient means the graph is positive for large x.
191IBMA3H-2-19r8-112.13 Rational functions with oblique and horizontal asymptotesPaper 1Hard12 marks
No calculator
Let \(f(x)=\frac{x^2-x-6}{x-1}\), for \(x\neq 1\).
(a)
Find the coordinates of the points where the graph of \(y=f(x)\) crosses the axes.
(2)
(b)
Show that \(f(x)=x-\frac{6}{x-1}\) and hence write down the equations of the vertical asymptote and the oblique asymptote.
(3)
(c)
Solve \(f(x)\geq x\).
(3)
(d)
Solve \(f(x)\leq 2\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x\)-intercepts \((3,0)\) and \((-2,0)\)
✓ 1
A1 \(y\)-intercept \((0,6)\)
✓ 1
Part (b)
M1 polynomial division or rewriting \(x^2-x-6=x(x-1)-6\) to obtain \(x-\frac{6}{x-1}\)
✓ 1
A1 vertical asymptote \(x=1\)
✓ 1
A1 oblique asymptote \(y=x\)
✓ 1
Part (c)
M1 \(f(x)-x=-\frac{6}{x-1}\)
✓ 1
R1 \(-\frac{6}{x-1}\geq 0\) requires \(x-1<0\)
✓ 1
A1 \(x<1\)
✓ 1
Part (d)
M1 form \(\frac{x^2-x-6}{x-1}-2\leq 0\) as a single fraction
✓ 1
A1 \(\frac{(x-4)(x+1)}{x-1}\leq 0\)
✓ 1
M1 sign analysis using the critical values \(-1\), \(1\) and \(4\)
✓ 1
A1 \(x\leq -1\) or \(1<x\leq 4\)
✓ 1
Common error: The value \(x=1\) is never part of the solution because \(f\) is undefined there, even when the inequality is non-strict.
192IBMA3H-2-19r8-132.14 Odd and even functions, inverse and self-inverse functionsPaper 1Hard10 marks
No calculator
The function \(f\) is defined by \(f(x)=\frac{kx+5}{x-4}\), for \(x\neq 4\), where \(k\) is a constant. The function \(f\) is self-inverse, that is, \(f^{-1}(x)=f(x)\).
(a)
Find the value of \(k\).
(4)
(b)
Using this value of \(k\), solve \(f(x)\geq x\).
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rearrange \(y=\frac{kx+5}{x-4}\) to make \(x\) the subject: \(x(y-k)=4y+5\)
A1 \(x\leq 4-\sqrt{21}\) or \(4<x\leq 4+\sqrt{21}\)
✓ 1
Common error: When the inequality involves a fraction, include the value that makes the denominator zero as a critical value, but exclude it from the solution set.
193IBMA3H-2-19r13-132.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 1Hard10 marks
No calculator
Let \(p(x)=x^3+ax^2+bx-12\), where \(a\) and \(b\) are constants. \((x-2)\) is a factor of \(p(x)\). When \(p(x)\) is divided by \((x+1)\) the remainder is \(-24\).
(a)
Find the values of \(a\) and \(b\).
(5)
(b)
Show that \(p(x)=0\) has exactly one real root.
(3)
(c)
Find the remainder when \(p(x)\) is divided by \((x-5)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(p(2)=0\) giving \(8+4a+2b-12=0\)
✓ 1
A1 \(2a+b=2\)
✓ 1
M1 \(p(-1)=-24\) giving \(-1+a-b-12=-24\)
✓ 1
A1 \(a-b=-11\)
✓ 1
A1 \(a=-3,\ b=8\)
✓ 1
Part (b)
M1 divide by \((x-2)\) to obtain \(p(x)=(x-2)(x^2-x+6)\)
✓ 1
M1 evaluate the discriminant of the quadratic: \(1-24=-23\)
✓ 1
R1 discriminant negative so no real roots from the quadratic; the only real root is \(x=2\)
✓ 1
Part (c)
M1 evaluate \(p(5)\) with \(a=-3,\ b=8\)
✓ 1
A1 \(125-75+40-12=78\)
✓ 1
Exam tip: Two unknown constants need two conditions. Write each condition as its own equation before solving them together.
194IBMA3H-2-19r21-122.7 Quadratic equations, inequalities and the discriminantPaper 1Hard7 marks
No calculator
The roots of the equation \(x^2-4x+2=0\) are \(\alpha\) and \(\beta\).
(a)
Write down the value of \(\alpha+\beta\) and the value of \(\alpha\beta\).
(2)
(b)
Find a quadratic equation, with integer coefficients, whose roots are \(\dfrac{\alpha^2}{\beta}\) and \(\dfrac{\beta^2}{\alpha}\).
A1 sum of new roots \(=\frac{\alpha^3+\beta^3}{\alpha\beta}=20\)
✓ 1
A1 product of new roots \(=\alpha\beta=2\)
✓ 1
A1 \(x^2-20x+2=0\)
✓ 1
Common error: A quadratic with roots \(r_1,r_2\) is \(x^2-(r_1+r_2)x+r_1r_2=0\); do not forget the minus sign on the sum.
195IBMA3H-2-19r21-132.14 Odd and even functions, inverse and self-inverse functionsPaper 1Hard13 marks
No calculator
The function \(f\) is defined by \(f(x)=\dfrac{x+2}{x-1}\), for \(x\neq 1\).
(a)
Write down the equations of the vertical and horizontal asymptotes of the graph of \(y=f(x)\).
(2)
(b)
Find the coordinates of the axes intercepts of the graph.
(2)
(c)
Find \(f^{-1}(x)\) and hence explain why \(f\) is a self-inverse function.
(3)
(d)
Solve the inequality \(f(x)\le x\), giving exact answers.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=1\)
✓ 1
A1 \(y=1\)
✓ 1
Part (b)
A1 \(y\)-intercept \((0,-2)\)
✓ 1
A1 \(x\)-intercept \((-2,0)\)
✓ 1
Part (c)
M1 \(y(x-1)=x+2\) so \(x(y-1)=y+2\)
✓ 1
A1 \(f^{-1}(x)=\frac{x+2}{x-1}\)
✓ 1
R1 \(f^{-1}(x)=f(x)\) so \(f\) is self-inverse
✓ 1
Part (d)
M1 \(\frac{x+2}{x-1}-x\le 0\) written as a single fraction
✓ 1
A1 \(\frac{-x^2+2x+2}{x-1}\le 0\), i.e. \(\frac{x^2-2x-2}{x-1}\ge 0\)
✓ 1
M1 critical values \(x=1\) and \(x=1\pm\sqrt{3}\) (\(2.73205\) and \(-0.73205\))
✓ 1
M1 sign analysis (table or sketch) with \(x=1\) excluded
✓ 1
A1 \(1-\sqrt{3}\le x<1\)
✓ 1
A1 \(x\ge 1+\sqrt{3}\)
✓ 1
Common error: Do not multiply both sides of an inequality by \((x-1)\): its sign changes with \(x\). Move everything to one side and analyse the sign of a single fraction.
196IBMA3H-2-19r22-092.5 Composite functions and inverse functionsPaper 1Hard11 marks
No calculator
A fashion designer controls a laser cutter with the function \(f(x)=\dfrac{3x+2}{x-1}\), \(x\neq 1\).
(a)
Find \(f^{-1}(x)\), and state the domain and the range of \(f^{-1}\).
(4)
(b)
Solve the equation \(f(x)=f^{-1}(x)\), giving your answers in exact form.
(5)
(c)
By calculating \(f(f(0))\), show that \(f\) is not a self-inverse function.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 swap variables and attempt to make the subject, for example \(x(y-3)=y+2\)
✓ 1
A1 \(f^{-1}(x)=\dfrac{x+2}{x-3}\)
✓ 1
A1 domain \(x\neq 3\)
✓ 1
A1 range \(f^{-1}(x)\neq 1\)
✓ 1
Part (b)
M1 equate \(\dfrac{3x+2}{x-1}=\dfrac{x+2}{x-3}\)
✓ 1
M1 cross-multiply to form a quadratic equation
✓ 1
A1 \(x^2-4x-2=0\)
✓ 1
M1 use the quadratic formula or complete the square
✓ 1
A1 \(x=2\pm\sqrt6\)
✓ 1
Part (c)
M1 \(f(0)=-2\) then \(f(-2)\) attempted
✓ 1
A1 \(f(f(0))=\dfrac43\neq 0\), so \(f\) is not self-inverse
✓ 1
Common error: The range of \(f^{-1}\) equals the domain of \(f\). Students often give the domain of \(f^{-1}\) twice by mistake.
197IBMA3H-2-19r22-112.14 Odd and even functions, inverse and self-inverse functionsPaper 1Hard10 marks
No calculator
The catch-rate model for a fishery is \(f(x)=\dfrac{2x+a}{x+b}\), \(x\neq -b\), where \(a\) and \(b\) are constants. The function \(f\) is self-inverse and \(f(5)=7\).
(a)
Find the values of \(a\) and \(b\).
(5)
(b)
For these values of \(a\) and \(b\), write down the range of \(f\).
(2)
(c)
Find the exact positive solution of the equation \(f(x)=x\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 form an equation from \(f(5)=7\), for example \(10+a=7(5+b)\)
✓ 1
R1 use \(f\) self-inverse to deduce \(f(7)=5\)
✓ 1
M1 form the second equation \(14+a=5(7+b)\)
✓ 1
A1 \(b=-2\)
✓ 1
A1 \(a=11\)
✓ 1
Part (b)
M1 write \(f(x)=2+\dfrac{15}{x-2}\) or identify the horizontal asymptote \(y=2\)
✓ 1
A1 range is \(f(x)\in\mathbb{R}\), \(f(x)\neq 2\)
✓ 1
Part (c)
M1 set \(\dfrac{2x+11}{x-2}=x\) and rearrange to \(x^2-4x-11=0\)
✓ 1
M1 use the quadratic formula or complete the square
✓ 1
A1 \(x=2+\sqrt{15}\)
✓ 1
Exam tip: If a function is self-inverse and \(f(p)=q\), then \(f(q)=p\) as well. This gives a second equation without needing to find the inverse.
198IBMA3H-2-20r30-132.14 Odd and even functions, inverse and self-inverse functionsPaper 1Hard12 marks
No calculator
The function f is defined by f(x) = (ex − e−x) ⁄ 2 for x ∈ ℝ.
(a)
Show that f is an odd function.
(2)
(b)
Find an expression for f−1(x).
(5)
(c)
Show that f−1 is also an odd function.
(3)
(d)
Find the exact value of f−1(3⁄4).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 f(−x) = (e−x − ex) ⁄ 2
✓ 1
A1 = −f(x), so f is odd (AG)
✓ 1
Part (b)
M1 set y = (ex − e−x) ⁄ 2 and multiply through by ex
M1 √(9⁄16 + 1) = 5⁄4, so the argument is 3⁄4 + 5⁄4 = 2
✓ 1
A1 ln 2 (≈ 0.693)
✓ 1
Common error: When using the quadratic formula for e^x, remember that e^x is always positive, so only one root can be valid.
199IBMA3H-2-18r34-122.13 Rational functions with oblique and horizontal asymptotesPaper 1Hard11 marks
No calculator
Let \(f(x)=\dfrac{x^2+1}{x-1}\), for \(x\ne1\).
(a)
Show that \(f(x)=x+1+\dfrac{2}{x-1}\), and write down the equations of the vertical and oblique asymptotes of the graph of \(y=f(x)\).
(4)
(b)
Let \(y=f(x)\). Show that \(x^2-yx+(y+1)=0\), and hence show that \(y^2-4y-4\ge0\) for every value of y in the range of f.
(3)
(c)
Hence find the range of f.
(3)
(d)
Explain why the graph of \(y=f(x)\) never meets its oblique asymptote.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(x^2+1=(x+1)(x-1)+2\)
✓ 1
A1 \(f(x)=x+1+\frac{2}{x-1}\)
✓ 1
A1 vertical asymptote \(x=1\)
✓ 1
A1 oblique asymptote \(y=x+1\)
✓ 1
Part (b)
A1 \(y(x-1)=x^2+1\), so \(x^2-yx+(y+1)=0\)
✓ 1
M1 there is a real x only if the discriminant \(\ge0\): \(y^2-4(y+1)\ge0\)
✓ 1
A1 \(y^2-4y-4\ge0\)
✓ 1
Part (c)
M1 solve \(y^2-4y-4=0\)
✓ 1
A1 \(y=2\pm2\sqrt2\)
✓ 1
A1 range: \(y\le2-2\sqrt2\) or \(y\ge2+2\sqrt2\) (that is, \(y\le-0.83\) or \(y\ge4.83\))
✓ 1
Part (d)
R1 \(f(x)-(x+1)=\frac{2}{x-1}\), which is never zero
✓ 1
Exam tip: The discriminant method finds the range of a rational function: write \(y=f(x)\) as a quadratic in x and require real solutions. The range is the set of y values for which the discriminant is not negative.
200IBMA3H-2-18r34-132.6 The quadratic function and its formsPaper 1Hard11 marks
No calculator
A designer models the cross-section of a curved glass panel by the curve \(y=kx^2-4x+k-3\), where k is a constant. A horizontal support beam lies along the x-axis. Lengths are in metres.
(a)
Find all values of k for which the curve meets the x-axis at two distinct points.
(5)
(b)
The panel lies entirely below the beam, so the curve does not touch the x-axis. Find the range of values of k.
(4)
(c)
For \(k=-2\), find the height of the highest point of the panel relative to the beam. State your answer as a negative number if it is below the beam.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 require discriminant \(>0\)
✓ 1
A1 \(16-4k(k-3)>0\)
✓ 1
M1 simplify to \(k^2-3k-4<0\), so \((k-4)(k+1)<0\)
✓ 1
A1 \(-1<k<4\)
✓ 1
R1 when \(k=0\) the curve is the line \(y=-4x-3\), which meets the axis only once, so exclude \(k=0\): \(-1<k<0\) or \(0<k<4\)
✓ 1
Part (b)
R1 the curve must open downwards, so \(k<0\)
✓ 1
M1 discriminant \(<0\)
✓ 1
A1 \(k^2-3k-4>0\), so \(k<-1\) or \(k>4\)
✓ 1
A1 combine with \(k<0\): \(k<-1\)
✓ 1
Part (c)
M1 the vertex is at \(x=-\frac{-4}{2(-2)}=-1\)
✓ 1
A1 \(y=-2(-1)^2-4(-1)-5=-3\), so the highest point is 3 m below the beam (answer \(-3\) m)
✓ 1
Common error: When a quadratic has a parameter as its leading coefficient, check the case where that coefficient is zero separately. The equation is then linear, and the discriminant no longer applies.
201IBMA3H-2-18r34-142.5 Composite functions and inverse functionsPaper 1Hard11 marks
No calculator
The functions f and g are defined by \(f(x)=\sqrt{x+3}\) for \(x\ge-3\), and \(g(x)=\dfrac{1}{x-1}\) for \(x\ne1\).
(a)
Write down the range of f and the range of g.
(2)
(b)
Find the largest possible domain of \(f\circ g\).
(4)
(c)
Find \((g\circ f)(x)\) and state its largest possible domain.
(3)
(d)
Solve \((f\circ g)(x)=2\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 range of f: \(f(x)\ge0\)
✓ 1
A1 range of g: \(g(x)\ne0\)
✓ 1
Part (b)
M1 need \(g(x)\ge-3\), that is \(\frac{1}{x-1}+3\ge0\)
✓ 1
A1 \(\frac{3x-2}{x-1}\ge0\)
✓ 1
M1 critical values \(x=\frac23\) and \(x=1\), with a sign analysis
✓ 1
A1 domain: \(x\le\frac23\) or \(x>1\)
✓ 1
Part (c)
A1 \((g\circ f)(x)=\frac{1}{\sqrt{x+3}-1}\)
✓ 1
M1 need \(x+3\ge0\) and \(\sqrt{x+3}\ne1\)
✓ 1
A1 domain: \(x\ge-3\), \(x\ne-2\)
✓ 1
Part (d)
M1 \(\frac{1}{x-1}+3=4\) (squaring both sides)
✓ 1
A1 \(x=2\) (this lies in the domain found in part (b))
✓ 1
Common error: Finding the domain of a composite function from the final formula only. First restrict the inner function's domain, then ensure its output lies in the domain of the outer function.
202IBMA3H-2-19r36-122.7 Quadratic equations, inequalities and the discriminantPaper 1Hard9 marks
No calculator
In a video game, a curved wall has equation y = x2 + 2x + 5. A laser beam is fired along the line y = mx − 4, where m is a constant.
(a)
Show that the x-coordinates of any intersection points satisfy x2 + (2 − m)x + 9 = 0.
(2)
(b)
Given that the beam hits the wall at two distinct points, write down an inequality in m.
(2)
(c)
Solve your inequality to find the possible values of m.
(3)
(d)
When m = 10, find the exact x-coordinates of the two points where the beam meets the wall.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates x2 + 2x + 5 = mx − 4
✓ 1
A1 rearranges to the given equation
✓ 1
Part (b)
M1 states that the discriminant must be positive
✓ 1
A1 (2 − m)2 − 36 > 0
✓ 1
Part (c)
M1 attempt to solve (2 − m)2 = 36
✓ 1
A1 critical values m = −4 and m = 8
✓ 1
A1 m < −4 or m > 8
✓ 1
Part (d)
M1 uses the quadratic formula on x2 − 8x + 9 = 0
✓ 1
A1 x = 4 ± √7
✓ 1
Common error: The solution of (m − 2)² > 36 is two separate intervals, not −4 < m < 8; check by testing a value such as m = 0.
203IBMA3H-2-19r36-132.16 Modulus, reciprocal and other graph transformationsPaper 1Hard7 marks
No calculator
Let f(x) = x and g(x) = 12/(x − 1), for x ≠ 1.
(a)
Solve the equation f(x) = g(x).
(3)
(b)
Hence solve the inequality g(x) ≥ f(x).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 multiplies by (x − 1) to get x2 − x − 12 = 0
✓ 1
M1 factorises (x − 4)(x + 3) = 0
✓ 1
A1 x = 4 and x = −3
✓ 1
Part (b)
M1 considers the sign of g(x) − f(x) using the critical values −3 and 4 and the vertical asymptote at x = 1 (ECF from (a))
✓ 1
A1 x ≤ −3
✓ 1
A1 1 < x ≤ 4
✓ 1
R1 x = 1 is excluded because g is undefined there
✓ 1
Common error: Do not multiply both sides of an inequality by an expression such as (x − 1) without considering its sign; a sign diagram or a sketch is safer.
204IBMA3H-2-19r36-142.16 Modulus, reciprocal and other graph transformationsPaper 1Hard8 marks
No calculator
In a video game, a character walks along a straight corridor, where x metres is the position along it. The danger level is f(x) = |x − 2| and the shelter rating is g(x) = 6 − |x|. The character is safe where g(x) ≥ f(x).
(a)
Find the set of values of x for which the character is safe.
(5)
(b)
Find the greatest value of g(x) − f(x) and the values of x at which it occurs.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 identifies x = 0 and x = 2 as the points where the modulus expressions change, and considers three intervals
✓ 1
A1 for x < 0: 6 + x ≥ 2 − x gives −2 ≤ x < 0
✓ 1
A1 for 0 ≤ x < 2: 6 − x ≥ 2 − x is always true
✓ 1
A1 for x ≥ 2: 6 − x ≥ x − 2 gives 2 ≤ x ≤ 4
✓ 1
A1 −2 ≤ x ≤ 4
✓ 1
Part (b)
M1 finds g(x) − f(x) on each interval: 4 + 2x, 4 and 8 − 2x
✓ 1
A1 greatest value 4
✓ 1
A1 occurs for 0 ≤ x ≤ 2
✓ 1
Exam tip: For inequalities with modulus signs, split the number line at the points where each expression inside a modulus is zero, then solve on each piece and combine.
205IBMA3H-2-18r43-122.16 Modulus, reciprocal and other graph transformationsPaper 1Hard9 marks
No calculator
Let \(f(x)=x^2-2x-3\) and let \(g(x)=|f(x)|\), for \(x\in\mathbb{R}\).
(a)
Find the coordinates of the local maximum point of the graph of \(y=g(x)\).
(2)
(b)
State the number of solutions of \(g(x)=k\) when (i) \(k=4\), (ii) \(k=6\).
(2)
(c)
Solve \(g(x)=3\).
(3)
(d)
Find the range of values of \(k\) for which \(g(x)=k\) has exactly four solutions.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertex of \(y=f(x)\) is \((1,-4)\)
✓ 1
A1 reflecting the negative part in the \(x\)-axis gives local maximum \((1,4)\)
✓ 1
Part (b)
A1 three solutions
✓ 1
A1 two solutions
✓ 1
Part (c)
M1 \(f(x)=3\) and \(f(x)=-3\)
✓ 1
A1 \(x=1\pm\sqrt7\)
✓ 1
A1 \(x=0\) and \(x=2\)
✓ 1
Part (d)
A1 \(k>0\)
✓ 1
A1 \(k<4\)
✓ 1
Exam tip: Sketch \(y=|f(x)|\) by reflecting the part of \(y=f(x)\) below the \(x\)-axis; the number of solutions of \(g(x)=k\) is the number of times the line \(y=k\) meets this graph.
206IBMA3H-2-20r48-122.13 Rational functions with oblique and horizontal asymptotesPaper 1Hard11 marks
No calculator
The function \(f\) is defined by \(f(x)=x^2-6x+8\), for \(x\in\mathbb{R}\). Consider the function \(h(x)=\frac{1}{f(x)}\).
(a)
Write down the equations of the two vertical asymptotes of the graph of \(y=h(x)\).
(2)
(b)
Find the coordinates of the local maximum point on the graph of \(y=h(x)\).
(2)
(c)
Find the range of \(h\).
(3)
(d)
Solve the inequality \(h(x)>\frac{1}{3}\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=2\)
✓ 1
A1 \(x=4\)
✓ 1
Part (b)
A1 \(f(x)=(x-3)^2-1\), so \(f\) has minimum \(-1\) at \(x=3\)
✓ 1
A1 \(h\) has a local maximum at \((3,-1)\)
✓ 1
Part (c)
M1 where \(f(x)>0\), \(h(x)>0\); where \(-1\le f(x)<0\), \(h(x)\le -1\)
✓ 1
A1 \(h(x)\le -1\)
✓ 1
A1 \(h(x)>0\)
✓ 1
Part (d)
M1 \(h(x)>\frac{1}{3}>0\) requires \(0<f(x)<3\)
✓ 1
A1 \(f(x)<3\): \(x^2-6x+5<0\), so \(1<x<5\)
✓ 1
A1 \(f(x)>0\): \(x<2\) or \(x>4\)
✓ 1
A1 \(1<x<2\) or \(4<x<5\)
✓ 1
Common error: Never multiply an inequality by \(f(x)\) without knowing its sign; split into cases, or deduce the sign of \(f(x)\) first as above.
207AAHL2-q00112.4 Key features of graphs and intersections using technologyPaper 2Easy7 marks
The number of fish in a lake, \(t\) years after the lake was stocked, is modelled by \(N(t)=\dfrac{6000}{1+19\mathrm{e}^{-0.4t}}\), for \(t\ge0\).
(a)
Find the number of fish initially put into the lake.
(1)
(b)
Write down the equation of the horizontal asymptote of the graph of \(N\), and interpret it in context.
(2)
(c)
Find the time taken for the number of fish to reach 90% of the value in part (b).
(2)
(d)
The number of fish increases most rapidly when \(N=3000\). Find the time at which this occurs.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(N(0)=\dfrac{6000}{20}=300\)
A1
(b)
\(N=6000\)
A1
in the long term the number of fish approaches (but does not exceed) 6000
Answers: (a) 300 · (b) \(N=6000\) · (c) 12.9 years · (d) 7.36 years
208AAHL2-q00232.8 The reciprocal function and rational functionsPaper 2Easy8 marks
For a converging lens with focal length 10 cm, the distance of an object from the lens, \(u\) cm, and the distance of its image from the lens, \(v\) cm, are related by \(\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{10}\), where \(u\gt 10\).
(a)
Show that \(v=\dfrac{10u}{u-10}\).
(2)
(b)
Find \(v\) when \(u=15\).
(1)
(c)
Write down the equations of the vertical and horizontal asymptotes of the graph of \(v\) against \(u\).
(2)
(d)
Interpret the horizontal asymptote in the context of the question.
209AAHL2-q00402.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Easy9 marks
The graph of the cubic function \(f(x)=ax^{3}+bx^{2}+cx+d\) passes through the points \((-1,\,0.5)\), \((0,\,4)\), \((2,\,2)\) and \((4,\,8)\).
(a)
Write down the value of \(d\).
(1)
(b)
Find the values of \(a\), \(b\) and \(c\).
(3)
(c)
Find the real zero of \(f\).
(2)
(d)
Find the coordinates of the local minimum point of the graph of \(f\).
(2)
(e)
Hence explain why \(f\) has only one real zero.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(d=4\)
A1
(b)
forming a system, e.g. \(-a+b-c+4=0.5,\ 8a+4b+2c+4=2,\ 64a+16b+4c+4=8\)
M1
use of technology to solve the system, e.g. matrix method or simultaneous-equation solver
(M1)
\(a=0.5,\ b=-2,\ c=1\)
A1
All three values.
(c)
solving \(0.5x^{3}-2x^{2}+x+4=0\) using technology
(M1)
\(x=-1.07\)
A1
\(-1.07394\ldots\)
(d)
\(x=2.39\)
A1
\(2.38742\ldots\)
\(y=1.79\)
A1
\(1.79174\ldots\)
(e)
the local minimum value \(1.79\gt 0\) (and the local maximum, at \((0.279,\,4.13)\), is also above the \(x\)-axis), so the graph crosses the \(x\)-axis only once, for \(x\lt 0.279\)
210AAHL2-q00502.13 Rational functions with oblique and horizontal asymptotesPaper 2Easy8 marks
The cost, in euros, of producing \(x\) items in a week is \(C(x)=0.02x^{2}+3x+500\). The average cost per item is \(A(x)=\dfrac{C(x)}{x}\), for \(x\gt 0\).
(a)
Find \(A(50)\).
(1)
(b)
Write down the equation of the oblique asymptote of the graph of \(y=A(x)\).
(1)
(c)
Interpret the oblique asymptote in the context of the question.
(1)
(d)
Find the number of items, \(x\in\mathbb{Z}^{+}\), that minimizes the average cost, and the corresponding average cost.
(3)
(e)
Find the range of values of \(x\) for which the average cost is at most €10 per item.
(2)
Show mark scheme
Working
Mark
Notes
(a)
€14
A1
\(\dfrac{50+150+500}{50}\)
(b)
\(y=0.02x+3\)
A1
(c)
for large production levels the fixed cost of €500 shared per item becomes negligible, and the average cost is approximately \(0.02x+3\)
R1
(d)
minimum on the graph of \(y=A(x)\) at \(x=158.11\ldots\)
(M1)
\(158\) items
A1
\(A(158)=9.32456\ldots\lt A(159)=9.32465\ldots\)
€9.32
A1
(e)
\(0.02x^{2}+3x+500\le10x\), or intersection of \(y=A(x)\) with \(y=10\)
214IBMA2H-047-01-112.10 Solving equations graphically and analyticallyPaper 2Easy5 marks
Calculator
The fuel consumption of a hybrid car, F litres per 100 km, at a steady speed of v km h−1 is modelled by \(F(v)=0.0025v^2-0.3v+14\), for \(30\le v\le 120\).
(a)
Find the speed at which fuel consumption is lowest, and the minimum fuel consumption.
(2)
(b)
Find the range of speeds for which the fuel consumption is at most 6.5 litres per 100 km.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(v=60\) km h−1
✓ 1
A1 \(F=5\) litres per 100 km
✓ 1
Part (b)
M1 forms \(0.0025v^2-0.3v+7.5\le 0\)
✓ 1
A1 finds the roots \(v=35.5\) and \(v=84.5\)
✓ 1
A1 \(35.5\le v\le 84.5\) (both within the domain)
✓ 1
Exam tip: Check that your critical values lie inside the stated domain before writing the final interval. Here both 35.5 and 84.5 lie between 30 and 120.
215IBMA2H-047-04-042.15 Solving inequalities g(x) ≥ f(x)Paper 2Easy7 marks
Calculator
After an injection, the concentration of a drug in a patient's blood is modelled by \(C(t)=12te^{-0.4t}\), where \(C\) is in mg L−1 and \(t\ge0\) is the time in hours.
(a)
Find the interval of time during which the drug is effective.
(3)
(b)
Use calculus to find the exact time of maximum concentration, and the exact maximum concentration.
(3)
(c)
Describe the behaviour of the graph as \(t\to\infty\) and interpret this in context.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts to solve \(12te^{-0.4t}=8\), e.g. intersection of graphs on GDC
✓ 1
A1 \(t=0.991\)
✓ 1
A1 \(t=5.07\), so effective for \(0.991<t<5.07\) hours
A1 \(C\to0\) (horizontal asymptote \(C=0\)): the drug is eventually almost completely removed from the blood
✓ 1
Exam tip: Give both ends of the interval: a single crossing point does not describe when the drug is effective.
216IBMA2H-047-04-092.4 Key features of graphs and intersections using technologyPaper 2Easy7 marks
Calculator
Phone A is charging. Its battery charge, as a percentage, is modelled by \(B_A(t)=100-80e^{-0.025t}\), where \(t\ge0\) is the time in minutes. Phone B starts charging at the same time and its charge is modelled by \(B_B(t)=15+1.1t\) until it is full.
(a)
Write down the charge of phone A at \(t=0\).
(1)
(b)
State the horizontal asymptote and interpret it.
(2)
(c)
Find the time at which the two phones have the same charge.
(2)
(d)
Find the time taken for phone A to reach 90% charge.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 20%
✓ 1
Part (b)
A1 \(B=100\)
✓ 1
R1 the charge approaches 100% but the model never quite reaches full charge
✓ 1
Part (c)
M1 attempts to solve \(100-80e^{-0.025t}=15+1.1t\), e.g. graph intersection on GDC
✓ 1
A1 \(t=61.7\) minutes
✓ 1
Part (d)
M1 \(80e^{-0.025t}=10\), so \(t=\frac{\ln8}{0.025}\)
✓ 1
A1 \(t=83.2\) minutes
✓ 1
Common error: Solving \(80e^{-0.025t}=90\) instead of \(=10\) mixes up the charge with the charge still missing; rearrange before taking logs.
217IBMA2H-047-03-022.8 The reciprocal function and rational functionsPaper 2Easy5 marks
Calculator
A smartphone is put on charge. Its battery level, \(Q\) percent, \(t\) minutes after charging starts is modelled by \(Q(t)=\frac{100t+600}{t+30}\), for \(t\geq 0\).
(a)
Write down the battery level when charging starts.
(1)
(b)
Find the time taken for the battery level to reach 80 %.
(2)
(c)
Write down the equation of the horizontal asymptote of the graph of \(Q\) and interpret it in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(Q(0)=\frac{600}{30}=20\) %
✓ 1
Part (b)
M1 sets \(\frac{100t+600}{t+30}=80\) and rearranges, e.g. \(100t+600=80t+2400\)
✓ 1
A1 \(t=90\) minutes
✓ 1
Part (c)
A1 \(Q=100\)
✓ 1
R1 the battery level gets closer and closer to 100 % but, according to the model, never reaches full charge
✓ 1
Exam tip: For \(\frac{ax+b}{cx+d}\) the horizontal asymptote is \(y=\frac{a}{c}\). When a context is given, the interpretation must mention what the quantity approaches, not just state the equation.
218IBMA2H-047-03-042.5 Composite functions and inverse functionsPaper 2Easy6 marks
Calculator
The gravitational field strength, \(g\) m s−2, at a height \(h\) km above the surface of the Earth is modelled by \(g(h)=9.81\left(\frac{6371}{6371+h}\right)^2\), for \(h\geq 0\).
(a)
A space station orbits at a height of 400 km. Find the value of \(g\) at this height.
(2)
(b)
Write down the range of \(g\).
(2)
(c)
Find the height at which \(g\) is half of its value at the surface.
Common error: Halving \(g\) does not double the distance from the centre: because of the square, the distance increases by a factor of \(\sqrt{2}\), so \(h\neq 6371\) km.
219IBMA2H-047-06-032.15 Solving inequalities g(x) ≥ f(x)Paper 2Easy6 marks
Calculator
After an injection, the concentration of a drug in a patient's blood is modelled by \(C(t)=12t\,e^{-0.4t}\) mg L−1, where \(t\) is the time in hours after the injection, \(0\le t\le 12\). The drug is effective when its concentration is at least 8 mg L−1.
(a)
Solve the equation \(C(t)=8\).
(3)
(b)
Find the length of time for which the drug is effective.
(1)
(c)
Find the maximum concentration of the drug and the time at which it occurs.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the GDC to find the intersection of \(y=12te^{-0.4t}\) and \(y=8\) (or the zeros of \(C(t)-8\))
✓ 1
A1 \(t=0.991\)
✓ 1
A1 \(t=5.07\)
✓ 1
Part (b)
A1 \(5.074-0.991=4.08\) hours
✓ 1
Part (c)
A1 \(t=2.5\) hours
✓ 1
A1 maximum concentration \(=30e^{-1}=11.0\) mg L−1
✓ 1
Common error: Subtracting rounded roots can change the third significant figure. Use the stored GDC values (0.99096… and 5.0740…) when finding the time interval.
220IBMA2H-047-07-022.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Easy6 marks
Calculator
A delivery drone takes off from a platform on a rooftop and later lands on the ground. Its height above the ground, in metres, t seconds after take-off is modelled by \(h(t)=-0.5t^3+4t^2+3\), for \(0\le t\le T\), where T is the time at which the drone reaches the ground.
(a)
Write down the height of the platform.
(1)
(b)
Find the value of T.
(2)
(c)
Find the maximum height of the drone and the time at which it occurs.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(h(0)=3\) m
✓ 1
Part (b)
M1 attempt to find the positive zero of \(h\) using technology
✓ 1
A1 \(T=8.09\) s
✓ 1
Part (c)
M1 use of GDC maximum feature (or \(h'(t)=0\))
✓ 1
A1 \(t=5.33\) s (accept \(\frac{16}{3}\))
✓ 1
A1 maximum height \(=40.9\) m
✓ 1
Common error: Setting the GDC window to the default range can hide the landing time near \(t=8.09\); choose a window that covers the whole domain before reading off zeros.
221IBMA2H-047-07-092.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Easy4 marks
Calculator
A basketball player releases the ball 2.1 m above the floor. The height of the ball, in metres, when it is a horizontal distance x metres from the player is modelled by \(h(x)=-0.12x^2+0.9x+2.1\). The ball passes through the hoop, which is 3.05 m above the floor, while descending.
(a)
Find the maximum height of the ball and the horizontal distance at which it occurs.
(2)
(b)
Find the horizontal distance from the player to the hoop.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=3.75\) m
✓ 1
A1 maximum height \(3.79\) m
✓ 1
Part (b)
M1 solves \(h(x)=3.05\), e.g. intersection with \(y=3.05\), giving \(1.27\) and \(6.23\)
✓ 1
A1 \(6.23\) m, rejecting \(1.27\) as the ball is still rising there
✓ 1
Exam tip: When an equation has two valid-looking roots, use the context (rising or falling) to select one and say why the other is rejected.
222IBMA2H-047-02-032.15 Solving inequalities g(x) ≥ f(x)Paper 2Easy6 marks
Calculator
A patient receives a single intravenous dose of an antibiotic. The concentration of the antibiotic in the blood plasma, \(C\) mg L−1, \(t\) hours after the dose is modelled by \(C = C_0 e^{-kt}\), where \(C_0\) and \(k\) are positive constants. Blood tests give \(C = 7.5\) at \(t = 3\) and \(C = 4.8\) at \(t = 6\).
(a)
Find the value of \(k\).
(2)
(b)
Find the value of \(C_0\).
(2)
(c)
The antibiotic stops being effective when the concentration falls below 2 mg L−1. Find the time after the dose when this first happens.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 divides the two equations: \(e^{-3k} = \frac{4.8}{7.5} = 0.64\)
M1 solves \(11.71875\,e^{-kt} = 2\), e.g. \(t = \frac{\ln(11.71875/2)}{k}\)
✓ 1
A1 \(t = 11.9\) hours (11.885…)
✓ 1
Exam tip: When two data points are given, divide one equation by the other to eliminate \(C_0\) first. Then store the unrounded \(k\) in your calculator for the later parts.
223IBMA2H-047-05-032.15 Solving inequalities g(x) ≥ f(x)Paper 2Easy6 marks
Calculator
A hospital dosing chart gives the dose, \(D\) mg, of a drug as a linear function of a patient's body mass, \(w\) kg. A patient of mass 60 kg receives 180 mg and a patient of mass 90 kg receives 255 mg.
(a)
Find the gradient of the line and interpret its value in this context.
(2)
(b)
Find an equation for \(D\) in terms of \(w\).
(2)
(c)
A patient is prescribed a dose of 220 mg. Find the patient's body mass according to the chart.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 gradient \(=\frac{255-180}{90-60}=2.5\)
✓ 1
R1 the dose increases by 2.5 mg for each additional kilogram of body mass
✓ 1
Part (b)
M1 substitutes a data point, e.g. \(D-180=2.5(w-60)\)
✓ 1
A1 \(D=2.5w+30\)
✓ 1
Part (c)
M1 \(220=2.5w+30\)
✓ 1
A1 \(w=76\) kg
✓ 1
Exam tip: When asked to interpret a gradient, give its units and say what changes per unit increase of the other variable, e.g. 'mg per kg'.
224IBMA3H-2-19r8-022.10 Solving equations graphically and analyticallyPaper 2Easy6 marks
Calculator
A cycling route is 8 km long. The height of the road above sea level, \(h\) metres, at a distance \(x\) km from the start is modelled by \(h=-2x^2+16x+40\), for \(0\leq x\leq 8\).
(a)
Find the greatest height reached by the road and the value of \(x\) at which it occurs.
(3)
(b)
Find the distance from the start, in the second half of the route (\(x>4\)), at which the road is at a height of 60 m. Give your answer to 3 significant figures.
(3)
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Mark
Notes
Part (a)
M1 attempt to find the vertex, e.g. \(x=-\frac{16}{2(-2)}\) or completing the square
✓ 1
A1 \(x=4\) km
✓ 1
A1 greatest height \(72\) m
✓ 1
Part (b)
M1 equate \(-2x^2+16x+40=60\) and rearrange to \(x^2-8x+10=0\)
✓ 1
A1 \(x=4\pm\sqrt{6}\) (accept \(6.45\) or \(1.55\) seen)
✓ 1
A1 \(x=6.45\) km
✓ 1
Common error: Both roots satisfy the equation, so check which one lies in the interval the question asks about.
225IBMA3H-2-19r8-042.7 Quadratic equations, inequalities and the discriminantPaper 2Easy5 marks
Calculator
A cyclist rides a 48 km route at a constant speed of \(v\) km/h, and also takes a rest stop of 0.5 hours. The total time for the trip is \(T\) hours, where \(T=\frac{48}{v}+0.5\), for \(v>0\).
(a)
Find the total time when \(v=24\).
(1)
(b)
Find the speed needed for the total time to be exactly 2 hours.
(2)
(c)
State the equation of the horizontal asymptote of the graph of \(T\) against \(v\), and interpret it in context.
(2)
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Notes
Part (a)
A1 \(T=\frac{48}{24}+0.5=2.5\) hours
✓ 1
Part (b)
M1 \(\frac{48}{v}+0.5=2\) so \(\frac{48}{v}=1.5\)
✓ 1
A1 \(v=32\) km/h
✓ 1
Part (c)
A1 \(T=0.5\)
✓ 1
R1 however fast the cyclist rides, the total time can never be less than the 0.5 hour rest stop (time tends to the rest time)
✓ 1
Exam tip: When you interpret an asymptote, say what the quantity approaches and why, using the units of the context.
226IBMA3H-2-19r13-022.16 Modulus, reciprocal and other graph transformationsPaper 2Easy5 marks
Calculator
The power drawn by a server rack in a data centre is modelled by \(P(n)=0.4n+2.5\), where \(P\) is in kW and \(n\) is the number of servers installed, with \(0\le n\le 40\).
(a)
Find \(P^{-1}(x)\).
(2)
(b)
State the domain of \(P^{-1}\).
(1)
(c)
Find \(P^{-1}(14.5)\) and state what this value represents in context.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempt to rearrange \(x=0.4n+2.5\) (or swap variables)
R1 the number of servers installed when the rack draws 14.5 kW
✓ 1
Exam tip: The domain of an inverse function is the range of the original function, so evaluate \(P\) at both ends of the allowed values of \(n\).
227IBMA3H-2-19r13-032.10 Solving equations graphically and analyticallyPaper 2Easy6 marks
Calculator
A cyclist rides at a constant speed of 18 km h−1, so the distance ridden after \(t\) hours is \(d=f(t)=18t\). The energy used, in kcal, for a ride of distance \(d\) km is \(g(d)=42d+150\). The cyclist rides for at most 6 hours.
(a)
Find an expression for \((g\circ f)(t)\).
(2)
(b)
Interpret the value 756 in context.
(1)
(c)
State the domain of \(g\circ f\) in this context.
(1)
(d)
Find the time, in hours, needed to use 2000 kcal. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
M1 substitute \(f(t)\) into \(g\), i.e. \(42(18t)+150\)
✓ 1
A1 \(756t+150\)
✓ 1
Part (b)
R1 the energy used, in kcal, for each hour ridden
✓ 1
Part (c)
A1 \(0\le t\le 6\)
✓ 1
Part (d)
M1 set \(756t+150=2000\)
✓ 1
A1 \(t=2.45\) hours
✓ 1
Common error: Reading \(g\circ f\) as \(g\) multiplied by \(f\). For \(g\circ f\) the output of \(f\) becomes the input of \(g\).
228IBMA3H-2-19r21-022.8 The reciprocal function and rational functionsPaper 2Easy6 marks
Calculator
The concentration, \(C\) ppm, of sulphur dioxide measured downwind of a volcanic vent at a distance of \(x\) km is modelled by \(C(x)=\dfrac{6x+5}{2x-3}\), for \(x>2\).
(a)
Show that \(C(x)=3+\dfrac{14}{2x-3}\).
(2)
(b)
Write down the equation of the horizontal asymptote of the graph of \(y=C(x)\) and state what it represents.
(2)
(c)
Find the distance from the vent at which the concentration is 10 ppm.
(2)
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Notes
Part (a)
M1 writes \(6x+5=3(2x-3)+14\) or uses polynomial division
✓ 1
A1 \(\frac{6x-9+14}{2x-3}=3+\frac{14}{2x-3}\) AG
✓ 1
Part (b)
A1 \(C=3\)
✓ 1
A1 far from the vent the concentration levels off at (approaches) 3 ppm
✓ 1
Part (c)
M1 \(\frac{6x+5}{2x-3}=10\) so \(6x+5=20x-30\)
✓ 1
A1 \(x=\frac{35}{14}=2.5\) km
✓ 1
Exam tip: Rewriting a rational function as a constant plus a fraction shows the horizontal asymptote immediately.
229IBMA3H-2-19r21-042.14 Odd and even functions, inverse and self-inverse functionsPaper 2Easy6 marks
Calculator
A cargo ship's fuel use is modelled by \(f(v)=0.012v^2\) tonnes per hour, where \(v\) is its speed in knots and \(8\le v\le 20\). The fuel costs €620 per tonne, so the cost in euros per hour of burning \(F\) tonnes per hour is \(c(F)=620F\).
(a)
Find \((c\circ f)(v)\) in the form \(kv^2\).
(2)
(b)
Find an expression for the inverse of \(c\circ f\) and state what it gives.
(2)
(c)
The company allows an hourly fuel cost of at most €2380. Find the greatest speed the ship may travel at, to 3 significant figures.
(2)
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Notes
Part (a)
M1 \(620\times 0.012v^2\)
✓ 1
A1 \(7.44v^2\)
✓ 1
Part (b)
M1 sets \(x=7.44v^2\) and rearranges, taking the positive root
✓ 1
A1 \(v=\sqrt{\frac{x}{7.44}}\); it gives the speed in knots for an hourly fuel cost of €\(x\)
✓ 1
Part (c)
M1 \(7.44v^2=2380\) (or uses the inverse with \(x=2380\))
✓ 1
A1 \(v=17.9\) knots
✓ 1
Common error: Take the positive root only when the domain restricts the variable to positive values, and check the answer lies in the stated domain.
230IBMA3H-2-20r30-012.1 Equations of straight lines, parallel and perpendicular linesPaper 2Easy5 marks
Calculator
The air pressure P kilopascals (kPa) at an altitude of h metres above sea level is modelled by P = 101 e−0.000125h. A light aircraft climbs through the air described by this model.
(a)
Find the pressure at an altitude of 2500 m, to 3 significant figures.
(2)
(b)
Find the altitude at which the pressure is 60 kPa, giving your answer to 3 significant figures.
(3)
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Notes
Part (a)
M1 substitute h = 2500 into the model
✓ 1
A1 P = 73.9 kPa
✓ 1
Part (b)
M1 e−0.000125h = 60⁄101
✓ 1
M1 −0.000125h = ln(60⁄101)
✓ 1
A1 h = 4170 m
✓ 1
Exam tip: To undo an exponential, take natural logarithms of both sides after isolating the exponential term.
231IBMA3H-2-20r30-022.8 The reciprocal function and rational functionsPaper 2Easy5 marks
Calculator
The acidity of a soil is measured by its pH, where pH = −log10[H+] and [H+] is the hydrogen ion concentration in mol dm−3. A farmer tests two fields. Field A has pH 5.2 and field B has pH 6.7.
(a)
Find [H+] for field A, giving your answer in the form a × 10k with 1 ≤ a < 10 and a to 3 significant figures.
(2)
(b)
Find how many times greater [H+] is in field A than in field B, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 [H+] = 10−5.2
✓ 1
A1 6.31 × 10−6 mol dm−3
✓ 1
Part (b)
M1 ratio = 10−5.2 ⁄ 10−6.7
✓ 1
M1 = 101.5
✓ 1
A1 31.6
✓ 1
Common error: A lower pH means a higher concentration; the ratio must be larger than 1 when comparing the more acidic field with the less acidic one.
232IBMA3H-2-18r34-022.10 Solving equations graphically and analyticallyPaper 2Easy4 marks
Calculator
A courier charges a fixed fee of €12 plus €0.80 per kilometre. The cost, €C, of delivering a parcel a distance of d km is \(C=f(d)=12+0.8d\), for \(d\ge0\).
(a)
Find \(f^{-1}(x)\).
(2)
(b)
Find \(f^{-1}(52)\) and explain what this value represents.
(2)
Show mark scheme
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Notes
Part (a)
M1 write \(x=12+0.8d\) and rearrange for d (or swap variables)
R1 it is the distance, in km, that can be delivered for €52
✓ 1
Common error: Do not confuse \(f^{-1}(x)\) with \(\frac{1}{f(x)}\). The inverse takes a cost as its input and returns a distance.
233IBMA3H-2-19r34-012.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Easy6 marks
Calculator
During a volcanic eruption, a lava fragment is thrown upward from a vent that is 10 m above a flat lava plain. Its height above the plain is modelled by \(h(t) = -4.9t^2 + 29.4t + 10\), where \(h\) is in metres and \(t\) is the time in seconds after launch.
(a)
Write \(h(t)\) in the form \(a(t-p)^2 + q\).
(3)
(b)
Find the time at which the fragment lands on the plain. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to complete the square or to use \(p = -\frac{b}{2a}\)
✓ 1
A1 \(a = -4.9\) and \(p = 3\)
✓ 1
A1 \(q = 54.1\), so \(h = -4.9(t-3)^2 + 54.1\)
✓ 1
Part (b)
M1 set \(h = 0\) and attempt to solve (formula, graph or from the vertex form)
✓ 1
A1 roots \(t = 6.32\ldots\) and \(t = -0.323\ldots\)
✓ 1
A1 reject the negative root: \(t = 6.32\) s
✓ 1
Common error: When a time is asked for, always check both roots of the quadratic and reject the one that is negative.
234IBMA3H-2-19r34-032.4 Key features of graphs and intersections using technologyPaper 2Easy5 marks
Calculator
A shipping company charges a cost per container that falls as more containers share the fixed costs of a voyage. For \(n\) containers, the cost per container, in £, is modelled by \(C(n) = \dfrac{120n + 3000}{n + 10}\), where \(n \ge 1\).
(a)
Find the cost per container when 20 containers are shipped.
(1)
(b)
Write down the equation of the horizontal asymptote of the graph of \(C\), and interpret it in this context.
(2)
(c)
Find the number of containers for which the cost per container is £135.
(2)
Show mark scheme
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Notes
Part (a)
A1 $C(20) = $ £180
✓ 1
Part (b)
A1 \(C = 120\)
✓ 1
A1 however many containers are shipped, the cost per container approaches £120 but never falls to it
✓ 1
Part (c)
M1 \(120n + 3000 = 135(n + 10)\)
✓ 1
A1 \(n = 110\)
✓ 1
Common error: Do not equate the horizontal asymptote with a value the function reaches; it is a limiting value as \(n\) becomes large.
235IBMA3H-2-19r36-022.3 Graphs of functions and sketching from a contextPaper 2Easy7 marks
Calculator
The concentration of a drug in the blood, C mg per litre, t hours after an injection is modelled by C = −0.05t2 + 1.2t + 2, for values of t up to the time the concentration falls to zero.
(a)
Write C in the form C = −0.05(t − p)2 + q, finding the values of p and q.
(4)
(b)
Find the time at which the concentration falls to zero. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses t = −b/(2a) or completes the square to find p
✓ 1
A1 p = 12
✓ 1
M1 substitutes t = 12 into the model (or expands −0.05(t − 12)2 to find q)
✓ 1
A1 q = 9.2
✓ 1
Part (b)
M1 sets −0.05(t − 12)2 + 9.2 = 0 (ECF from (a)) or uses the quadratic formula
✓ 1
M1 t = 12 ± √184
✓ 1
A1 t = 25.6 hours (negative root rejected as t must be positive)
✓ 1
Common error: Writing p = −12 instead of p = 12: in the form a(t − p)² + q, the vertex is at t = p, so a vertex at t = 12 gives (t − 12).
236IBMA3H-2-19r36-032.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Easy4 marks
Calculator
The ground along a proposed route for a water pipe is modelled by the curve y = 0.5x2 − 3x + 4, where x is the horizontal distance in hundreds of metres and y is the height in tens of metres. A straight pipe is laid along the line y = x + 1 and meets the ground at two points.
(a)
Find the values of x at the two points where the pipe meets the ground. Give each answer to 3 significant figures.
(2)
(b)
Find the height of the pipe at each of these points, to 3 significant figures. Use the unrounded values of x from part (a).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates 0.5x2 − 3x + 4 = x + 1 (or graphs both functions)
✓ 1
A1 x = 0.838 and x = 7.16
✓ 1
Part (b)
A1 y = 1.84 (at the smaller x)
✓ 1
A1 y = 8.16 (at the larger x) (ECF from (a))
✓ 1
Exam tip: On a calculator question, state the equation you are solving or the graphs you are intersecting so method marks are still available if a value is misread.
237IBMA3H-2-18r43-032.4 Key features of graphs and intersections using technologyPaper 2Easy4 marks
Calculator
Two robots move along the same straight rail. Their positions, in metres from a marker, at time \(t\) seconds are \(s_A=0.5t^2+1\) and \(s_B=4t-2\), for \(0\le t\le10\).
(a)
Show that the robots are at the same position when \(t^2-8t+6=0\).
(2)
(b)
Hence find the times, to 3 significant figures, at which the robots are at the same position.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(0.5t^2+1=4t-2\)
✓ 1
A1 \(t^2-8t+6=0\)
✓ 1
Part (b)
A1 \(t=0.838\) s
✓ 1
A1 \(t=7.16\) s
✓ 1
Exam tip: Both roots lie inside the stated domain here, so both are valid; always check each against the domain given.
238IBMA3H-2-20r48-032.15 Solving inequalities g(x) ≥ f(x)Paper 2Easy6 marks
Calculator
The concentration of chlorine in a swimming pool is modelled by \(C(t)=5e^{-0.12t}\), where \(C\) is in mg per litre and \(t\) is the time in hours after the pool is treated.
(a)
Write down the concentration at the moment the pool is treated.
(1)
(b)
Find the concentration 6 hours after treatment. Give your answer to 3 significant figures.
(2)
(c)
The pool must be treated again when the concentration falls to 1.5 mg per litre. Find the time at which this happens, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(5\) mg per litre
✓ 1
Part (b)
M1 \(5e^{-0.12\times 6}\)
✓ 1
A1 \(2.43\) mg per litre
✓ 1
Part (c)
M1 \(5e^{-0.12t}=1.5\)
✓ 1
M1 take natural logarithms: \(t=\frac{\ln(5/1.5)}{0.12}\)
✓ 1
A1 \(t=10.0\) hours
✓ 1
Exam tip: To undo \(e^{kt}\), take \(\ln\) of both sides after isolating the exponential; do not take logs term by term.
239AAHL2-q00072.3 Graphs of functions and sketching from a contextPaper 2Medium8 marks
Let \(f(x)=2^{x}\) and \(g(x)=x^{2}\), for \(x\in\mathbb{R}\). The function \(h\) is defined by \(h(x)=f(x)-g(x)\). The following diagram shows part of the graph of \(y=h(x)\).
(a)
Verify that \(x=2\) and \(x=4\) are zeros of \(h\).
(1)
(b)
Find the third zero of \(h\).
(1)
(c)
Find the coordinates of the local maximum point and of the local minimum point on the graph of \(h\).
(2)
(d)
Find the set of values of \(k\) for which the equation \(h(x)=k\) has exactly three solutions.
(2)
(e)
Hence, or otherwise, solve the inequality \(2^{x}\gt x^{2}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(2^{2}-2^{2}=0\) and \(2^{4}-4^{2}=16-16=0\)
A1
(b)
\(x=-0.767\)
A1
\(-0.766664\ldots\)
(c)
maximum \((0.485,\,1.16)\)
A1
\((0.485089\ldots,\ 1.16436\ldots)\)
minimum \((3.21,\,-1.05)\)
A1
\((3.21243\ldots,\ -1.05060\ldots)\)
(d)
horizontal line between the minimum and maximum values
(M1)
The graph increases without bound as \(x\to\infty\) and decreases without bound as \(x\to-\infty\).
\(-1.05\lt k\lt 1.16\)
A1
FT from (c). Award A0 for non-strict inequalities.
(e)
\(-0.767\lt x\lt 2\)
A1
FT from (b).
\(x\gt 4\)
A1
Answers: (b) \(-0.767\) · (c) max \((0.485,\,1.16)\), min \((3.21,\,-1.05)\) · (d) \(-1.05\lt k\lt 1.16\) · (e) \(-0.767\lt x\lt 2\) or \(x\gt 4\)
240AAHL2-q00092.4 Key features of graphs and intersections using technologyPaper 2Medium9 marks
A mass on a spring is pulled down and released. Its displacement from its rest position, \(y\) cm, \(t\) seconds after it is released is modelled by \(y(t)=12\mathrm{e}^{-0.25t}\cos(2t)\), for \(t\ge0\).
(a)
Find the first time at which the mass passes through its rest position.
(2)
(b)
Find the coordinates of the first local minimum point on the graph of \(y\), for \(t\gt 0\).
(2)
(c)(i)
Explain why \(|y(t)|\le12\mathrm{e}^{-0.25t}\) for all \(t\ge0\).
(1)
(c)(ii)
Hence find a time \(T_{1}\) such that \(|y(t)|\lt 1\) for all \(t\gt T_{1}\).
(2)
(d)
Use your graphic display calculator to find the least value of \(T\) such that \(|y(t)|\lt 1\) for all \(t\gt T\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\cos(2t)=0\), \(2t=\dfrac{\pi}{2}\)
(M1)
\(t=\dfrac{\pi}{4}\;(=0.785)\) seconds
A1
(b)
\(t=1.51\)
A1
\(1.50862\ldots\)
\(y=-8.17\)
A1
\(-8.16617\ldots\)
(c)(i)
\(|\cos(2t)|\le1\) and \(12\mathrm{e}^{-0.25t}\gt 0\)
R1
(c)(ii)
\(12\mathrm{e}^{-0.25t}\lt 1\)
M1
\(T_{1}=4\ln12=9.94\)
A1
Accept any value \(T_{1}\ge9.94\) with a correct justification.
(d)
finding the largest solution of \(y(t)=1\) or \(y(t)=-1\)
(M1)
Solutions of \(|y|=1\) near the end are 8.28, 9.11 and 9.62.
241AAHL2-q00102.4 Key features of graphs and intersections using technologyPaper 2Medium9 marks
A company sells \(x\) hundred units of a product each month, where \(0\le x\le60\). The monthly revenue, \(R\) thousand euros, and the monthly cost, \(C\) thousand euros, are modelled by
\(R(x)=60x\,\mathrm{e}^{-0.05x}\) and \(C(x)=150+10x\).
(a)
Find the maximum monthly revenue and the value of \(x\) at which it occurs.
(2)
(b)
Find the values of \(x\) for which the company breaks even.
(2)
(c)
Find the maximum monthly profit and the value of \(x\) at which it occurs.
(3)
(d)
The fixed cost of 150 thousand euros increases to \(F\) thousand euros, while the rest of the model is unchanged. Find the value of \(F\) above which the company cannot make a profit.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(x=20\)
A1
441 thousand euros
A1
\(441.455\ldots\;(=1200\mathrm{e}^{-1})\)
(b)
\(x=3.78\)
A1
\(3.78191\ldots\)
\(x=27.0\)
A1
\(26.9980\ldots\)
(c)
\(P(x)=60x\,\mathrm{e}^{-0.05x}-150-10x\), maximum from the graph
(M1)
\(x=13.5\)
A1
\(13.4647\ldots\)
127 thousand euros
A1
\(127.417\ldots\) Accept €127 000 or €127 418.
(d)
the profit decreases by \(F-150\) for every \(x\), so the greatest profit is \(127.417\ldots-(F-150)\)
242AAHL2-q00262.9 Exponential and logarithmic functionsPaper 2Medium9 marks
The sound level, \(L\) decibels (dB), of a sound of intensity \(I\) W m−2 is given by \(L=10\log_{10}\left(\dfrac{I}{I_{0}}\right)\), where \(I_{0}=10^{-12}\) W m−2.
(a)
Find the sound level of a sound with intensity \(3.2\times10^{-5}\) W m−2.
(2)
(b)
Find an expression for \(I\) in terms of \(L\).
(2)
(c)
Show that doubling the intensity of a sound increases its sound level by approximately 3.01 dB.
(2)
(d)
At a point in a workshop, one machine produces a sound level of 72 dB. When \(n\) identical machines are running, the intensity at this point is \(n\) times the intensity from one machine. Find the least number of machines that produce a sound level of at least 85 dB at this point.
243AAHL2-q00272.10 Solving equations graphically and analyticallyPaper 2Medium9 marks
The intensity of the light from a lighthouse, measured at a distance of \(x\) km from the lighthouse, is modelled by
\(I(x)=\dfrac{5000\,\mathrm{e}^{-kx}}{x^{2}}\), for \(x\gt 0\),
where \(k\ge0\) is a constant that depends on the weather. The crew of a ship can see the light when \(I(x)\ge2\). For each value of \(k\), \(I\) is a decreasing function of \(x\).
(a)
On a perfectly clear night, \(k=0\). Find the greatest distance from the lighthouse at which the light can be seen.
(2)
(b)
On a misty night, \(k=0.4\). Find the greatest distance from the lighthouse at which the light can be seen.
(2)
(c)
On another night, the light can just be seen at a distance of 12 km. Find the value of \(k\) on that night.
(3)
(d)
On the misty night, the light is uncomfortably bright when \(I(x)\gt 20\). Find the set of distances at which the light can be seen but is not uncomfortably bright.
Answers: (a) 50 km · (b) 8.73 km · (c) \(k=0.238\) · (d) \(5.39\le x\le8.73\)
244AAHL2-q00392.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Medium8 marks
The profile of a ski slope is modelled by \(h(x)=-0.01x^{3}+0.2x^{2}-1.5x+10\), for \(0\le x\le L\), where \(h\) is the height in metres above level ground and \(x\) is the horizontal distance in metres from the top of the slope. The slope ends where it meets the level ground, at \(x=L\).
(a)
Find the value of \(L\).
(2)
(b)
Show that \((x-10)\) is a factor of \(h(x)-5\).
(2)
(c)
Hence find the quadratic polynomial \(q(x)\) such that \(h(x)-5=-0.01(x-10)\,q(x)\).
(2)
(d)
Use your answer to part (c) to explain why a skier on the slope is at a height of 5 m at exactly one point.
(2)
Show mark scheme
Working
Mark
Notes
(a)
solving \(h(x)=0\), e.g. graph or polynomial solver
(M1)
\(L=14.4\)
A1
\(14.4061\ldots\) The other two roots are not real.
(b)
\(h(10)=-10+20-15+10\)
M1
\(h(10)=5\), so \(h(10)-5=0\) and by the factor theorem \((x-10)\) is a factor
A1
(c)
division of \(-0.01x^{3}+0.2x^{2}-1.5x+5\) by \((x-10)\), or comparing coefficients
(M1)
\(q(x)=x^{2}-10x+50\)
A1
(d)
\(\Delta=(-10)^{2}-4(1)(50)=-100\)
M1
\(\Delta\lt 0\), so \(q(x)=0\) has no real roots; the only solution of \(h(x)=5\) is \(x=10\)
R1
Answers: (a) \(L=14.4\) · (c) \(x^{2}-10x+50\)
245AAHL2-q00482.13 Rational functions with oblique and horizontal asymptotesPaper 2Medium8 marks
Let \(f(x)=\dfrac{3x-1}{x^{2}-2x+5}\), for \(x\in\mathbb{R}\).
(a)
Explain why the graph of \(f\) has no vertical asymptotes.
(2)
(b)
Write down the equation of the horizontal asymptote of the graph of \(f\).
(1)
(c)
Find the coordinates of the local maximum point and the local minimum point of the graph of \(f\).
(2)
(d)
Find the set of values of \(k\) for which the equation \(f(x)=k\) has exactly two solutions.
(3)
Show mark scheme
Working
Mark
Notes
(a)
discriminant of the denominator: \((-2)^{2}-4(1)(5)=-16\)
M1
Or completing the square: \((x-1)^{2}+4\).
\(\Delta\lt 0\), so the denominator is never zero
R1
(b)
\(y=0\)
A1
(c)
maximum \((2.44,\,1.04)\)
A1
\((2.44151\ldots,\ 1.04056\ldots)\)
minimum \((-1.77,\,-0.541)\)
A1
\((-1.77485\ldots,\ -0.540569\ldots)\)
(d)
considering horizontal lines \(y=k\) on the graph, noting that \(y=0\) meets the graph only at \(x=\tfrac{1}{3}\)
(M1)
\(-0.541\lt k\lt 0\)
A1
\(0\lt k\lt 1.04\)
A1
Award A1A0 for a single interval \(-0.541\lt k\lt 1.04\).
Answers: (b) \(y=0\) · (c) max \((2.44,\,1.04)\), min \((-1.77,\,-0.541)\) · (d) \(-0.541\lt k\lt 0\) or \(0\lt k\lt 1.04\)
246AAHL2-q00492.13 Rational functions with oblique and horizontal asymptotesPaper 2Medium10 marks
Let \(f(x)=\dfrac{2x^{2}-x-3}{x+2}\), for \(x\neq-2\).
(a)
Find the equation of the oblique asymptote of the graph of \(f\).
(2)
(b)
Write down the equation of the vertical asymptote of the graph of \(f\).
(1)
(c)
Find the \(x\)-intercepts of the graph of \(f\).
(2)
(d)
Find the coordinates of the local maximum point and of the local minimum point of the graph of \(f\).
(2)
(e)
Find the set of values of \(x\) for which the vertical distance between the graph of \(f\) and its oblique asymptote is less than \(0.1\).
248AAHL2-q00702.15 Solving inequalities g(x) ≥ f(x)Paper 2Medium8 marks
A company’s monthly profit, \(P\) thousand euros, from producing \(x\) hundred units is modelled by
\(P(x)=-0.02x^{3}+0.9x^{2}+3x-40\), for \(0\le x\le50\).
(a)
Find the values of \(x\) for which the company does not make a loss.
(3)
(b)
Find the values of \(x\) for which the profit is at least €150 000.
(2)
(c)
Production is planned in whole hundreds of units, so \(x\in\mathbb{Z}^{+}\). The company wants the profit per hundred units, \(\dfrac{P(x)}{x}\), to be at least €8000. Find the number of possible production levels.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(P(x)\ge0\), e.g. sketch with zeros
(M1)
zeros \(x=5.46407\ldots\) and \(x=47.2779\ldots\)
(A1)
\(5.46\le x\le47.3\)
A1
Accept strict inequalities.
(b)
\(P(x)\ge150\), intersections \(15.6063\ldots\) and \(43.4148\ldots\)
(M1)
\(15.6\le x\le43.4\)
A1
(c)
\(P(x)\ge8x\) (since \(x\gt 0\))
M1
\(12.6151\ldots\le x\le36.7042\ldots\)
A1
So \(13\le x\le36\); e.g. \(P(12)-96=-4.96\), \(P(13)-104=3.16\).
250AAHL2-q00862.16 Modulus, reciprocal and other graph transformationsPaper 2Medium6 marks
(a)
Solve the inequality \(|\ln x|\lt 2-x\), for \(x\gt 0\).
(3)
(b)
Write down the value of \(k\) for which the equation \(|\ln x|=k-x\) has exactly one solution.
(1)
(c)
Explain why the equation \(|\ln x|=k-x\) has no solution when \(k\) is less than the value in part (b).
(2)
Show mark scheme
Working
Mark
Notes
(a)
sketch of \(y=|\ln x|\) and \(y=2-x\)
(M1)
intersections at \(x=0.158594\ldots\) and \(x=1.55714\ldots\)
(A1)
\(0.159\lt x\lt 1.56\)
A1
(b)
\(k=1\)
A1
The line passes through the cusp \((1,\,0)\).
(c)
\(h(x)=|\ln x|+x\) is decreasing on \(0\lt x\lt 1\) and increasing for \(x\gt 1\), so \(h(x)\ge h(1)=1\); that is, \(|\ln x|\ge1-x\) for all \(x\gt 0\)
R1
Accept an equivalent argument, e.g. the graph of \(y=|\ln x|\) never lies below the line \(y=1-x\), which touches it only at \((1,0)\).
for \(k\lt 1\): \(k-x\lt 1-x\le|\ln x|\), so the graphs never meet
R1
Answers: (a) \(0.159\lt x\lt 1.56\) · (b) \(k=1\)
251AAHL2-q00882.16 Modulus, reciprocal and other graph transformationsPaper 2Medium7 marks
The temperature, \(T\) °C, in an office \(t\) hours after midnight is modelled by
\(T(t)=19+4\sin\left(\dfrac{\pi(t-8)}{12}\right)\), for \(0\le t\le24\).
The office is comfortable when \(|T(t)-21|\le1\).
(a)
Write down the maximum and minimum temperatures in the office.
(2)
(b)
Find the values of \(t\) for which the office is comfortable.
(4)
(c)
Find the total time, in hours, for which the office is comfortable.
(1)
Show mark scheme
Working
Mark
Notes
(a)
maximum 23 °C
A1
minimum 15 °C
A1
(b)
\(20\le T(t)\le22\)
M1
Or sketch of \(y=|T(t)-21|\) and \(y=1\).
\(T=20\) at \(t=8.96523\ldots\) and \(t=19.0348\ldots\)
(A1)
\(T=22\) at \(t=11.2393\ldots\) and \(t=16.7606\ldots\)
(A1)
\(8.97\le t\le11.2\) or \(16.8\le t\le19.0\)
A1
(c)
4.55 hours
A1
\(4.54838\ldots\). Accept 4.54 to 4.55.
Answers: (a) 23 °C, 15 °C · (b) \(8.97\le t\le11.2\) or \(16.8\le t\le19.0\) · (c) 4.55 h
252IBMA2H-047-01-062.13 Rational functions with oblique and horizontal asymptotesPaper 2Medium10 marks
Calculator
In a free-kick training drill, the height, h metres, of a football above the ground is modelled by \(h(x)=-0.0125x^2+0.6x\), where x metres is the horizontal distance from the kicking point. The crossbar of the goal is 2.44 m above the ground.
(a)
Find the maximum height of the ball.
(2)
(b)
Find the set of values of x for which the ball is higher than the crossbar.
(4)
(c)
For a different kick, \(h(x)=-kx^2+0.6x\), where \(k>0\). Find the value of k for which the maximum height of the ball is exactly 2.44 m.
(3)
(d)
Write down the set of values of k for which the ball never rises above the crossbar height.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(x=-\frac{b}{2a}=24\) or a GDC
✓ 1
A1 7.2 m
✓ 1
Part (b)
M1 sets up \(-0.0125x^2+0.6x-2.44>0\) or \(h(x)=2.44\)
✓ 1
A1 \(x=4.49\) (4.486…)
✓ 1
A1 \(x=43.5\) (43.514…)
✓ 1
A1 \(4.49<x<43.5\)
✓ 1
Part (c)
M1 forms \(-kx^2+0.6x-2.44=0\), or finds the vertex at \(x=\frac{0.3}{k}\)
✓ 1
M1 sets the discriminant to zero, \(0.36-4k(2.44)=0\), or sets the maximum height to 2.44: \(\frac{0.09}{k}=2.44\)
Exam tip: A maximum height of exactly a given value means the horizontal line just touches the parabola, so the discriminant is zero. Substituting the vertex \(x=-\frac{b}{2a}\) gives the same result.
253IBMA2H-047-01-122.15 Solving inequalities g(x) ≥ f(x)Paper 2Medium12 marks
Calculator
In a hospital laboratory, two bacterial cultures are grown. The number of bacteria in each culture t hours after the start of observation is modelled by \(N_A(t)=500e^{0.2t}\) and \(N_B(t)=\dfrac{8000}{1+15e^{-0.5t}}\), for \(t\ge 0\).
(a)
Show that the two cultures contain the same number of bacteria at \(t=0\).
(2)
(b)
Write down the limiting number of bacteria in culture B.
(1)
(c)
Find the exact time at which \(N_A\) reaches 8000.
(2)
(d)
Solve the inequality \(N_A(t)\ge N_B(t)\).
(4)
(e)
For \(0<t<13.8\), find the maximum value of \(N_B(t)-N_A(t)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(t=0\) into both models
✓ 1
A1 \(N_A(0)=500\) and \(N_B(0)=\frac{8000}{16}=500\)
✓ 1
Part (b)
A1 8000
✓ 1
Part (c)
M1 \(e^{0.2t}=16\)
✓ 1
A1 \(t=5\ln 16\ (=20\ln 2)\) hours
✓ 1
Part (d)
M1 attempts to find the intersections of the graphs on a GDC
✓ 1
A1 \(t=13.8\) (13.787…)
✓ 1
A1 \(t\ge 13.8\)
✓ 1
A1 also includes \(t=0\), since \(N_A(0)=N_B(0)\) and \(N_B>N_A\) for \(0<t<13.8\)
✓ 1
Part (e)
M1 attempts to maximise \(N_B(t)-N_A(t)\) on a GDC
✓ 1
A1 occurs at \(t=8.61\)
✓ 1
A1 maximum difference is 3850 bacteria
✓ 1
Common error: An isolated point where the graphs meet, here at t = 0, satisfies '≥' and must be included. Answers that give only \(t\ge 13.8\) miss it.
254IBMA2H-047-04-052.2 Functions, domain, range and inverse functionsPaper 2Medium7 marks
Calculator
A thermistor in a car engine sensor has resistance \(R\) kΩ at temperature \(T\) kelvin given by \(R(T)=10e^{3500\left(\frac{1}{T}-\frac{1}{298}\right)}\), \(273\le T\le373\). The car's computer uses the inverse function to convert a resistance reading into a temperature.
(a)
Find an expression for \(R^{-1}(x)\).
(3)
(b)
Find the domain of \(R^{-1}\).
(2)
(c)
The sensor reads 4.00 kΩ. Find the engine temperature.
A1 \(R^{-1}(x)=\frac{1}{\frac{\ln(x/10)}{3500}+\frac{1}{298}}\) or equivalent, e.g. \(\frac{1043000}{298\ln(x/10)+3500}\)
✓ 1
Part (b)
M1 evaluates \(R(373)=0.943\) and \(R(273)=29.3\) (as \(R\) is decreasing)
✓ 1
A1 \(0.943\le x\le29.3\)
✓ 1
Part (c)
M1 substitutes \(x=4\) into \(R^{-1}\) (or solves \(R(T)=4\))
✓ 1
A1 \(T=323\) K
✓ 1
Common error: The domain of the inverse is the range of the original function. Because \(R\) decreases, \(R(373)\) gives the lower end, not the upper end.
255IBMA2H-047-04-102.14 Odd and even functions, inverse and self-inverse functionsPaper 2Medium13 marks
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A tank initially holds 200 L of water containing 10 g of dissolved salt. From \(t=0\), brine is pumped in at 6 L per minute, carrying 30 g of salt per minute, and the tank is kept well mixed. Nothing leaves the tank. The concentration of salt, in g L−1, is modelled by \(C(t)=\frac{10+30t}{200+6t}\), where \(t\) is in minutes. The tank holds 500 L, so the model applies for \(0\le t\le50\).
(a)
Find the initial concentration.
(1)
(b)
If the model were extended to all \(t\ge0\), write down the equation of the horizontal asymptote of \(y=C(t)\) and explain its meaning in context.
(2)
(c)
Show that \(C\) is increasing for \(0\le t\le50\).
(3)
(d)
Find an expression for \(C^{-1}(x)\).
(3)
(e)
Find the domain of \(C^{-1}\).
(2)
(f)
Find the time at which the concentration reaches 2.5 g L−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(C(0)=\frac{10}{200}=0.05\) g L−1
✓ 1
Part (b)
A1 \(C=5\)
✓ 1
R1 the concentration would approach that of the incoming brine, \(30\div6=5\) g L−1
R1 the numerator is positive and the denominator is a positive square, so \(C'(t)>0\) and \(C\) is increasing
✓ 1
Part (d)
M1 sets \(x=\frac{10+30t}{200+6t}\) (or interchanges variables) and multiplies out: \(200x+6tx=10+30t\)
✓ 1
M1 collects \(t\) terms: \(t(30-6x)=200x-10\)
✓ 1
A1 \(C^{-1}(x)=\frac{200x-10}{30-6x}\)
✓ 1
Part (e)
M1 \(C(50)=\frac{1510}{500}=3.02\)
✓ 1
A1 \(0.05\le x\le3.02\)
✓ 1
Part (f)
M1 \(C^{-1}(2.5)=\frac{500-10}{30-15}\)
✓ 1
A1 \(t=32.7\) minutes
✓ 1
Exam tip: Showing that \(C\) is increasing justifies that \(C\) is one-to-one, so an inverse exists. The domain of \(C^{-1}\) is the range of \(C\); link these parts in your reasoning.
256IBMA2H-047-04-122.3 Graphs of functions and sketching from a contextPaper 2Medium7 marks
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The period \(T\) hours of a satellite in a circular orbit around the Earth is modelled by \(T(r)=0.0874r^{3/2}\), where \(r\) is the orbital radius measured from the centre of the Earth in thousands of kilometres.
(a)
Find the period of the space station.
(2)
(b)
Find an expression for \(T^{-1}(t)\).
(2)
(c)
Find the orbital radius of the geostationary satellite.
Common error: Raising to the power \(\frac32\) instead of \(\frac23\) when inverting gives an impossible radius; the inverse of \(r^{3/2}\) is \(t^{2/3}\).
257IBMA2H-047-03-062.13 Rational functions with oblique and horizontal asymptotesPaper 2Medium11 marks
Calculator
A small workshop builds custom carbon-fibre bicycle frames. When it builds \(x\) frames in a month, the mean cost per frame, in dollars, is modelled by \(C(x)=\frac{2x^2+30x+4500}{x+10}\), for \(x>0\).
(a)
Show that \(C(x)=2x+10+\frac{4400}{x+10}\).
(2)
(b)
Write down the equation of the oblique asymptote of the graph of \(C\) and interpret it in context.
(2)
(c)
Find the numbers of frames for which the mean cost per frame is less than $200.
(3)
(d)
Find the number of frames that minimises the mean cost per frame, and the corresponding minimum mean cost.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts division or expands \((x+10)(2x+10)=2x^2+30x+100\)
✓ 1
A1 remainder \(4500-100=4400\), giving the required form
✓ 1
Part (b)
A1 \(y=2x+10\)
✓ 1
R1 for large outputs the mean cost per frame increases approximately linearly (by about $2 per extra frame), as the fixed-cost term becomes negligible
✓ 1
Part (c)
M1 solves \(C(x)=200\), e.g. \(x^2-85x+1250=0\) or uses GDC
✓ 1
A1 \(x=18.9\) or \(x=66.1\)
✓ 1
A1 \(19\leq x\leq 66\), \(x\in\mathbb{Z}\)
✓ 1
Part (d)
M1 sets \(C'(x)=2-\frac{4400}{(x+10)^2}=0\) or uses GDC minimum
✓ 1
A1 \(x=36.9\)
✓ 1
R1 compares integers: \(C(36)=177.65\), \(C(37)=177.62\), so 37 frames
✓ 1
A1 minimum mean cost $177.62
✓ 1
Exam tip: When the variable counts whole objects, a non-integer optimum must be checked against the integers either side; the nearest integer is not automatically the best.
258IBMA2H-047-03-082.4 Key features of graphs and intersections using technologyPaper 2Medium9 marks
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After an injection, the concentration of an anaesthetic in a patient's blood, \(C\) mg L−1, \(t\) hours later is modelled by \(C(t)=\frac{20t}{t^2+4}\), for \(t\geq 0\).
(a)
Find the maximum concentration and the time at which it occurs.
(3)
(b)
Hence write down the range of \(C\).
(1)
(c)
Write down the equation of the horizontal asymptote of the graph of \(C\) and interpret it in context.
(2)
(d)
The anaesthetic is effective while \(C\geq 2\). Find the length of time for which it is effective.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 quotient rule: \(C'(t)=\frac{20(t^2+4)-20t(2t)}{(t^2+4)^2}\) or uses GDC
✓ 1
A1 \(C'(t)=0\) when \(t=2\) hours
✓ 1
A1 maximum concentration 5 mg L−1
✓ 1
Part (b)
A1 \(0\leq C\leq 5\)
✓ 1
Part (c)
A1 \(C=0\)
✓ 1
R1 in the long term the anaesthetic is (almost) entirely cleared from the blood
✓ 1
Part (d)
M1 solves \(\frac{20t}{t^2+4}=2\), e.g. \(t^2-10t+4=0\)
✓ 1
A1 \(t=0.417\) and \(t=9.58\)
✓ 1
A1 duration \(=2\sqrt{21}=9.17\) hours
✓ 1
Exam tip: When a numerator has lower degree than the denominator, the horizontal asymptote is \(y=0\). Combining this with the maximum gives the range immediately.
259IBMA2H-047-03-102.8 The reciprocal function and rational functionsPaper 2Medium8 marks
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A camera lens has a focal length of 50 mm. When an object is a distance \(u\) mm from the lens, a sharp image forms at a distance \(v\) mm behind the lens, where \(v(u)=\frac{50u}{u-50}\), for \(u>50\).
(a)
Write down the equations of the asymptotes of the graph of \(v\) and interpret each in context.
(3)
(b)
Find an expression for \(u\) in terms of \(v\), and comment on your answer.
(2)
(c)
The sensor can be positioned between 52 mm and 60 mm behind the lens. Find the range of object distances that can be brought into sharp focus.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertical asymptote \(u=50\) and horizontal asymptote \(v=50\)
✓ 1
R1 as the object approaches the focal point, the image distance becomes arbitrarily large
✓ 1
R1 for very distant objects, the image forms (approximately) 50 mm behind the lens
✓ 1
Part (b)
A1 \(u=\frac{50v}{v-50}\)
✓ 1
R1 this has the same form as \(v(u)\), so the function is self-inverse
✓ 1
Part (c)
M1 substitutes \(v=52\) and \(v=60\) into the inverse
✓ 1
A1 \(u=1300\) and \(u=300\)
✓ 1
A1 \(300\leq u\leq 1300\) mm
✓ 1
Exam tip: \(v\) is a decreasing function of \(u\), so the larger image distance corresponds to the smaller object distance; write the final interval with its endpoints in increasing order.
260IBMA2H-047-06-082.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Medium7 marks
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A footballer kicks a ball from the bottom of a hill. Taking the kicking point as the origin, with \(x\) the horizontal distance in metres, the path of the ball is modelled by \(y=1.2x-0.04x^2\) and the surface of the hill by \(y=3\ln(1+0.1x)\), for \(x\ge0\). Heights \(y\) are in metres.
(a)
Find the horizontal distance from the kicking point to where the ball lands on the hill.
(3)
(b)
Find the height of the landing point above the kicking point.
(1)
(c)
Find the greatest vertical distance between the ball and the surface of the hill during the flight.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up \(1.2x-0.04x^2=3\ln(1+0.1x)\) or equivalent, e.g. \(1.2x-0.04x^2-3\ln(1+0.1x)=0\)
✓ 1
M1 solves using GDC (intersection or zero), choosing the non-zero solution
✓ 1
A1 \(x=26.3\) m
✓ 1
Part (b)
A1 \(y=3\ln(1+0.1\times26.325)=3.87\) m
✓ 1
Part (c)
M1 considers \(d(x)=1.2x-0.04x^2-3\ln(1+0.1x)\) and finds its maximum (by GDC or by solving \(d'(x)=0\))
✓ 1
A1 \(x=13.4\)
✓ 1
A1 maximum vertical distance \(=6.35\) m
✓ 1
Common error: Finding the maximum height of the ball alone (at \(x=15\)) does not answer part (c). The hill is also rising, so you need the maximum of the difference between the two heights.
261IBMA2H-047-06-102.15 Solving inequalities g(x) ≥ f(x)Paper 2Medium10 marks
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An open-topped packaging tray is made from a rectangular sheet of card measuring 30 cm by 20 cm. A square of side \(x\) cm is cut from each corner and the sides are folded up.
Corner squares of side x cm are cut out (dashed). Diagram not to scale
(a)
Show that the volume of the tray is \(V(x)=4x^3-100x^2+600x\) cm3.
(2)
(b)
Write down the set of possible values of \(x\).
(1)
(c)
The tray must hold at least 900 cm3. Find the set of possible values of \(x\).
R1 rejects third root \(x=16.9\) as outside \(0<x<10\), giving \(2.30\le x\le5.78\)
✓ 1
Part (d)
M1 finds maximum of \(V\) (GDC or \(V'(x)=12x^2-200x+600=0\))
✓ 1
A1 \(x=3.92\)
✓ 1
A1 \(V=1060\) cm3
✓ 1
Exam tip: A cubic can have three real roots; check each against the domain from part (b) and state clearly why any is rejected.
262IBMA2H-047-00-072.10 Solving equations graphically and analyticallyPaper 2Medium8 marks
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A garden fountain sends a jet of water from a nozzle 0.5 m above the surface of a pool. The jet reaches a maximum height of 4.5 m above the pool at a horizontal distance of 3.2 m from the nozzle. The height of the jet, y m, at horizontal distance x m from the nozzle is modelled by \(y = a(x - h)^2 + k\).
(a)
Find the value of a.
(2)
(b)
Find the horizontal distance from the nozzle at which the jet hits the pool surface.
(3)
(c)
Find the length of the horizontal interval over which the jet is more than 3 m above the pool.
Common error: In vertex form \(a(x-h)^2 + k\), a vertex at \(x = 3.2\) means \(h = +3.2\), giving \((x - 3.2)^2\), not \((x + 3.2)^2\).
263IBMA2H-047-00-092.6 The quadratic function and its formsPaper 2Medium9 marks
Calculator
A quadratic function is given by \(f(x) = a(x + 1)(x - 5)\), where \(a \ne 0\). The graph of \(y = \frac{1}{f(x)}\) passes through the point \(\left(2, -\frac{1}{18}\right)\).
(a)
Find the value of a and write \(f(x)\) in the form \(px^2 + qx + r\).
(3)
(b)
Write down the coordinates of the local maximum point of \(y = \frac{1}{f(x)}\), justifying your answer.
(2)
(c)
Solve the equation \(\frac{1}{f(x)} = f(x)\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(f(2) = -18\): \(a(3)(-3) = -18\)
✓ 1
A1 \(a = 2\)
✓ 1
A1 \(f(x) = 2x^2 - 8x - 10\)
✓ 1
Part (b)
R1 f has a minimum value of −18 at \(x = 2\) (the vertex), and between the asymptotes \(x=-1\) and \(x=5\) the values of \(f\) lie in \([-18, 0)\), so \(\frac{1}{f}\) is greatest there
✓ 1
A1 \(\left(2, -\frac{1}{18}\right)\)
✓ 1
Part (c)
M1 obtains \([f(x)]^2 = 1\), so \(f(x) = 1\) or \(f(x) = -1\)
✓ 1
A1 \(2x^2 - 8x - 11 = 0 \Rightarrow x = -1.08\) or \(x = 5.08\)
✓ 1
A1 \(2x^2 - 8x - 9 = 0 \Rightarrow x = -0.915\) or \(x = 4.92\)
✓ 1
A1 all four solutions stated: \(x = -1.08, -0.915, 4.92, 5.08\)
✓ 1
Common error: Squaring gives \(f(x) = \pm 1\). Taking only \(f(x) = 1\) loses half the solutions; the curves \(y = f\) and \(y = 1/f\) also meet where both equal −1.
264IBMA2H-047-00-112.10 Solving equations graphically and analyticallyPaper 2Medium7 marks
Calculator
A ground-station satellite dish has a parabolic cross-section modelled by \(y = ax^2\), where x and y are in metres and the vertex is at the origin. The dish has diameter 3.0 m and depth 0.40 m. The receiver must be placed at the focus, a distance \(\frac{1}{4a}\) m above the vertex.
Diagram not to scale
(a)
Find the exact value of a.
(2)
(b)
Find the distance of the receiver above the vertex.
(1)
(c)
A wider dish is designed by applying a horizontal stretch, scale factor 1.2, to the original curve. Write down the equation of the new cross-section and state its diameter and depth.
(2)
(d)
Find the distance of the receiver above the vertex for the new dish.
M1 new coefficient \(\frac{8}{45 \times 1.44}\), so distance \(= \frac{1}{4} \times \frac{45 \times 1.44}{8}\)
✓ 1
A1 2.025 m (2.03 m to 3 s.f.)
✓ 1
Common error: A horizontal stretch by factor 1.2 replaces x with \(\frac{x}{1.2}\), not \(1.2x\). Check: the rim must move from \(x = 1.5\) to \(x = 1.8\).
265IBMA2H-047-07-042.11 Transformations of graphsPaper 2Medium10 marks
Calculator
The concentration of a painkiller in a patient's blood, in mg L−1, t hours after taking a tablet is modelled by \(C(t)=12te^{-0.5t}\), \(t\ge 0\). The drug is effective when its concentration is at least 5 mg L−1.
(a)
Find the maximum concentration and the time at which it occurs.
(2)
(b)
Find the values of t for which \(C(t)=5\).
(3)
(c)
Hence find the length of time for which the drug is effective.
(1)
(d)
A slow-release version of the tablet has concentration \(P(t)=1.5\,C\!\left(\frac{t}{2}\right)\). Describe the two transformations that map the graph of \(y=C(t)\) onto the graph of \(y=P(t)\).
(2)
(e)
Find the maximum concentration for the slow-release version and the time at which it occurs.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(t=2\) hours
✓ 1
A1 maximum \(=8.83\) mg L−1
✓ 1
Part (b)
M1 attempt to solve \(12te^{-0.5t}=5\), e.g. intersection of \(y=C(t)\) and \(y=5\)
✓ 1
A1 \(t=0.548\)
✓ 1
A1 \(t=4.95\)
✓ 1
Part (c)
A1 \(4.9495-0.5480=4.40\) hours
✓ 1
Part (d)
A1 horizontal stretch, scale factor 2
✓ 1
A1 vertical stretch, scale factor 1.5
✓ 1
Part (e)
M1 applies transformations to \((2,\,8.83)\), or uses GDC on \(P\)
✓ 1
A1 maximum \(13.2\) mg L−1 at \(t=4\) hours
✓ 1
Common error: Replacing t by \(\frac{t}{2}\) stretches horizontally by factor 2, not \(\frac{1}{2}\): the peak moves later, from 2 h to 4 h.
266IBMA2H-047-07-062.16 Modulus, reciprocal and other graph transformationsPaper 2Medium10 marks
Calculator
Let \(f(x)=x^3-4x^2+x+6\), for \(x\in\mathbb{R}\).
(a)
Show that \(f(-1)=0\) and hence factorise \(f(x)\) fully.
(3)
(b)
Find the coordinates of the local maximum and local minimum points of the graph of \(y=f(x)\).
(2)
(c)
Solve \(f(|x|)=0\).
(2)
(d)
Find the set of values of k for which the equation \(|f(x)|=k\) has exactly four solutions.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(f(-1)=-1-4-1+6=0\)
✓ 1
M1 division or comparing coefficients to obtain quadratic factor \(x^2-5x+6\)
✓ 1
A1 \(f(x)=(x+1)(x-2)(x-3)\)
✓ 1
Part (b)
A1 local maximum \((0.131,\,6.06)\)
✓ 1
A1 local minimum \((2.54,\,-0.879)\)
✓ 1
Part (c)
M1 recognises only the non-negative zeros of \(f\) are used, reflected in the y-axis
✓ 1
A1 \(x=\pm 2,\ \pm 3\)
✓ 1
Part (d)
M1 reflects the part of the graph with \(f(x)<0\) in the x-axis, giving a local maximum of \(|f|\) at \((2.54,\,0.879)\)
✓ 1
R1 for \(k>0\) each outer branch gives one solution; the hump on \(-1<x<2\) gives two solutions when \(k<6.06\), the hump on \(2<x<3\) gives none when \(k>0.879\)
✓ 1
A1 \(0.879<k<6.06\)
✓ 1
Exam tip: For \(|f(x)|=k\), count intersections branch by branch with a horizontal line; the critical values of k are the heights of the turning points of \(|f|\).
267IBMA2H-047-07-102.4 Key features of graphs and intersections using technologyPaper 2Medium10 marks
Calculator
A phone and a tablet are put on charge at the same time. Their charge levels, as percentages, t minutes later are modelled by Phone: \(B(t)=100-85e^{-0.025t}\) Tablet: \(T(t)=30+0.6t\) for \(0\le t\le 110\).
(a)
Write down \(B(0)\) and the equation of the horizontal asymptote of \(y=B(t)\), interpreting the asymptote in context.
(2)
(b)
State the range of B for \(0\le t\le 110\).
(1)
(c)
Find the values of t at which the two devices have the same charge level.
(3)
(d)
Find the length of time for which the phone has a higher charge level than the tablet.
(1)
(e)
With a fast charger, the phone's charge is given by \(F(t)=B(2t)\). Describe the transformation that maps \(y=B(t)\) onto \(y=F(t)\), and find the time taken for the phone to reach 90% with the fast charger.
(3)
Show mark scheme
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Notes
Part (a)
A1 \(B(0)=15\) (%)
✓ 1
A1 \(B=100\) (accept \(y=100\)): the phone's charge approaches but never reaches 100%
✓ 1
Part (b)
A1 \(15\le B\le 94.6\) (since \(B(110)=94.57\))
✓ 1
Part (c)
M1 attempt to find intersections of \(y=B(t)\) and \(y=T(t)\) using technology
M1 solves \(100-85e^{-0.05t}=90\), i.e. \(e^{-0.05t}=\frac{10}{85}\)
✓ 1
A1 \(t=42.8\) minutes
✓ 1
Common error: Rounding the intersection times before subtracting can change the third significant figure of the duration; subtract the full GDC values.
268IBMA2H-047-02-062.4 Key features of graphs and intersections using technologyPaper 2Medium7 marks
Calculator
The change in velocity \(\Delta v\) m s−1 that a rocket stage can achieve is modelled by \(\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)\). Here \(v_e\) m s−1 is the exhaust speed, \(m_0\) kg is the initial mass including fuel, and \(m_f\) kg is the final mass after the fuel is burned.
(a)
A stage has \(v_e = 3200\), \(m_0 = 550\,000\) and \(m_f = 150\,000\). Find \(\Delta v\).
(2)
(b)
The mission requires \(\Delta v = 4800\). With \(v_e = 3200\) and \(m_0 = 550\,000\) unchanged, find the maximum possible final mass \(m_f\).
(3)
(c)
Engineers instead keep \(m_0 = 550\,000\) and \(m_f = 150\,000\) and redesign the engine. Find the exhaust speed needed to achieve \(\Delta v = 4800\).
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 substitutes: \(\Delta v = 3200\ln\left(\frac{550\,000}{150\,000}\right)\)
Common error: When you rearrange for \(m_f\), the exponent is negative: \(m_f = m_0 e^{-\Delta v/v_e}\). A larger required \(\Delta v\) must give a smaller final mass. Use this to check your answer.
269IBMA2H-047-02-082.2 Functions, domain, range and inverse functionsPaper 2Medium8 marks
Calculator
In a 100 m race, a sprinter's speed \(v\) m s−1 at time \(t\) seconds after the start is modelled by \(v = 11.8\left(1 - e^{-t/\tau}\right)\), \(t \ge 0\), where \(\tau\) is a positive constant. The sprinter reaches a speed of 10 m s−1 when \(t = 2.1\).
(a)
Find the value of \(\tau\).
(3)
(b)
Express \(t\) as a function of \(v\), and state the set of values of \(v\) for which this function is defined in the context.
(3)
(c)
Hence find the time taken for the sprinter to reach 99% of the maximum speed.
M1 \(e^{-2.1/\tau} = \frac{1.8}{11.8}\), so \(\tau = \frac{2.1}{\ln(11.8/1.8)}\)
✓ 1
A1 \(\tau = 1.12\) s (1.116837…)
✓ 1
Part (b)
M1 rearranges to \(e^{-t/\tau} = 1 - \frac{v}{11.8}\)
✓ 1
A1 \(t = \tau\ln\left(\frac{11.8}{11.8 - v}\right)\), with \(\tau \approx 1.12\)
✓ 1
A1 \(0 \le v < 11.8\)
✓ 1
Part (c)
M1 substitutes \(v = 0.99 \times 11.8\) to get \(t = \tau\ln 100\)
✓ 1
A1 \(t = 5.14\) s (5.1432…), using the unrounded value of \(\tau\); accept 5.16 s from \(\tau = 1.12\)
✓ 1
Common error: \(11.8\) is a horizontal asymptote and is never reached, so \(v = 11.8\) must be excluded from the domain of the inverse. Writing \(0 \le v \le 11.8\) loses the mark. Keep \(\tau\) unrounded in your calculator for later parts.
270IBMA2H-047-02-092.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Medium13 marks
Calculator
The profile of a water slide is modelled by \(h(x) = ax^3 + bx^2 + cx + d\), \(0 \le x \le 20\). Here \(h\) metres is the height above the pool deck at a horizontal distance of \(x\) metres from the start. The slide starts horizontally at a height of 12 m when \(x = 0\), and it meets the deck horizontally when \(x = 20\).
Diagram not to scale
(a)
Show that \(d = 12\) and \(c = 0\), and find the value of \(a\) and the value of \(b\).
(4)
(b)
Show that \(h(x) = 0.003(x - 20)^2(x + 10)\), and explain the significance of the repeated factor in context.
(3)
(c)
Find the horizontal distance from the start at which the slide is 6 m above the deck.
(2)
(d)
Find the maximum angle that the slide makes with the horizontal.
A1 multiplies by 0.003 to obtain \(0.003x^3 - 0.09x^2 + 12\)
✓ 1
R1 the repeated root at \(x = 20\) means the curve touches the deck there without crossing it, so the slide meets the deck tangentially
✓ 1
Part (c)
M1 solves \(0.003x^3 - 0.09x^2 + 12 = 6\), i.e. \(x^3 - 30x^2 + 2000 = 0\), for \(0 \le x \le 20\)
✓ 1
A1 \(x = 10\) m
✓ 1
Part (d)
M1 \(h'(x) = 0.009x^2 - 0.18x\)
✓ 1
M1 sets \(h''(x) = 0.018x - 0.18 = 0\), giving \(x = 10\); the gradient at the endpoints is 0
✓ 1
A1 \(h'(10) = -0.9\)
✓ 1
A1 angle \(= \arctan 0.9 = 42.0°\) (41.987…)
✓ 1
Exam tip: "Meets horizontally" gives two conditions at the same point: the height is zero and the gradient is zero. Two conditions at one root signal a repeated factor.
271IBMA2H-047-05-062.4 Key features of graphs and intersections using technologyPaper 2Medium7 marks
Calculator
The charge level of an electric car battery, \(P\) %, is modelled by \(P(t)=\frac{100t+12}{t+0.6}\), \(t \geq 0\), where \(t\) is the time in hours since charging started.
(a)
Write down the equation of the horizontal asymptote of the graph of \(P\) and interpret it in context.
(2)
(b)
Find the time taken for the battery to reach 90% charge.
(2)
(c)
Find the time, in minutes, that the battery takes to charge from 50% to 80%.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(P=100\)
✓ 1
R1 the charge level approaches 100% but never reaches it according to the model
✓ 1
Part (b)
M1 solves \(\frac{100t+12}{t+0.6}=90\), e.g. \(10t=42\)
✓ 1
A1 \(t=4.2\) hours
✓ 1
Part (c)
M1 solves \(P(t)=50\) and \(P(t)=80\)
✓ 1
A1 \(t=0.36\) and \(t=1.8\)
✓ 1
A1 \(1.44\) hours \(=86.4\) minutes
✓ 1
Exam tip: Check the units asked for: the model uses hours, so convert the time difference to minutes at the very end.
272IBMA2H-047-05-092.8 The reciprocal function and rational functionsPaper 2Medium7 marks
Calculator
In a classroom optics kit, a converging lens has focal length 5 cm. When an object is placed \(u\) cm from the lens, a sharp image forms \(v\) cm from the lens, where \(v=\frac{5u}{u-5}\), \(u>5\).
(a)
Find \(v\) when \(u=12\).
(2)
(b)
Write down the equation of the vertical asymptote of the graph of \(v\) against \(u\) and interpret it in context.
(2)
(c)
The magnification is defined as \(\frac{v}{u}\). Find the value of \(u\) for which the magnification is 3.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 substitutes \(u=12\): \(\frac{60}{7}\)
✓ 1
A1 \(v=8.57\) cm
✓ 1
Part (b)
A1 \(u=5\)
✓ 1
R1 as the object moves towards 5 cm from the lens, the image distance increases without bound
✓ 1
Part (c)
M1 \(\frac{v}{u}=\frac{5}{u-5}\)
✓ 1
M1 \(\frac{5}{u-5}=3\)
✓ 1
A1 \(u=\frac{20}{3}=6.67\) cm
✓ 1
Exam tip: Simplify \(\frac{v}{u}\) algebraically before setting it equal to 3; cancelling \(u\) turns the equation into a simple linear one.
273IBMA3H-2-18r7-032.16 Modulus, reciprocal and other graph transformationsPaper 2Medium6 marks
Calculator
The number of new weekly cases of an infection in two towns is modelled, for 0 ≤ t ≤ 30, where t is the number of weeks since monitoring began, by Town A: A(t) = 100e0.07t Town B: B(t) = 500 − 12t.
(a)
Find the value of t for which the two towns have the same number of new weekly cases. Give your answer to 3 significant figures.
(3)
(b)
State, with a reason, which town has more new weekly cases for 20 ≤ t ≤ 30.
(1)
(c)
Find how many more new weekly cases Town A has than Town B when t = 30. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up \(100e^{0.07t}=500-12t\)
✓ 1
M1 solves using a graph or equation solver
✓ 1
A1 \(t=16.0\) weeks
✓ 1
Part (b)
R1 Town A, since A is increasing and B is decreasing, and they are equal at \(t=16.0\)
✓ 1
Part (c)
M1 \(A(30)-B(30)=100e^{2.1}-(500-360)\)
✓ 1
A1 \(676.6\ldots\approx 677\)
✓ 1
Exam tip: Keep unrounded values on your calculator when substituting into later parts.
274IBMA3H-2-18r7-042.16 Modulus, reciprocal and other graph transformationsPaper 2Medium7 marks
Calculator
The function p is defined by p(x) = x3 − 7x2 + 8x + 6, for x ∈ ℝ. Give non-integer answers to 3 significant figures.
(a)
Write down the y-intercept of the graph of y = p(x).
(1)
(b)
Find the zeros of p.
(2)
(c)
Find the coordinates of the local maximum and the local minimum of the graph.
(2)
(d)
Find the set of values of k for which the equation p(x) = k has three distinct real solutions.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 6
✓ 1
Part (b)
M1 uses a graph or solver for \(p(x)=0\)
✓ 1
A1 \(x=-0.508,\ 2.24,\ 5.26\)
✓ 1
Part (c)
A1 local maximum \((0.667,\ 8.52)\)
✓ 1
A1 local minimum \((4,\ -10)\)
✓ 1
Part (d)
R1 the horizontal line \(y=k\) must lie between the local minimum and local maximum values
✓ 1
A1 \(-10<k<8.52\)
✓ 1
Common error: Use strict inequalities here: at k = −10 or k = 8.52 the line touches the graph and gives only two distinct solutions.
275IBMA3H-2-18r7-062.2 Functions, domain, range and inverse functionsPaper 2Medium6 marks
Calculator
The function g is defined by g(x) = x2 − 4x + 7, for x ∈ ℝ.
(a)
Write g(x) in the form (x − h)2 + k.
(2)
(b)
Write down the range of g.
(1)
(c)
The domain of g is now restricted to x ≥ a so that g has an inverse. Write down the smallest possible value of a.
(1)
(d)
With this restricted domain, find g−1(12).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 completes the square
✓ 1
A1 \((x-2)^2+3\)
✓ 1
Part (b)
A1 \(g(x)\ge 3\)
✓ 1
Part (c)
A1 \(a=2\)
✓ 1
Part (d)
M1 solves \(x^2-4x+7=12\) with \(x\ge 2\) (or \(x=2+\sqrt{12-3}\))
✓ 1
A1 \(x=5\) (rejects \(x=-1\))
✓ 1
Common error: Both roots of g(x) = 12 satisfy the equation, but only the one in the restricted domain is the inverse value.
276IBMA3H-2-18r7-112.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Medium6 marks
Calculator
The polynomial p(x) = 2x3 − 5x2 + ax + b leaves a remainder of 7 when divided by (x − 1) and a remainder of −2 when divided by (x + 2).
(a)
Write down two equations connecting a and b.
(2)
(b)
Find the values of a and b.
(2)
(c)
Find the remainder when p(x) is divided by (2x − 1).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 using the remainder theorem at \(x=1\): \(a+b=10\)
✓ 1
A1 using the remainder theorem at \(x=-2\): \(b-2a=34\)
✓ 1
Part (b)
M1 solves the equations simultaneously
✓ 1
A1 \(a=-8\) and \(b=18\)
✓ 1
Part (c)
M1 evaluates p at \(x=\frac{1}{2}\)
✓ 1
A1 \(\frac{1}{4}-\frac{5}{4}-4+18=13\)
✓ 1
Common error: For a divisor (2x − 1), evaluate at x = ½, not at x = 1.
277IBMA3H-2-18r7-122.4 Key features of graphs and intersections using technologyPaper 2Medium7 marks
Calculator
A retailer's online revenue and in-store revenue, both in thousands of euros per month, are modelled for 0 ≤ t ≤ 20, where t is the number of months since the start of the year, by Online: R(t) = 20 × 1.15t In-store: S(t) = 60 + 4t.
(a)
Solve the inequality R(t) ≥ S(t) for 0 ≤ t ≤ 20. Give the critical value to 3 significant figures.
(3)
(b)
Hence find the first whole number of months t for which online revenue is at least in-store revenue. Justify your answer.
(2)
(c)
Find, in thousands of euros to 3 significant figures, by how much online revenue exceeds in-store revenue when t = 20.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up \(20(1.15)^t\ge 60+4t\)
✓ 1
A1 critical value \(t=12.1\) from the intersection of the graphs
✓ 1
A1 \(12.1\le t\le 20\)
✓ 1
Part (b)
A1 \(t=13\)
✓ 1
R1 at \(t=12\) online revenue is about 107.0 and in-store revenue is 108, so online revenue is still lower
✓ 1
Part (c)
M1 \(20(1.15)^{20}-(60+80)\)
✓ 1
A1 \(187\)
✓ 1
Exam tip: Interpret the interval in context: the revenue models only run to month 20, so state the upper limit of the solution.
278IBMA3H-2-19r8-052.16 Modulus, reciprocal and other graph transformationsPaper 2Medium7 marks
Calculator
A bronze sample is removed from a kiln and cools. Its temperature, \(T\) °C, after \(t\) hours is modelled by \(T=20+75e^{-0.12t}\), for \(t\geq 0\).
(a)
Write down the initial temperature of the sample.
(1)
(b)
Sketch the graph of \(T\) against \(t\), labelling the intercept with the \(T\)-axis and any asymptote.
(3)
(c)
Find the time taken for the sample to cool to 50 °C. Give your answer in hours to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(95\) °C
✓ 1
Part (b)
A1 decreasing curve that is concave up (gets flatter) starting on the vertical axis
✓ 1
A1 intercept \((0,95)\) labelled
✓ 1
A1 horizontal asymptote \(T=20\) labelled, curve approaching but never reaching it
✓ 1
Part (c)
M1 \(75e^{-0.12t}=30\) so \(e^{-0.12t}=0.4\)
✓ 1
M1 take natural logarithms: \(-0.12t=\ln 0.4\)
✓ 1
A1 \(t=7.64\) hours
✓ 1
Model sketch
Exam tip: On a sketch, mark the asymptote as a labelled dashed line and make sure the curve flattens towards it rather than crossing it.
279IBMA3H-2-19r8-082.4 Key features of graphs and intersections using technologyPaper 2Medium7 marks
Calculator
A data centre invests \(x\) thousand euros (€) in cooling upgrades, for \(x\geq 0\). The annual saving is \(S(x)=8\ln(x+1)\) thousand euros and the cost is \(C(x)=2x\) thousand euros.
(a)
Use your calculator to find the values of \(x\) at which the graphs of \(y=S(x)\) and \(y=C(x)\) intersect. Give non-zero answers to 3 significant figures.
(2)
(b)
Hence solve \(S(x)\geq C(x)\) and interpret your answer in context.
(2)
(c)
Find the greatest value of \(S(x)-C(x)\). Give your answer to 3 significant figures.
(3)
Show mark scheme
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Mark
Notes
Part (a)
A1 \(x=0\)
✓ 1
A1 \(x=9.35\)
✓ 1
Part (b)
A1 \(0\leq x\leq 9.35\)
✓ 1
R1 the saving is at least the cost for an investment of up to about €9350
✓ 1
Part (c)
M1 differentiate: \(\frac{8}{x+1}-2=0\)
✓ 1
A1 \(x=3\)
✓ 1
A1 \(8\ln 4-6=5.09\)
✓ 1
Exam tip: Check the inequality direction by testing a value inside and outside the interval you found, for example one with \(x\) just above 0 and one beyond the larger root.
280IBMA3H-2-19r8-092.8 The reciprocal function and rational functionsPaper 2Medium6 marks
Calculator
The number of terabytes stored in a data centre is modelled by \(D(t)=500e^{0.04t}\), where \(t\) is the time in months after installation, for \(t\geq 0\).
(a)
Find \(D^{-1}(x)\).
(3)
(b)
Interpret \(D^{-1}(1500)\) in context and find its value to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(x=500e^{0.04t}\) so \(e^{0.04t}=\frac{x}{500}\)
✓ 1
M1 take natural logarithms: \(0.04t=\ln\frac{x}{500}\)
✓ 1
A1 \(D^{-1}(x)=25\ln\frac{x}{500}\)
✓ 1
Part (b)
R1 the number of months after installation at which 1500 terabytes are stored
✓ 1
M1 \(25\ln\frac{1500}{500}=25\ln 3\)
✓ 1
A1 \(27.5\) months
✓ 1
Common error: The inverse of an exponential model is a logarithm: \(\ln\) undoes \(e^{(\ldots)}\), and the coefficient \(0.04\) must be divided out, giving the factor \(25\).
281IBMA3H-2-19r13-012.13 Rational functions with oblique and horizontal asymptotesPaper 2Medium7 marks
Calculator
An archaeologist on a platform tosses a finds bag upwards. The height of the bag above the trench floor is modelled by \(h(t)=1.2+6t-4.9t^2\), where \(h\) is in metres and \(t\ge 0\) is the time in seconds after release. The model applies until the bag reaches the trench floor.
(a)
State the height of the bag at the moment of release.
(1)
(b)
Find the maximum height of the bag above the trench floor, giving your answer to 3 significant figures.
(2)
(c)
Find the time the bag takes to reach the trench floor, giving your answer to 3 significant figures.
(2)
(d)
Describe a sketch of \(h\) against \(t\) for the full flight, giving the coordinates of the intercepts and the maximum.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 1.2 m
✓ 1
Part (b)
M1 attempt to find the vertex / maximum (e.g. \(t=\frac{6}{9.8}\) or technology)
✓ 1
A1 3.04 m
✓ 1
Part (c)
M1 set \(h=0\), i.e. \(4.9t^2-6t-1.2=0\), and attempt to solve
✓ 1
A1 \(t=1.40\) s (reject the negative root)
✓ 1
Part (d)
A1 downward-opening parabola starting at (0, 1.2) and ending on the \(t\)-axis at (1.40, 0)
✓ 1
A1 maximum at about (0.612, 3.04), with the curve not extending below \(h=0\) or to negative \(t\)
✓ 1
Model sketch
Common error: Choosing the negative root of the quadratic. Always check that each solution fits the domain stated in the model.
282IBMA3H-2-19r13-042.4 Key features of graphs and intersections using technologyPaper 2Medium5 marks
Calculator
A company compares two monthly pricing plans for cloud storage. For \(x\) terabytes (TB) stored, with \(0\le x\le 100\), the monthly costs in euros are Plan A: \(A(x)=12+0.8x\) Plan B: \(B(x)=8+12\ln(x+1)\).
(a)
Use technology to find the values of \(x\) for which the two plans cost the same. Give each answer to 3 significant figures.
(3)
(b)
Hence state the amounts of storage, in TB, for which Plan A is cheaper than Plan B. Use the answers from (a) to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equate \(A(x)=B(x)\) or graph both functions
✓ 1
A1 \(x=0.437\)
✓ 1
A1 \(x=55.5\)
✓ 1
Part (b)
M1 identifies the interval between the intersections
✓ 1
A1 \(0.437<x<55.5\)
✓ 1
Exam tip: After finding intersections on a graphing display, check which graph is higher between them by testing a value (for example \(x=10\)) rather than guessing from the picture.
283IBMA3H-2-19r13-052.10 Solving equations graphically and analyticallyPaper 2Medium7 marks
Calculator
A sample from an archaeological site contains carbon-14. The number of carbon-14 decays detected per minute is modelled by \(A(t)=96\times\left(\frac{1}{2}\right)^{t/5730}\), where \(t\) is the time in years since the organism died.
(a)
State the value of \(A(0)\) and the equation of the horizontal asymptote of the graph of \(A\).
(2)
(b)
Describe a sketch of \(A\) for \(t\ge 0\).
(2)
(c)
A sample is measured to give 40 decays per minute. Find the age of the sample, giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(A(0)=96\)
✓ 1
A1 \(A=0\)
✓ 1
Part (b)
A1 decreasing curve starting at (0, 96), concave up
✓ 1
A1 approaches but never reaches the \(t\)-axis
✓ 1
Part (c)
M1 set \(96\times(0.5)^{t/5730}=40\)
✓ 1
M1 attempt to solve using logarithms or technology (e.g. \(t=5730\log_2 2.4\))
✓ 1
A1 \(t=7240\) years
✓ 1
Model sketch
Common error: Giving the asymptote as \(t=0\). A horizontal asymptote is a value of \(A\), so here it is the line \(A=0\).
284IBMA3H-2-19r21-052.11 Transformations of graphsPaper 2Medium7 marks
Calculator
The graph of \(y=g(x)\), where \(g(x)=\dfrac{a}{x-p}+q\), has vertical asymptote \(x=2\) and horizontal asymptote \(y=-3\). The graph passes through the point \((5,0)\).
(a)
Write down the value of \(p\) and the value of \(q\).
(2)
(b)
Find the value of \(a\).
(2)
(c)
Solve \(g(x)=x-1\), giving your answers to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(p=2\)
✓ 1
A1 \(q=-3\)
✓ 1
Part (b)
M1 \(0=\frac{a}{5-2}-3\)
✓ 1
A1 \(a=9\)
✓ 1
Part (c)
M1 \(\frac{9}{x-2}-3=x-1\) so \(\frac{9}{x-2}=x+2\)
✓ 1
A1 \(9=x^2-4\) so \(x^2=13\)
✓ 1
A1 \(x=\pm 3.61\) (3.60555...)
✓ 1
Exam tip: Asymptotes of \(y=\frac{a}{x-p}+q\) are \(x=p\) and \(y=q\); read them off before using a point to find \(a\).
285IBMA3H-2-19r21-062.5 Composite functions and inverse functionsPaper 2Medium10 marks
Calculator
Functions \(f\) and \(g\) are defined by \(f(x)=3e^{x}-2\) for \(x\in\mathbb{R}\) and \(g(x)=\ln(x+5)\) for \(x>-5\).
(a)
Find the range of \(f\).
(2)
(b)
Find an expression for \(f^{-1}(x)\) and state its domain.
(3)
(c)
Show that \((g\circ f)(x)=\ln(3e^{x}+3)\).
(2)
(d)
Solve \((g\circ f)(x)=3\), giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(e^x>0\)
✓ 1
A1 \(f(x)>-2\)
✓ 1
Part (b)
M1 \(y+2=3e^x\) and takes logarithms
✓ 1
A1 \(f^{-1}(x)=\ln\left(\frac{x+2}{3}\right)\)
✓ 1
A1 domain \(x>-2\) (ECF from part (a))
✓ 1
Part (c)
M1 \(g(3e^x-2)=\ln(3e^x-2+5)\)
✓ 1
A1 \(\ln(3e^x+3)\) AG
✓ 1
Part (d)
M1 \(3e^x+3=e^3\)
✓ 1
A1 \(e^x=\frac{e^3-3}{3}=5.695...\)
✓ 1
A1 \(x=1.74\)
✓ 1
Exam tip: Check that the output of the inner function lies in the domain of the outer function before composing.
286IBMA3H-2-19r21-092.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Medium11 marks
Calculator
Two models for the thickness of a lava flow, in metres, at a point \(x\) units from a vent are \(f(x)=x^2+x-2\) and \(g(x)=x^3+x^2-6x+4\), treated as functions for \(x\in\mathbb{R}\).
(a)
Show that \(g(x)-f(x)=x^3-7x+6\).
(1)
(b)
Factorise \(x^3-7x+6\) completely.
(3)
(c)
Hence solve \(g(x)\ge f(x)\).
(4)
(d)
Solve \(g(x)-f(x)\ge 6\), giving your answers to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x^3+x^2-6x+4-x^2-x+2=x^3-7x+6\) AG
✓ 1
Part (b)
M1 shows \(x=1\) is a root since \(1-7+6=0\) (factor theorem)
✓ 1
A1 \(x^3-7x+6=(x-1)(x^2+x-6)\)
✓ 1
A1 \((x-1)(x-2)(x+3)\)
✓ 1
Part (c)
A1 critical values \(x=-3,\ 1,\ 2\)
✓ 1
M1 sign analysis or sketch of a cubic with positive leading coefficient
Exam tip: For a polynomial inequality, find the roots first, then decide the sign in each interval from a quick sketch.
287IBMA3H-2-19r22-042.4 Key features of graphs and intersections using technologyPaper 2Medium6 marks
Calculator
The stock of fish in a protected lake, in tonnes, is modelled by \(S(t)=\dfrac{500}{1+4e^{-0.3t}}\), where \(t\) is the number of years after protection began. A processing plant opens only once the stock reaches 300 tonnes.
(a)
Find the initial stock.
(2)
(b)
State the value that \(S(t)\) approaches as \(t\) becomes very large.
(1)
(c)
Find, to 3 significant figures, the number of years after protection began that the plant opens.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(t=0\) into \(S\)
✓ 1
A1 \(S(0)=\dfrac{500}{5}=100\) tonnes
✓ 1
Part (b)
A1 500 tonnes
✓ 1
Part (c)
M1 set \(\dfrac{500}{1+4e^{-0.3t}}=300\) (or find the intersection with \(y=300\) using technology)
✓ 1
M1 correct manipulation, for example \(e^{-0.3t}=\dfrac{1}{6}\), or correct graph intersection
✓ 1
A1 \(t=5.97\) years
✓ 1
Exam tip: For equations with \(t\) in an exponent you may solve algebraically with logarithms or find the intersection of two graphs on your calculator. Either method earns full marks if the working is shown.
288IBMA3H-2-19r22-062.6 The quadratic function and its formsPaper 2Medium6 marks
Calculator
A fashion label sells scarves. At a price of £\(p\) the weekly demand, in hundreds of scarves, is \(D(p)=200e^{-0.08p}\) and the weekly supply is \(S(p)=4p+20\), also in hundreds of scarves.
(a)
Write down the demand when the scarves are given away free of charge.
(1)
(b)
Find the equilibrium price, where demand equals supply. Give your answer to 3 significant figures.
(3)
(c)
Find the equilibrium quantity, in hundreds of scarves, to 3 significant figures. Use the unrounded value of \(p\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 200 hundred scarves (20 000 scarves)
✓ 1
Part (b)
M1 set \(200e^{-0.08p}=4p+20\) or sketch both graphs
✓ 1
M1 use technology to find the intersection
✓ 1
A1 \(p=12.9\) (£12.90)
✓ 1
Part (c)
M1 substitute the unrounded \(p=12.8648\ldots\) into \(S\) or \(D\)
✓ 1
A1 71.5 hundred scarves
✓ 1
Exam tip: Store the unrounded intersection value in your calculator memory and use it in later parts; rounding too early can change the third significant figure.
289IBMA3H-2-19r22-142.4 Key features of graphs and intersections using technologyPaper 2Medium7 marks
Calculator
A fishing co-operative sends \(x\) boats to sea in a week. Its weekly profit, in hundreds of pounds, is modelled by \(P(x)=-2x^2+36x-100\).
(a)
Find the number of boats that gives the greatest profit, and the value of that profit in pounds.
(2)
(b)
Solve \(P(x)>0\), giving the end points to 3 significant figures.
(3)
(c)
Hence find the number of different whole numbers of boats for which the co-operative makes a profit.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(x=-\dfrac{b}{2a}\) or technology to find the vertex
M1 solve \(P(x)=0\) using technology or the quadratic formula
✓ 1
A1 \(x=3.43\) and \(x=14.6\) (exact \(9\pm\sqrt{31}\))
✓ 1
A1 \(3.43<x<14.6\)
✓ 1
Part (c)
A1 the whole numbers from 4 to 14 inclusive
✓ 1
A1 11
✓ 1
Common error: The number of boats must be a whole number, so list the integers inside the interval rather than rounding the end points.
290IBMA3H-2-20r30-062.2 Functions, domain, range and inverse functionsPaper 2Medium7 marks
Calculator
In a standard tuning, the frequency of a note n semitones above the note A4 is f = 440 × 2n⁄12 hertz (Hz), where n can be negative for notes below A4.
(a)
Show that the frequency of the note 7 semitones above A4 is 659 Hz to 3 significant figures.
(2)
(b)
Find the inverse function, giving n in terms of f.
(3)
(c)
A note has frequency 311 Hz. Find n to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 f = 440 × 27⁄12
✓ 1
A1 659.25… = 659 Hz (AG)
✓ 1
Part (b)
M1 f⁄440 = 2n⁄12
✓ 1
M1 take log2 (or ln) of both sides
✓ 1
A1 n = 12 log2(f⁄440)
✓ 1
Part (c)
M1 n = 12 log2(311⁄440)
✓ 1
A1 n = −6.01
✓ 1
Common error: Writing log₂(f/440) as log₂f − 440 is invalid; keep the quotient together or use log₂f − log₂440.
291IBMA3H-2-20r30-072.4 Key features of graphs and intersections using technologyPaper 2Medium6 marks
Calculator
The mass M kg of a crop on a test plot, t weeks after planting, is modelled by M = a × bt, where a and b are positive constants and b > 1. The mass is 120 kg after 1 week and 405 kg after 4 weeks.
(a)
Find the values of a and b.
(3)
(b)
Find the number of weeks after planting at which the mass first exceeds 2000 kg, giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 divide the two equations: b3 = 405⁄120 = 3.375
✓ 1
A1 b = 1.5
✓ 1
A1 a = 80
✓ 1
Part (b)
M1 80 × 1.5t = 2000, so 1.5t = 25
✓ 1
M1 t = ln 25 ⁄ ln 1.5
✓ 1
A1 t = 7.94 weeks
✓ 1
Exam tip: Dividing two equations of the form a·b^t removes the unknown a and leaves an equation in b alone.
292IBMA3H-2-18r34-062.16 Modulus, reciprocal and other graph transformationsPaper 2Medium8 marks
Calculator
The vertical position of a submersible relative to sea level, in metres (positive upwards), is modelled by \(f(x)=x^2-4x-5\) for \(0\le x\le6\), where x is the time in minutes. The distance of the submersible from sea level is \(g(x)=|f(x)|\).
(a)
Sketch the graph of \(y=g(x)\) for \(0\le x\le6\), labelling the end points, the local maximum and the point where it meets the x-axis.
(4)
(b)
Solve \(|f(x)|=5\) for \(0\le x\le6\). Give non-integer answers to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 starts at \((0,5)\) and increases to a local maximum at \((2,9)\)
✓ 1
A1 then decreases to touch the x-axis at \((5,0)\) with a sharp corner (the part of \(f\) below the axis is reflected)
✓ 1
A1 then increases to the end point \((6,7)\)
✓ 1
A1 curved sections are smooth parabolic arcs and the graph never goes below the x-axis
✓ 1
Part (b)
M1 consider both \(f(x)=5\) and \(f(x)=-5\)
✓ 1
A1 \(f(x)=-5\): \(x^2-4x=0\), so \(x=0\) and \(x=4\)
✓ 1
M1 \(f(x)=5\): \(x^2-4x-10=0\), so \(x=2\pm\sqrt{14}\)
✓ 1
A1 \(x=2+\sqrt{14}=5.74\) (reject \(2-\sqrt{14}<0\)); solutions are \(x=0,\ 4,\ 5.74\)
✓ 1
Model sketch
Exam tip: To solve \(|f(x)|=k\), solve \(f(x)=k\) and \(f(x)=-k\) separately, then check each solution lies in the stated domain.
293IBMA3H-2-18r34-072.8 The reciprocal function and rational functionsPaper 2Medium6 marks
Calculator
The daily handling cost, in thousands of euros, of a logistics depot that processes n hundred pallets is modelled by \(C(n)=0.5(n-4)^2+2\), for \(n\ge4\).
(a)
Write down the range of C.
(1)
(b)
Find \(C^{-1}(x)\).
(3)
(c)
The depot has a daily budget of €10 000. Find the greatest number of pallets it can process in a day.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(C(n)\ge2\)
✓ 1
Part (b)
M1 \(x=0.5(n-4)^2+2\) (swap variables)
✓ 1
M1 \((n-4)^2=2x-4\), taking the positive root since \(n\ge4\)
✓ 1
A1 \(C^{-1}(x)=4+\sqrt{2x-4}\)
✓ 1
Part (c)
A1 \(C^{-1}(10)=4+\sqrt{16}=8\)
✓ 1
R1 800 pallets, since n is measured in hundreds and the cost increases for \(n\ge4\)
✓ 1
Common error: Taking the negative square root when finding the inverse. The domain \(n\ge4\) means only the positive root gives the correct inverse.
294IBMA3H-2-18r34-092.13 Rational functions with oblique and horizontal asymptotesPaper 2Medium9 marks
Calculator
Part of the profile of a curved roof truss is modelled by \(f(x)=\dfrac{2x^2+3x-4}{x-1}\), for \(x\ne1\), where x and \(f(x)\) are in metres.
(a)
Find the equations of the two asymptotes of the graph of \(y=f(x)\).
(3)
(b)
Find the coordinates of the axis intercepts of the graph. Give the x-intercepts to 3 significant figures.
(3)
(c)
Sketch the graph of \(y=f(x)\), showing the asymptotes and intercepts. Describe your sketch.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 vertical asymptote \(x=1\)
✓ 1
M1 divide: \(2x^2+3x-4=(x-1)(2x+5)+1\), so \(f(x)=2x+5+\frac{1}{x-1}\)
✓ 1
A1 oblique asymptote \(y=2x+5\)
✓ 1
Part (b)
A1 y-intercept \((0,4)\)
✓ 1
M1 solve \(2x^2+3x-4=0\) using the quadratic formula
✓ 1
A1 \(x=0.851\) and \(x=-2.35\)
✓ 1
Part (c)
A1 both asymptotes shown as dashed lines: \(x=1\) and \(y=2x+5\)
✓ 1
A1 the branch for \(x>1\) lies above the oblique asymptote, rising to \(+\infty\) as \(x\to1^+\) and approaching \(y=2x+5\) from above as \(x\to\infty\)
✓ 1
A1 the branch for \(x<1\) lies below the oblique asymptote, passes through \((-2.35,0)\), \((0,4)\) and \((0.851,0)\), and falls to \(-\infty\) as \(x\to1^-\)
✓ 1
Model sketch
Exam tip: For a rational function with a quadratic numerator and linear denominator, use polynomial division to find the oblique asymptote. The remainder term tells you which side of the asymptote the curve lies.
295IBMA3H-2-19r34-052.8 The reciprocal function and rational functionsPaper 2Medium7 marks
Calculator
The concentration of a gas in a volcanic plume, in parts per million, is modelled by \(C(x) = \dfrac{60x + 40}{2x + 5}\), where \(x \ge 0\) is the distance in metres downwind from the vent.
(a)
Write down the equations of the horizontal and vertical asymptotes of the graph of \(y = C(x)\).
(2)
(b)
Find the distance from the vent at which the concentration is 25 ppm.
(2)
(c)
Find an expression for \(C^{-1}(x)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 horizontal asymptote \(y = 30\)
✓ 1
A1 vertical asymptote \(x = -2.5\)
✓ 1
Part (b)
M1 \(60x + 40 = 25(2x + 5)\)
✓ 1
A1 \(x = 8.5\) m
✓ 1
Part (c)
M1 write \(y = \frac{60x+40}{2x+5}\) and interchange the variables, or make \(x\) the subject: \(y(2x+5) = 60x + 40\)
Exam tip: For \(\frac{ax+b}{cx+d}\), the horizontal asymptote is \(y = \frac{a}{c}\) and the vertical asymptote is where the denominator is zero.
296IBMA3H-2-19r34-062.2 Functions, domain, range and inverse functionsPaper 2Medium12 marks
Calculator
The illuminance of a stage spotlight, in lux, at a point \(x\) metres along a line across the stage is modelled by \(g(x) = \dfrac{1}{f(x)}\), where \(f(x) = x^2 - 4x + 5\). Treat the model as valid for all real \(x\).
(a)
Write \(f(x)\) in the form \((x-p)^2 + q\).
(2)
(b)
Hence find the coordinates of the maximum point of the graph of \(y = g(x)\).
(2)
(c)
Write down the range of \(g\).
(2)
(d)
The illuminance must be at least 0.25 lux for a performer to be clearly lit. Find the width of the section of the line for which \(g(x) \ge 0.25\). Give your answer to 3 significant figures.
(3)
(e)
Sketch the graph of \(y = g(x)\), describing its key features in words.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to complete the square
✓ 1
A1 \(f(x) = (x-2)^2 + 1\)
✓ 1
Part (b)
M1 the smallest value of \(f\) is 1, so the largest value of \(g\) is \(\frac{1}{1}\)
✓ 1
A1 \((2, 1)\)
✓ 1
Part (c)
A1 \(g(x) > 0\)
✓ 1
A1 \(g(x) \le 1\)
✓ 1
Part (d)
M1 \(g(x) \ge 0.25\) means \(f(x) \le 4\), so \((x-2)^2 \le 3\)
✓ 1
A1 \(2 - \sqrt{3} \le x \le 2 + \sqrt{3}\)
✓ 1
A1 width \(= 2\sqrt{3} = 3.46\) m
✓ 1
Part (e)
A1 bell-shaped curve, symmetric about \(x = 2\), with maximum point \((2, 1)\)
✓ 1
A1 passes through \((0, 0.2)\) (and \((4, 0.2)\))
✓ 1
A1 curve lies above the \(x\)-axis and approaches the horizontal asymptote \(y = 0\) on both sides
✓ 1
Model sketch
Exam tip: The reciprocal of a function turns its minimum into a maximum, and any zero of the denominator would give a vertical asymptote; check whether \(f\) has real zeros.
297IBMA3H-2-19r34-082.13 Rational functions with oblique and horizontal asymptotesPaper 2Medium7 marks
Calculator
A volcano observatory installs \(n\) monitoring sensors, where \(n\) is a positive whole number. The average cost per sensor, in thousands of euros, is modelled by \(A(n) = \dfrac{0.5n^2 + 30n + 400}{n}\).
(a)
Write \(A(n)\) in the form \(pn + q + \dfrac{r}{n}\), where \(p\), \(q\) and \(r\) are constants.
(2)
(b)
Write down the equation of the oblique asymptote of the graph of \(y = A(n)\).
(1)
(c)
Find the number of sensors that minimises the average cost per sensor, and state this minimum average cost. Give the cost to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 divide each term of the numerator by \(n\)
✓ 1
A1 \(A(n) = 0.5n + 30 + \dfrac{400}{n}\)
✓ 1
Part (b)
A1 \(y = 0.5n + 30\)
✓ 1
Part (c)
M1 attempt to minimise, e.g. solve \(A'(n) = 0.5 - \frac{400}{n^2} = 0\) or use a graphing tool
A1 \(n = 28\) and minimum average cost \(=\) €58.3 thousand
✓ 1
Common error: When the variable must be a whole number, finding the turning point is not enough; evaluate the function at the whole numbers either side.
298IBMA3H-2-19r34-102.13 Rational functions with oblique and horizontal asymptotesPaper 2Medium7 marks
Calculator
The fuel cost per nautical mile, in £, of a cargo ship travelling at \(v\) knots is modelled for \(5 \le v \le 25\) by \(F(v) = av^2 + bv + c\). The cost is £8.75 at 5 knots, £7.00 at 10 knots and £8.00 at 20 knots.
(a)
Find the values of \(a\), \(b\) and \(c\).
(4)
(b)
Find the speed that minimises the fuel cost per nautical mile, and the minimum cost. Give both answers to 3 significant figures, using the unrounded speed to find the cost.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use the three data points to write three equations in \(a\), \(b\), \(c\) and attempt to solve
✓ 1
A1 \(a = 0.03\)
✓ 1
A1 \(b = -0.8\)
✓ 1
A1 \(c = 12\)
✓ 1
Part (b)
M1 \(v = -\frac{b}{2a} = \frac{0.8}{0.06}\)
✓ 1
A1 \(v = 13.3\) knots
✓ 1
A1 minimum cost \(=\) £6.67 per nautical mile
✓ 1
Exam tip: Three data points give three simultaneous equations for a quadratic; a graphing calculator's equation solver is the quickest check.
299IBMA3H-2-19r36-092.7 Quadratic equations, inequalities and the discriminantPaper 2Medium6 marks
Calculator
In a role-playing game, the strength of enemies at level x is modelled by f(x) = e0.3x, and the damage a player can deal at level x is modelled by g(x) = 2x + 2, for x ≥ 0.
(a)
Use technology to find the value of x, for x > 0, at which f(x) = g(x). Give your answer to 3 significant figures.
(2)
(b)
Hence solve g(x) ≥ f(x) for x ≥ 0.
(2)
(c)
The levels in the game are whole numbers 1, 2, 3, … Find the highest level at which the player’s damage is at least the enemy’s strength.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 graphs y = e0.3x and y = 2x + 2 (or solves the equation numerically)
✓ 1
A1 x = 10.4
✓ 1
Part (b)
M1 uses g(0) = 2 > f(0) = 1, so the line is above the curve for small x and the curves cross once for x > 0
✓ 1
A1 0 ≤ x ≤ 10.4 (ECF from (a))
✓ 1
Part (c)
M1 selects the greatest integer in the solution set from (b)
✓ 1
A1 level 10 (ECF from (b))
✓ 1
Exam tip: To solve g(x) ≥ f(x) with technology, find the intersection points first, then use the graph to decide which side of them the graph of g lies above the graph of f.
300IBMA3H-2-18r43-042.4 Key features of graphs and intersections using technologyPaper 2Medium5 marks
Calculator
The stock of fish in a managed fishery is modelled by \(S(t)=\dfrac{900t+400}{t+4}\) tonnes, where \(t\ge0\) is the time in years since monitoring began.
(a)
Find the stock at the start of monitoring.
(1)
(b)
Find the equation of the horizontal asymptote of the graph of \(y=S(t)\) and interpret it in context.
(2)
(c)
Find the time at which the stock is 700 tonnes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 100 tonnes
✓ 1
Part (b)
A1 \(y=900\)
✓ 1
A1 in the long term the stock approaches, but does not reach, 900 tonnes
✓ 1
Part (c)
M1 \(900t+400=700(t+4)\)
✓ 1
A1 \(t=12\) years
✓ 1
Exam tip: An asymptote of a model describes long-term behaviour; say what the quantity approaches, in context.
301IBMA3H-2-18r43-072.13 Rational functions with oblique and horizontal asymptotesPaper 2Medium7 marks
Calculator
A fashion designer makes \(n\) scarves. The total cost, in euros, is \(C=1200+8n\). The average cost per scarf is \(A=\dfrac{C}{n}\), for \(n>0\).
(a)
Find the equation of the horizontal asymptote of the graph of \(A\) against \(n\), and state what it means in this context.
(2)
(b)
Find the number of scarves for which the average cost per scarf is €14.
(2)
(c)
Find the smallest number of scarves for which the average cost per scarf is less than €9.50.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(A=8\)
✓ 1
A1 as more scarves are made, the average cost per scarf approaches €8
✓ 1
Part (b)
M1 \(8+\frac{1200}{n}=14\)
✓ 1
A1 \(n=200\)
✓ 1
Part (c)
M1 \(8+\frac{1200}{n}<9.5\), so \(\frac{1200}{n}<1.5\)
✓ 1
A1 \(n>800\)
✓ 1
A1 801 scarves
✓ 1
Common error: The inequality is strict: \(n=800\) gives exactly €9.50, so the smallest whole number that works is 801.
302IBMA3H-2-18r43-102.7 Quadratic equations, inequalities and the discriminantPaper 2Medium7 marks
Calculator
The numbers of fish (in tonnes) in two fish farms at time \(t\) years are \(A(t)=1200-50t\) and \(B(t)=400+\dfrac{3200}{t+1}\), for \(0\le t\le20\).
(a)
Show that \(A(t)=B(t)\) leads to the equation \(t^2-15t+48=0\).
(3)
(b)
Hence find the times at which the two farms hold equal stock. Give your answers to 3 significant figures.
(2)
(c)
Find the stock, to the nearest tonne, at the later of these two times. Use the unrounded value of \(t\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1200-50t=400+\frac{3200}{t+1}\)
✓ 1
M1 multiply by \((t+1)\): \((800-50t)(t+1)=3200\)
✓ 1
A1 \(-50t^2+750t-2400=0\), so \(t^2-15t+48=0\)
✓ 1
Part (b)
A1 \(t=4.63\)
✓ 1
A1 \(t=10.4\)
✓ 1
Part (c)
M1 substitute \(t=\frac{15+\sqrt{33}}{2}\) into \(A(t)\) (or \(B(t)\))
✓ 1
A1 681 tonnes
✓ 1
Exam tip: When a later part says to use the unrounded value, store the earlier answer on your calculator rather than retyping the rounded one.
303IBMA3H-2-20r48-072.4 Key features of graphs and intersections using technologyPaper 2Medium8 marks
Calculator
The number of digital loans made by a library in a year is modelled by \(L(t)=Ae^{kt}\), where \(t\) is the number of years after 1 January 2020. The library had 1200 loans in 2020 (\(t=0\)) and 1950 loans in 2024 (\(t=4\)).
(a)
Find the value of \(A\) and the value of \(k\), giving \(k\) to 3 significant figures.
(3)
(b)
Use the unrounded value of \(k\) to predict the number of loans when \(t=7\), to the nearest whole number.
(2)
(c)
Use the unrounded value of \(k\) to find the value of \(t\) at which the model predicts 5000 loans. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(A=1200\)
✓ 1
M1 \(1950=1200e^{4k}\)
✓ 1
A1 \(k=\frac{\ln 1.625}{4}=0.121\)
✓ 1
Part (b)
M1 \(L(7)=1200e^{7k}\)
✓ 1
A1 \(2807\)
✓ 1
Part (c)
M1 \(5000=1200e^{kt}\)
✓ 1
M1 \(t=\frac{1}{k}\ln\frac{5000}{1200}\)
✓ 1
A1 \(t=11.8\)
✓ 1
Common error: Rounding \(k\) to 3 s.f. before predicting changes the answer noticeably; keep the full value in your calculator.
304IBMA3H-2-20r48-102.1 Equations of straight lines, parallel and perpendicular linesPaper 2Medium8 marks
Calculator
On a map with units in kilometres, wind turbines are at \(A(-2,3)\) and \(B(6,7)\). A cable runs along the perpendicular bisector of \(AB\). A substation \(S\) is on this cable and on the \(x\)-axis.
Units in kilometres
(a)
Find the coordinates of the midpoint of \(AB\).
(1)
(b)
Find the equation of the perpendicular bisector of \(AB\), in the form \(y=mx+c\).
(3)
(c)
Find the coordinates of \(S\).
(2)
(d)
Find the distance \(SA\), correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \((2,5)\)
✓ 1
Part (b)
M1 gradient of \(AB\) is \(\frac{7-3}{6+2}=\frac{1}{2}\)
✓ 1
A1 perpendicular gradient \(-2\)
✓ 1
A1 \(y=-2x+9\)
✓ 1
Part (c)
M1 set \(y=0\) in the equation from (b)
✓ 1
A1 \(S=(4.5,0)\)
✓ 1
Part (d)
M1 \(SA=\sqrt{(4.5+2)^2+(0-3)^2}\)
✓ 1
A1 \(7.16\) km
✓ 1
Exam tip: Every point on a perpendicular bisector is equidistant from the two endpoints, so \(SB\) gives the same distance as \(SA\) and provides a check.
305IBMA3H-2-20r48-112.11 Transformations of graphsPaper 2Medium8 marks
Calculator
The function \(g\) is defined by \(g(x)=2\ln(x-3)+1\), for \(x>3\). The graph of \(y=g(x)\) is obtained from the graph of \(y=\ln x\) by a sequence of three transformations.
(a)
Describe the three transformations.
(3)
(b)
Write down the equation of the vertical asymptote of the graph of \(y=g(x)\).
(1)
(c)
Find \(g^{-1}(x)\).
(2)
(d)
Hence solve \(g(x)=5\). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 translation of 3 units in the positive \(x\)-direction (to the right)
✓ 1
A1 vertical stretch with scale factor 2
✓ 1
A1 translation of 1 unit upwards (applied after the stretch)
✓ 1
Part (b)
A1 \(x=3\)
✓ 1
Part (c)
M1 \(x=2\ln(y-3)+1\), so \(\ln(y-3)=\frac{x-1}{2}\)
✓ 1
A1 \(g^{-1}(x)=3+e^{\frac{x-1}{2}}\)
✓ 1
Part (d)
M1 \(x=g^{-1}(5)=3+e^{2}\) (ECF from (c)) or \(\ln(x-3)=2\)
✓ 1
A1 \(x=10.4\)
✓ 1
Exam tip: The inverse graph is the reflection of \(y=g(x)\) in \(y=x\), so the vertical asymptote of \(g\) becomes the horizontal asymptote \(y=3\) of \(g^{-1}\).
306AAHL2-q00282.10 Solving equations graphically and analyticallyPaper 2Hard15 marks
The depth of water at a quay, \(d\) metres, \(t\) hours after midnight is modelled by \(d(t)=6+2.5\cos\left(\dfrac{\pi t}{6}\right)\), for \(0\le t\le24\).
(a)
Write down the maximum depth of the water and the values of \(t\) at which it occurs.
(2)
(b)
Describe a sequence of transformations that maps the graph of \(y=\cos t\) onto the graph of \(y=d(t)\).
(3)
(c)
Find the first time after midnight at which the depth of the water is 5 m.
(2)
(d)
A boat is moored at the quay while it is loaded. As it is loaded, the depth of water it needs to stay afloat safely, \(r\) metres, is modelled by \(r(t)=4.2+0.15t\), for \(0\le t\le24\). Find the values of \(t\) for which \(d(t)=r(t)\).
(3)
(e)
Hence find the total time between midnight and the end of the day during which the boat can stay at the quay safely.
(3)
(f)
Find the greatest value of \(d(t)-r(t)\) for \(8\lt t\lt 15\), and the value of \(t\) at which it occurs.
(2)
Show mark scheme
Working
Mark
Notes
(a)
8.5 m
A1
\(t=0,\ 12,\ 24\)
A1
(b)
horizontal stretch with scale factor \(\dfrac{6}{\pi}\)
A1
vertical stretch with scale factor 2.5
A1
followed by a vertical translation of 6 units up
A1
The translation must come after the vertical stretch. Accept a translation by \(\begin{pmatrix}0\\6\end{pmatrix}\).
(c)
\(6+2.5\cos\left(\dfrac{\pi t}{6}\right)=5\)
(M1)
\(t=3.79\) hours (about 03:47)
A1
\(3.78593\ldots\)
(d)
graphing \(y=d(t)\) and \(y=r(t)\), or \(y=d(t)-r(t)\)
(M1)
\(t=3.96,\ 8.61\)
A1
\(3.96124\ldots,\ 8.60864\ldots\)
\(t=14.7,\ 22.3\)
A1
\(14.6903\ldots,\ 22.2676\ldots\)
(e)
safe when \(d(t)\ge r(t)\): \(0\le t\le3.96\), \(8.61\le t\le14.7\) and \(22.3\le t\le24\)
Answers: (a) 8.5 m at \(t=0,\ 12,\ 24\) · (c) \(t=3.79\) · (d) \(t=3.96,\ 8.61,\ 14.7,\ 22.3\) · (e) 11.8 hours · (f) 2.52 m at \(t=11.8\)
307AAHL2-q00512.13 Rational functions with oblique and horizontal asymptotesPaper 2Hard16 marks
A straight line \(L\) passes through the point \(\mathrm{P}(2,\,1)\). It meets the positive \(x\)-axis at \(\mathrm{A}(a,\,0)\), where \(a\gt 2\), and the positive \(y\)-axis at B. The point O is the origin.
Diagram not to scale
(a)
Show that the area of triangle OAB is \(S(a)=\dfrac{a^{2}}{2(a-2)}\).
(2)
(b)
Find the values of \(p\), \(q\) and \(r\) such that \(S(a)=pa+q+\dfrac{r}{a-2}\).
(3)
(c)
Hence write down the equations of the asymptotes of the graph of \(y=S(a)\).
(2)
(d)
Find the least possible area of triangle OAB and the value of \(a\) for which it occurs.
(2)
(e)
Find the set of values of \(a\) for which the area of triangle OAB is at most \(5\).
(3)
(f)(i)
For a value \(k\) of the area, with \(k\) greater than the least possible area, there are two possible positions of A, with \(x\)-coordinates \(a_{1}\) and \(a_{2}\).
Show that \(a_{1}\) and \(a_{2}\) are the roots of \(a^{2}-2ka+4k=0\).
(1)
(f)(ii)
Given that \(a_{2}=3a_{1}\), find the value of \(k\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
gradient of \(L\) is \(\dfrac{1-0}{2-a}\), so \(y=-\dfrac{x-a}{a-2}\); or similar triangles
M1
\(y\)-coordinate of B is \(\dfrac{a}{a-2}\), and area \(=\dfrac{1}{2}\times a\times\dfrac{a}{a-2}\)
A1
\(S(a)=\dfrac{a^{2}}{2(a-2)}\)
AG
(b)
\(a^{2}=(a-2)(a+2)+4\), or division of \(a^{2}\) by \(2(a-2)\)
M1
\(p=\dfrac{1}{2},\ q=1\)
A1
\(r=2\)
A1
(c)
\(a=2\)
A1
\(S=\dfrac{a}{2}+1\)
A1
FT from their (b).
(d)
least area \(4\)
A1
Minimum point of the graph of \(y=S(a)\) for \(a\gt 2\).
Intervals \(0\le t\le0.305\) and \(7.14\le t\le10.5\).
3.63 hours
A1
\(3.62712\ldots\)
Answers: (a) \(t=\tfrac{10}{3}\), \(\tfrac{50}{3\mathrm{e}}\approx6.13\) mg l−1 · (b) \(0.752\le t\le9.04\) · (c) \(C\), by about 6.87 min · (d) \(0\lt t\lt 8.54\) · (e) 3.63 h
309IBMA2H-047-01-092.15 Solving inequalities g(x) ≥ f(x)Paper 2Hard12 marks
Calculator
A planetary rover is powered by solar panels. On one day, the power generated, in watts, t hours after sunrise is modelled by \(P(t)=400\sin\left(\frac{\pi t}{12}\right)\), for \(0\le t\le 12\). The power demand of the rover's instruments, in watts, is modelled by \(D(t)=100+1.5t^2\).
(a)
Find \(P(6)\) and \(D(6)\), and state which is larger.
(2)
(b)
Solve \(P(t)\ge D(t)\) for \(0\le t\le 12\).
(4)
(c)
Find the length of time for which the power generated is at least the demand.
(1)
(d)
Find the maximum value of the power surplus \(P(t)-D(t)\) and the time at which it occurs.
(3)
(e)
The demand is redesigned as \(D(t)=c+1.5t^2\), where \(c\) is a constant. Find the greatest value of c for which \(P(t)\ge D(t)\) for at least one value of t.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(P(6)=400\) W and \(D(6)=154\) W
✓ 1
A1 \(P(6)>D(6)\)
✓ 1
Part (b)
M1 attempts to find the intersections of \(P\) and \(D\), e.g. on a GDC
✓ 1
A1 \(t=0.979\)
✓ 1
A1 \(t=9.57\)
✓ 1
A1 \(0.979\le t\le 9.57\)
✓ 1
Part (c)
A1 8.59 hours
✓ 1
Part (d)
M1 attempts to maximise \(P(t)-D(t)\), e.g. with a GDC maximum or by solving \(\frac{d}{dt}(P-D)=0\)
✓ 1
A1 \(t=5.41\) hours
✓ 1
A1 251 W
✓ 1
Part (e)
R1 requires \(c\le\) the maximum value of \(400\sin\left(\frac{\pi t}{12}\right)-1.5t^2\). This equals the maximum surplus found in the previous part plus 100, attained at the same t
✓ 1
A1 \(c=351\) W to 3 s.f. (accept 351.3)
✓ 1
Exam tip: An inequality like \(P(t)\ge c+1.5t^2\) has a solution exactly when c is no more than the maximum of \(P(t)-1.5t^2\). Reusing the maximum surplus you found in the previous part, plus 100, saves time.
310IBMA2H-047-06-122.15 Solving inequalities g(x) ≥ f(x)Paper 2Hard12 marks
Calculator
In a laboratory culture, the number of bacteria \(t\) hours after the start of an experiment is modelled by \(P(t)=\frac{1000}{1+9e^{-0.5t}}\), \(t\ge0\). A simpler model proposed for the same culture is \(E(t)=100e^{0.4t}\), \(t\ge0\).
(a)
Find \(P(0)\).
(1)
(b)
The two models agree at \(t=0\). Find the value of \(t>0\) at which the models next give the same number of bacteria, and state that number.
(3)
(c)
Find \(P^{-1}(x)\), stating its domain in the context of the model.
(4)
(d)
Hence find the time taken for the population to reach 800 bacteria.
(2)
(e)
Compare the long-term behaviour of the two models and state, with a reason, which is more realistic for large \(t\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(P(0)=\frac{1000}{10}=100\)
✓ 1
Part (b)
M1 uses the GDC to find the intersection of \(y=P(t)\) and \(y=E(t)\) for \(t>0\)
✓ 1
A1 \(t=3.02\) hours
✓ 1
A1 335 bacteria
✓ 1
Part (c)
M1 rearranges \(x=\frac{1000}{1+9e^{-0.5t}}\) to \(1+9e^{-0.5t}=\frac{1000}{x}\)
R1 \(E(t)\) increases without bound, whereas \(P(t)\to1000\) as \(t\to\infty\)
✓ 1
R1 \(P\) is more realistic, since limited space and nutrients in the culture cap the population
✓ 1
Exam tip: The domain of \(P^{-1}\) is the range of \(P\). Since \(t\ge0\), the population starts at 100 and never reaches 1000, giving \(100\le x<1000\).
311IBMA2H-047-07-072.8 The reciprocal function and rational functionsPaper 2Hard13 marks
Calculator
A camera lens has focal length 10 cm. When an object is placed u cm from the lens, a sharp image forms v cm from the lens, where \(v=\dfrac{10u}{u-10}\), for \(u>10\).
(a)
Show that \(v=10+\dfrac{100}{u-10}\).
(2)
(b)
Describe a sequence of transformations that maps the graph of \(v=\frac{1}{u}\) onto the graph of \(v=10+\frac{100}{u-10}\).
(3)
(c)
Write down the equations of the asymptotes of the graph and interpret each in context.
(2)
(d)
The total object-to-image distance is \(S=u+v\). Use technology to find the minimum value of S and the value of u at which it occurs.
(2)
(e)
The camera body allows a total distance of less than 50 cm. Solve the inequality \(u+v<50\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(10u=10(u-10)+100\)
✓ 1
A1 divides by \(u-10\) to obtain \(v=10+\frac{100}{u-10}\)
✓ 1
Part (b)
A1 a stretch with scale factor 100: either a vertical stretch parallel to the v-axis, or a horizontal stretch parallel to the u-axis
✓ 1
A1 a translation of 10 units in the positive u-direction
✓ 1
A1 a translation of 10 units in the positive v-direction
✓ 1
Accept a single translation by \(\binom{10}{10}\) for the last two marks.
—
Accept any order that gives the correct final graph. With a vertical stretch, the vertical translation must come after the stretch, or be by \(\frac{1}{10}\) unit if it comes before.
—
Part (c)
A1 \(u=10\) and \(v=10\)
✓ 1
R1 both interpretations are needed for this mark: as the object approaches the focal point, the image distance becomes very large; for a very distant object, the image forms at the focal length, 10 cm
✓ 1
Part (d)
A1 \(u=20\) cm
✓ 1
A1 minimum \(S=40\) cm
✓ 1
Part (e)
M1 forms \(u+\frac{10u}{u-10}=50\), or the inequality
✓ 1
M1 multiplies by \(u-10>0\) and rearranges to \(u^2-50u+500=0\) (or \(<0\))
✓ 1
A1 \(u=25\pm5\sqrt{5}\), i.e. \(13.8\) and \(36.2\)
✓ 1
A1 \(13.8<u<36.2\)
✓ 1
Accept a correct graphical or technology solution that gives the same interval.
—
Exam tip: Multiplying an inequality by \(u-10\) is only safe because the domain guarantees \(u-10>0\). State this to justify keeping the inequality sign.
312IBMA2H-047-02-112.4 Key features of graphs and intersections using technologyPaper 2Hard11 marks
Calculator
A deep-space probe is powered by a radioisotope generator. The radioactive fuel has a half-life of 87.7 years. In addition, the electrical conversion system loses 0.8% of its efficiency each year. The electrical power output, \(P\) watts, \(t\) years after launch is modelled by \(P(t) = 470 \times 2^{-t/87.7} \times 0.992^{t}\).
(a)
Show that \(P(t)\) can be written as \(P(t) = 470e^{-ct}\), and find the value of \(c\) to 3 significant figures.
(3)
(b)
The probe's instruments need at least 300 W. Find the time after launch at which the power first falls below 300 W.
(2)
(c)
Find the time taken for the power output to halve.
(2)
(d)
A second probe, launched at the same time, has an initial output of 520 W. It uses the same fuel, but its conversion system loses 1.5% of its efficiency each year. Find the time after launch at which the two probes have the same power output.
Common error: A loss of 0.8% per year means multiplying by 0.992 each year, not by 0.008. Also, the common half-life factor cancels when you equate the two probes. Leaving it in makes the algebra harder but gives the same answer.
313IBMA3H-2-18r7-072.2 Functions, domain, range and inverse functionsPaper 2Hard7 marks
Calculator
During an epidemic the number of new cases per day, N, in a region is modelled by N(t) = 60te−0.2t, for t ≥ 0, where t is the number of days since the first case was recorded.
(a)
Find the values of t for which N(t) = 80. Give your answers to 4 significant figures.
(3)
(b)
Hence find the length of time, to 3 significant figures, for which there are at least 80 new cases per day. Use unrounded values.
(1)
(c)
Find the value of t at which the number of new cases per day is greatest.
(1)
(d)
Find the range of N for t ≥ 0. Give the upper bound to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets \(60te^{-0.2t}=80\)
✓ 1
M1 solves graphically or numerically for both solutions
✓ 1
A1 \(t=1.982\) and \(t=10.15\)
✓ 1
Part (b)
A1 \(10.1479-1.9819=8.17\) days
✓ 1
Part (c)
A1 \(t=5\)
✓ 1
Part (d)
A1 lower bound 0 (attained at \(t=0\))
✓ 1
A1 upper bound \(300e^{-1}=110\), so \(0\le N\le 110\)
✓ 1
Exam tip: Use the graph to check that there really are two solutions before solving: the line N = 80 must cross the curve once on each side of the maximum.
314IBMA3H-2-19r8-122.3 Graphs of functions and sketching from a contextPaper 2Hard11 marks
Calculator
The proportion of carbon-14 remaining in a bone sample \(t\) years after death is modelled by \(g(t)=2^{-t/5730}\), for \(t\geq 0\).
(a)
Find an expression for \(g^{-1}(p)\), where \(p\) is the proportion remaining.
(3)
(b)
A bone has \(p=0.30\). Find its age, to the nearest 10 years.
(2)
(c)
A laboratory instrument gives a reading \(m\) and the true proportion is \(k(m)=0.9m+0.05\). Find an expression for \((g^{-1}\circ k)(m)\).
(2)
(d)
Find the age, to the nearest 10 years, when the reading is \(m=0.40\).
(2)
(e)
Find the reading \(m\) for which the age is exactly 5730 years.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(p=2^{-t/5730}\), take logarithms to base 2 (or natural logs)
✓ 1
M1 \(\log_2 p=-\frac{t}{5730}\)
✓ 1
A1 \(g^{-1}(p)=-5730\log_2 p\)
✓ 1
Part (b)
M1 \(-5730\log_2 0.30\)
✓ 1
A1 \(9950\) years
✓ 1
Part (c)
M1 substitute \(k(m)\) into \(g^{-1}\)
✓ 1
A1 \(-5730\log_2(0.9m+0.05)\)
✓ 1
Part (d)
M1 \(-5730\log_2(0.41)\)
✓ 1
A1 \(7370\) years
✓ 1
Part (e)
M1 \(\log_2(0.9m+0.05)=-1\) so \(0.9m+0.05=0.5\)
✓ 1
A1 \(m=0.5\)
✓ 1
Exam tip: For a composite such as \(g^{-1}\circ k\), the inner function is applied first, so substitute \(k(m)\) into \(g^{-1}\).
315IBMA3H-2-19r13-142.16 Modulus, reciprocal and other graph transformationsPaper 2Hard11 marks
Calculator
The data throughput of a computing cluster is modelled by \(T(n)=\frac{600n}{n+15}\) Gbps, where \(n\ge 0\) is the number of servers in use (treat \(n\) as a continuous variable).
(a)
Find \(T(0)\) and the equation of the horizontal asymptote of the graph of \(T\).
(2)
(b)
Describe a sketch of \(T\) for \(n\ge 0\).
(2)
(c)
Find the number of servers needed for a throughput of 450 Gbps.
(2)
(d)
Find \(T^{-1}(x)\).
(3)
(e)
State the domain of \(T^{-1}\) in this context.
(1)
(f)
Use \(T^{-1}\) to find the number of servers needed for a throughput of 540 Gbps.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(T(0)=0\)
✓ 1
A1 \(T=600\)
✓ 1
Part (b)
A1 increasing curve from the origin, concave down
✓ 1
A1 approaching the line \(T=600\) without reaching it
✓ 1
Part (c)
M1 set \(\frac{600n}{n+15}=450\)
✓ 1
A1 \(n=45\)
✓ 1
Part (d)
M1 set \(x=\frac{600n}{n+15}\) and multiply out: \(xn+15x=600n\)
✓ 1
M1 collect and factorise: \(n(600-x)=15x\)
✓ 1
A1 \(T^{-1}(x)=\frac{15x}{600-x}\)
✓ 1
Part (e)
A1 \(0\le x<600\)
✓ 1
Part (f)
A1 \(\frac{15\times 540}{600-540}=135\)
✓ 1
Model sketch
Common error: Allowing the domain of \(T^{-1}\) to include 600. The throughput can approach 600 Gbps but never reach it, so 600 is not in the range of \(T\).
316IBMA3H-2-19r21-142.5 Composite functions and inverse functionsPaper 2Hard12 marks
Calculator
A shipping company models the change in a container-loading rate using the function \(h(x)=\dfrac{3x+4}{x+1}\), for \(x\neq -1\).
(a)
Write \(h(x)\) in the form \(a+\dfrac{b}{x+1}\).
(2)
(b)
State the range of \(h\).
(1)
(c)
Find \((h\circ h)(x)\) in the form \(\dfrac{px+q}{rx+s}\), where \(p,q,r,s\) are integers.
(3)
(d)
Find the values of \(x\) that must be excluded from the domain of \(h\circ h\).
(2)
(e)
Solve \((h\circ h)(x)=h(x)\), giving your answers to 3 significant figures.
M1 multiplies numerator and denominator by \((x+1)\)
✓ 1
A1 \(\frac{13x+16}{4x+5}\)
✓ 1
Part (d)
A1 \(x\neq -1\)
✓ 1
A1 \(h(x)\neq -1\) so \(3x+4=-x-1\) gives \(x\neq -\frac{5}{4}\)
✓ 1
Part (e)
M1 \(\frac{13x+16}{4x+5}=\frac{3x+4}{x+1}\)
✓ 1
M1 \((13x+16)(x+1)=(3x+4)(4x+5)\)
✓ 1
A1 \(x^2-2x-4=0\)
✓ 1
A1 \(x=3.24\) or \(x=-1.24\) (neither is excluded in part (d))
✓ 1
Exam tip: After solving, always check each answer against the excluded values of the domain.
317IBMA3H-2-19r22-102.12 Polynomial functions, factor and remainder theorems, sum and product of rootsPaper 2Hard11 marks
Calculator
The deflection, in millimetres, of a robot arm at a distance \(x\) decimetres from its base is modelled by \(d(x)=x^3-9x^2+24x-10\), for \(0\le x\le 6\).
(a)
Find \(d'(x)\) and hence find the values of \(x\) at which the graph of \(d\) has turning points.
(3)
(b)
Find the deflection at each turning point.
(2)
(c)
Hence find the set of values of \(k\) for which the equation \(d(x)=k\) has exactly three solutions in the domain \(0\le x\le 6\).
(2)
(d)
For \(k=8\), show that \(x=3\) is a solution of \(d(x)=k\) and find the other two solutions in exact form.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(d'(x)=3x^2-18x+24\)
✓ 1
M1 set \(d'(x)=0\) and solve (or use technology)
✓ 1
A1 \(x=2\) and \(x=4\)
✓ 1
Part (b)
M1 substitute \(x=2\) and \(x=4\) into \(d\)
✓ 1
A1 \(d(2)=10\) and \(d(4)=6\)
✓ 1
Part (c)
R1 three solutions occur when the horizontal line \(y=k\) lies between the local minimum and local maximum values
✓ 1
A1 \(6<k<10\)
✓ 1
Part (d)
M1 show \(d(3)=27-81+72-10=8\)
✓ 1
M1 factorise \(x^3-9x^2+24x-18=(x-3)(x^2-6x+6)\) or divide by \((x-3)\)
✓ 1
M1 solve \(x^2-6x+6=0\) by formula or completing the square
✓ 1
A1 \(x=3+\sqrt3\) and \(x=3-\sqrt3\)
✓ 1
Exam tip: For 'how many solutions' questions, sketch the cubic using its turning points and slide a horizontal line \(y=k\) up and down the graph. Mention the end points when the domain is restricted.
318IBMA3H-2-20r30-122.16 Modulus, reciprocal and other graph transformationsPaper 2Hard9 marks
Calculator
The functions g and f are defined for x ∈ ℝ by g(x) = |x² − 4x| and f(x) = x + 2.
(a)
Solve x² − 4x = x + 2, giving exact answers.
(2)
(b)
Solve −(x² − 4x) = x + 2.
(2)
(c)
Hence solve g(x) ≥ f(x), giving the endpoints to 3 significant figures where they are not integers.
(3)
(d)
Find the number of integers n with 0 ≤ n ≤ 20 that satisfy g(n) ≥ f(n).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 x² − 5x − 2 = 0
✓ 1
A1 x = (5 ± √33) ⁄ 2
✓ 1
Part (b)
M1 x² − 3x + 2 = 0
✓ 1
A1 x = 1 or x = 2
✓ 1
Part (c)
A1 x ≤ −0.372
✓ 1
A1 1 ≤ x ≤ 2
✓ 1
A1 x ≥ 5.37
✓ 1
Part (d)
M1 integers 1, 2 and 6 to 20 inclusive
✓ 1
A1 17
✓ 1
Exam tip: To solve |A| ≥ B, find where A = B and where −A = B, then test a value in each interval between the solutions.
319IBMA3H-2-20r30-142.1 Equations of straight lines, parallel and perpendicular linesPaper 2Hard10 marks
Calculator
A grower monitors two fungal infections on separate leaves. The areas, in cm², of infection A and infection B after t days are A and B, where ln A = 0.35t + 1.6 and ln B = 0.15t + 3.1.
(a)
Show that A can be written as A = aekt, and find a to 3 significant figures and the value of k.
(3)
(b)
Find the time at which the two infections have equal area.
(2)
(c)
Find the time at which infection A has twice the area of infection B, to 4 significant figures.
(3)
(d)
Find the area of infection A at this time, using the unrounded value of t and giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 A = e0.35t + 1.6 = e1.6 × e0.35t
✓ 1
A1 k = 0.35
✓ 1
A1 a = e1.6 = 4.95
✓ 1
Part (b)
M1 0.35t + 1.6 = 0.15t + 3.1
✓ 1
A1 t = 7.5 days
✓ 1
Part (c)
M1 ln A − ln B = ln 2, so 0.2t − 1.5 = ln 2
✓ 1
M1 t = (1.5 + ln 2) ⁄ 0.2
✓ 1
A1 t = 10.97 days
✓ 1
Part (d)
M1 A = e0.35 × 10.9657… + 1.6
✓ 1
A1 230 cm²
✓ 1
Exam tip: When two quantities are given as logarithms, subtract the logarithms to compare them directly instead of converting each back to an exponential.
320IBMA3H-2-19r34-122.16 Modulus, reciprocal and other graph transformationsPaper 2Hard12 marks
Calculator
The vertical displacement from its resting position, in cm, of a ground sensor on a volcano is modelled by \(d(x) = x^2 - 6x + 5\), where \(x\) is the time in hours and \(0 \le x \le 6\). The size of the displacement is \(|d(x)|\).
(a)
Find the zeros of \(d\), the value of \(|d(0)|\), and the greatest value of \(|d(x)|\) for \(1 \le x \le 5\).
(3)
(b)
Find the set of values of \(k\) for which the equation \(|d(x)| = k\) has exactly four solutions in the interval \(0 \le x \le 6\).
(3)
(c)
For \(k = 4\), find all three solutions of \(|d(x)| = 4\) in the interval. Give your answers in exact form.
(3)
(d)
Hence find the total length of time for which \(|d(x)| \le 4\). Give your answer in exact form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 zeros \(x = 1\) and \(x = 5\)
✓ 1
A1 \(|d(0)| = 5\)
✓ 1
A1 greatest value 4, at \(x = 3\)
✓ 1
Part (b)
M1 consider the horizontal line \(y = k\) meeting the graph of \(y = |d(x)|\)
A1 \(d(x) \ge -4\) always holds, and \(d(x) \le 4\) gives \(3 - 2\sqrt{2} \le x \le 3 + 2\sqrt{2}\)
✓ 1
A1 length \(= 4\sqrt{2}\ (\approx 5.66)\) hours
✓ 1
Exam tip: In an equation \(|f(x)| = k\), the number of solutions changes at the turning points of \(|f|\); sketching \(y = |f(x)|\) first shows where.
321IBMA3H-2-19r34-142.13 Rational functions with oblique and horizontal asymptotesPaper 2Hard13 marks
Calculator
A theatre designer uses the curve \(y = f(x)\), where \(f(x) = \dfrac{x^2 - x + 4}{x - 1}\) and \(x \ne 1\), to shape the edge of a backdrop.
(a)
Show that \(f(x) = x + \dfrac{4}{x - 1}\).
(2)
(b)
Write down the equations of the asymptotes of the graph of \(y = f(x)\).
(2)
(c)
Find the coordinates of the stationary points of the graph.
(3)
(d)
Hence write down the range of \(f\).
(2)
(e)
The straight line \(y = mx\) meets the graph of \(y = f(x)\) for some values of \(m\). Find the set of values of \(m\) for which this happens.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at polynomial division, or write the numerator as \(x(x-1) + 4\)
✓ 1
A1 \(f(x) = x + \dfrac{4}{x-1}\)
✓ 1
Part (b)
A1 vertical asymptote \(x = 1\)
✓ 1
A1 oblique asymptote \(y = x\)
✓ 1
Part (c)
M1 \(f'(x) = 1 - \dfrac{4}{(x-1)^2}\)
✓ 1
M1 set \(f'(x) = 0\): \((x-1)^2 = 4\), so \(x = 3\) or \(x = -1\)
✓ 1
A1 \((3, 5)\) and \((-1, -3)\)
✓ 1
Part (d)
A1 \(f(x) \ge 5\)
✓ 1
A1 or \(f(x) \le -3\)
✓ 1
Part (e)
M1 equate \(\dfrac{x^2 - x + 4}{x-1} = mx\) and multiply by \((x-1)\)
✓ 1
A1 \((1-m)x^2 + (m-1)x + 4 = 0\)
✓ 1
M1 real solutions need discriminant \(\ge 0\): \((m-1)^2 - 16(1-m) = (m-1)(m+15) \ge 0\)
✓ 1
A1 \(m \le -15\) or \(m > 1\) (\(m = 1\) is excluded because the equation becomes \(4 = 0\))
✓ 1
Exam tip: A rational function with a numerator one degree higher than the denominator has an oblique asymptote; polynomial division reveals its equation.
322IBMA3H-2-18r43-112.14 Odd and even functions, inverse and self-inverse functionsPaper 2Hard9 marks
Calculator
The function \(f\) is odd and is defined for all real \(x\). For \(x\ge0\), \(f(x)=x^2-2x\).
(a)
Find an expression for \(f(x)\) when \(x<0\).
(2)
(b)
Show that \(f(0)=0\) for any odd function.
(2)
(c)
Solve \(f(x)=-1.5\). Give your answer to 3 significant figures.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(f(x)=-f(-x)\) with \(-x>0\), so \(f(x)=-\big((-x)^2-2(-x)\big)\)
✓ 1
A1 \(f(x)=-x^2-2x\)
✓ 1
Part (b)
M1 \(f(0)=-f(-0)=-f(0)\)
✓ 1
R1 so \(2f(0)=0\), hence \(f(0)=0\)
✓ 1
Part (c)
M1 for \(x\ge0\): \(x^2-2x=-1.5\)
✓ 1
A1 discriminant \(4-6<0\), so no solution with \(x\ge0\)
✓ 1
M1 for \(x<0\): \(-x^2-2x=-1.5\), i.e. \(x^2+2x-1.5=0\)
✓ 1
A1 \(x=-1\pm\sqrt{2.5}\)
✓ 1
A1 \(x=-2.58\) only, since \(-1+\sqrt{2.5}=0.581\) is not less than 0
✓ 1
Common error: An odd function has two separate formulas here; solve on each domain separately and reject any root outside its own domain.
323IBMA3H-2-18r43-132.6 The quadratic function and its formsPaper 2Hard11 marks
Calculator
A robot's sensor reading is modelled by \(f(x)=\dfrac{8x-3}{2x+1}\), where \(x\ge0\) is the distance, in metres, from a wall.
(a)
Write down the equation of the horizontal asymptote of the graph of \(y=f(x)\).
(1)
(b)
Find an expression for \(f^{-1}(x)\).
(3)
(c)
Find the range of \(f\).
(2)
(d)
Find the distance from the wall at which the reading is 3.5.
(2)
(e)
Find the larger value of \(x\) for which \(f(x)=x\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(y=4\)
✓ 1
Part (b)
M1 \(y(2x+1)=8x-3\) (swap and rearrange)
✓ 1
M1 collect terms in \(x\) and factorise
✓ 1
A1 \(f^{-1}(x)=\dfrac{x+3}{8-2x}\)
✓ 1
Part (c)
A1 \(f(0)=-3\) and \(f\) is increasing
✓ 1
A1 \(-3\le y<4\)
✓ 1
Part (d)
M1 \(f^{-1}(3.5)=\frac{3.5+3}{8-7}\)
✓ 1
A1 6.5 m
✓ 1
Part (e)
M1 \(8x-3=x(2x+1)\)
✓ 1
A1 \(2x^2-7x+3=0\)
✓ 1
A1 \(x=3\) (the other solution is \(x=0.5\))
✓ 1
Exam tip: The range of \(f\) becomes the domain of \(f^{-1}\); the horizontal asymptote of \(f\) shows the value excluded from it.
324IBMA3H-2-20r48-132.4 Key features of graphs and intersections using technologyPaper 2Hard11 marks
Calculator
A library holds \(D(t)\) digital items and \(P(t)\) physical items, \(t\) years after it begins its catalogue review, where \(D(t)=800\times 1.18^t\) and \(P(t)=9200\times 0.97^t\).
(a)
Find the time taken for the number of digital items to double. Give your answer to 3 significant figures.
(3)
(b)
Find the value of \(t\) at which the two collections are the same size. Give your answer to 3 significant figures.
(4)
(c)
Hence find the smallest whole number value of \(t\) for which \(D(t)>P(t)\).
(2)
(d)
Find the number of digital items when \(t=13\), to the nearest 10.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1.18^t=2\)
✓ 1
M1 \(t=\frac{\ln 2}{\ln 1.18}\)
✓ 1
A1 \(4.19\) years
✓ 1
Part (b)
M1 \(800\times 1.18^t=9200\times 0.97^t\)
✓ 1
M1 \(\left(\frac{1.18}{0.97}\right)^t=11.5\)
✓ 1
M1 take logarithms: \(t=\frac{\ln 11.5}{\ln(1.18/0.97)}\)
✓ 1
A1 \(t=12.5\)
✓ 1
Part (c)
R1 \(D\) is increasing and \(P\) is decreasing, so \(D>P\) for \(t>12.46\ldots\)
✓ 1
A1 \(t=13\)
✓ 1
Part (d)
M1 \(800\times 1.18^{13}\)
✓ 1
A1 \(6880\)
✓ 1
Exam tip: Divide both sides by the same base power before taking logs, so the unknown appears in a single exponent.
325IBMA3H-2-20r48-142.14 Odd and even functions, inverse and self-inverse functionsPaper 2Hard12 marks
Calculator
The function \(f\) is defined by \(f(x)=x^2-6x+10\), for \(x\ge 3\). The function \(g\) is defined by \(g(x)=2x-7\), for \(x\in\mathbb{R}\).
(a)
Write \(f(x)\) in the form \((x-h)^2+k\).
(2)
(b)
Write down the range of \(f\).
(1)
(c)
Find \(f^{-1}(x)\) and state its domain.
(4)
(d)
Show that \(g(f(x))=2x^2-12x+13\).
(2)
(e)
Solve \(g(f(x))=5\). Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(h=3\)
✓ 1
A1 \(k=1\)
✓ 1
Part (b)
A1 \(f(x)\ge 1\)
✓ 1
Part (c)
M1 \(y=(x-3)^2+1\) and swap: \((y-3)^2=x-1\)
✓ 1
A1 \(y-3=\pm\sqrt{x-1}\)
✓ 1
R1 positive root chosen because the range of \(f^{-1}\) is \(y\ge 3\), so \(f^{-1}(x)=3+\sqrt{x-1}\)
✓ 1
A1 domain \(x\ge 1\)
✓ 1
Part (d)
M1 \(g(f(x))=2(x^2-6x+10)-7\)
✓ 1
A1 \(2x^2-12x+13\) (AG)
✓ 1
Part (e)
M1 \(2x^2-12x+8=0\), so \(x^2-6x+4=0\)
✓ 1
A1 \(x=3\pm\sqrt{5}\)
✓ 1
A1 \(x=3+\sqrt{5}=5.24\), as \(3-\sqrt{5}<3\) is outside the domain of \(f\)
✓ 1
Common error: When solving a composite equation, check each root against the domain of the inner function; here the smaller root lies outside \(x\ge 3\).
326AAHL2-q00742.15 Solving inequalities g(x) ≥ f(x)Paper 3Hard30 marks
This question investigates when repeated composition of a rational function returns the original input.
Throughout, \(f(x)=\dfrac{ax+b}{cx+d}\), where \(a,b,c,d\in\mathbb{R}\), \(c\neq0\) and \(ad-bc\neq0\), and \(x\) is restricted to values for which every expression is defined.
(a)(i)
Let \(f_{1}(x)=\dfrac{x+2}{x-1}\).
Write down the equations of the asymptotes of the graph of \(y=f_{1}(x)\).
(2)
(a)(ii)
Show that \(f_{1}(f_{1}(x))=x\).
(2)
(a)(iii)
State what the result of part (a)(ii) tells you about the graph of \(y=f_{1}(x)\).
(1)
(a)(iv)
Solve the inequality \(f_{1}(x)\gt x\), giving your answer in exact form.
(3)
(b)
Show that \(f(f(x))=\dfrac{(a^{2}+bc)x+b(a+d)}{c(a+d)x+bc+d^{2}}\).
(3)
(c)(i)
Hence show that if \(a+d=0\) then \(f\) is self-inverse.
(2)
(c)(ii)
By comparing coefficients, show that if \(f\) is self-inverse then \(a+d=0\).
(2)
(d)
Show that when \(a+d=0\) the asymptotes of the graph of \(y=f(x)\) intersect on the line \(y=x\).
(1)
(e)(i)
Let \(f_{2}(x)=\dfrac{x-3}{x+1}\).
Find \(f_{2}(f_{2}(x))\), giving your answer as a single fraction.
(2)
(e)(ii)
Find \(f_{2}^{-1}(x)\).
(2)
(e)(iii)
Hence deduce that \(f_{2}(f_{2}(f_{2}(x)))=x\).
(1)
(f)(i)
Let \(s=a+d\) and \(\Delta=ad-bc\). Suppose that \(s^{2}=\Delta\).
Show that \(a^{2}+bc=-sd\) and \(bc+d^{2}=-sa\).
(3)
(f)(ii)
Hence prove that \(f(f(x))=f^{-1}(x)\), and deduce that \(f(f(f(x)))=x\).
(3)
(g)
Let \(g(x)=\dfrac{kx-7}{x+2}\), where \(k\in\mathbb{R}\) and \(2k+7\neq0\). Use the result of part (f) to find two values of \(k\) for which \(g(g(g(x)))=x\).
\(a^{2}+bc=-(ad-bc)\neq0\), so \(f(f(x))=x\) and \(f\) is self-inverse
R1
hence the result
AG
(c)(ii)
\(f(f(x))=x\) for all valid \(x\) gives \(c(a+d)x^{2}+(d^{2}-a^{2})x-b(a+d)=0\) identically
M1
coefficient of \(x^{2}\): \(c(a+d)=0\) and \(c\neq0\), so \(a+d=0\)
R1
hence the result
AG
(d)
asymptotes \(x=-\dfrac{d}{c}\) and \(y=\dfrac{a}{c}=-\dfrac{d}{c}\), so they meet at \(\left(-\dfrac{d}{c},\,-\dfrac{d}{c}\right)\), which lies on \(y=x\)
327AAHL2-q00752.15 Solving inequalities g(x) ≥ f(x)Paper 3Hard28 marks
This question uses polynomial inequalities to prove that the arithmetic mean of \(n\) positive numbers is never less than their geometric mean.
(a)
By considering \(\left(\sqrt{x}-\sqrt{y}\right)^{2}\), show that \(\dfrac{x+y}{2}\ge\sqrt{xy}\) for \(x,y\ge0\), and state when equality holds.
(3)
(b)(i)
For \(n\in\mathbb{Z}\), \(n\ge2\), let \(P_{n}(x)=x^{n}-nx+n-1\), where \(x\ge0\).
Show that \(P_{2}(x)=(x-1)^{2}\).
(1)
(b)(ii)
Show that \(P_{3}(x)=(x-1)^{2}(x+2)\).
(2)
(b)(iii)
Use your graphic display calculator to sketch the graph of \(y=P_{4}(x)\), for \(0\le x\le2\), and write down the coordinates of its minimum point.
(2)
(b)(iv)
Solve the inequality \(P_{3}(x)\lt 4\), for \(x\ge0\).
(2)
(c)(i)
Show that \(P_{n}(1)=0\) and \(P_{n}'(1)=0\).
(2)
(c)(ii)
By considering the sign of \(P_{n}'(x)\), show that \(P_{n}(x)\ge0\) for all \(x\ge0\), with equality only when \(x=1\).
(3)
(d)
Hence write down an inequality connecting \(x^{n}\) and \(nx-(n-1)\), valid for all \(x\ge0\).
(1)
(e)(i)
Let \(a_{1},a_{2},\ldots,a_{n}\) be positive real numbers and let \(A_{k}=\dfrac{a_{1}+a_{2}+\cdots+a_{k}}{k}\), for \(1\le k\le n\).
Show that \(nA_{n}-(n-1)A_{n-1}=a_{n}\), for \(n\ge2\).
(1)
(e)(ii)
By substituting \(x=\dfrac{A_{n}}{A_{n-1}}\) into the result of part (d), show that \(A_{n}^{\,n}\ge a_{n}A_{n-1}^{\,n-1}\).
(4)
(e)(iii)
Hence prove by mathematical induction that \(A_{n}^{\,n}\ge a_{1}a_{2}\cdots a_{n}\) for all \(n\in\mathbb{Z}^{+}\).
(4)
(f)
A cuboid has edges of length \(x\), \(y\) and \(z\) cm, and the total length of its twelve edges is 48 cm. Use the result of part (e) to find the maximum possible volume of the cuboid and describe the cuboid for which it occurs.
multiplying by \(A_{n-1}^{\,n}\gt 0\): \(A_{n}^{\,n}\ge a_{n}A_{n-1}^{\,n-1}\)
M1
hence the result
AG
(e)(iii)
\(n=1\): \(A_{1}=a_{1}\), so \(A_{1}^{\,1}\ge a_{1}\) is true
A1
assume true for \(n=k\): \(A_{k}^{\,k}\ge a_{1}a_{2}\cdots a_{k}\)
M1
Must be stated as an assumption.
by (e)(ii) with \(n=k+1\): \(A_{k+1}^{\,k+1}\ge a_{k+1}A_{k}^{\,k}\ge a_{k+1}a_{1}a_{2}\cdots a_{k}\)
M1
true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\); hence true for all \(n\in\mathbb{Z}^{+}\)
R1
Award only if the first three marks are awarded.
(f)
\(4(x+y+z)=48\Rightarrow x+y+z=12\), so \(A_{3}=4\) and \(xyz\le4^{3}\)
M1
maximum volume 64 cm3
A1
when \(x=y=z=4\): a cube of edge 4 cm
A1
Answers: (b)(iii) \((1,0)\) · (b)(iv) \(0\le x\lt 2\) · (d) \(x^{n}\ge nx-(n-1)\) · (f) 64 cm3, a cube of edge 4 cm
Model sketch
328AAHL2-q00912.16 Modulus, reciprocal and other graph transformationsPaper 3Hard27 marks
This question investigates functions of the form \(S(x)=\displaystyle\sum_{i=1}^{n}|x-a_{i}|\), where \(a_{1}\lt a_{2}\lt\cdots\lt a_{n}\) are real constants, and uses them to find where a sum of distances is least.
(a)(i)
Let \(S_{2}(x)=|x-1|+|x-5|\).
Express \(S_{2}(x)\) without modulus signs on each of the intervals \(x\lt 1\), \(1\le x\le5\) and \(x\gt 5\).
(3)
(a)(ii)
Sketch the graph of \(y=S_{2}(x)\).
(2)
(a)(iii)
Solve \(S_{2}(x)=6\).
(2)
(b)(i)
Let \(S_{3}(x)=|x-1|+|x-4|+|x-6|\).
Use your graphic display calculator to find the minimum value of \(S_{3}(x)\) and the value of \(x\) at which it occurs.
(2)
(b)(ii)
Explain why the graph of \(y=S_{3}(x)\) consists of straight line segments.
(1)
(c)
Now consider the general function \(S(x)\). Show that, for \(a_{k}\lt x\lt a_{k+1}\), where \(1\le k\le n-1\), \(S'(x)=2k-n\).
(3)
(d)(i)
Hence, show that
if \(n=2m+1\) is odd, \(S(x)\) has a unique minimum at \(x=a_{m+1}\);
(3)
(d)(ii)
if \(n=2m\) is even, \(S(x)\) takes its minimum value at every point of the interval \(a_{m}\le x\le a_{m+1}\).
(2)
(e)(i)
Five friends live beside a straight road, at distances 0, 2, 3, 7 and 12 km from a road junction. They wish to meet at a point on the road at a distance \(x\) km from the junction.
Write down the value of \(x\) that minimizes the total distance travelled by the five friends, and find this minimum total distance.
(2)
(e)(ii)
Find the set of meeting points for which the total distance travelled is at most 20 km.
(4)
(f)(i)
A sixth friend, who lives \(p\) km from the junction, where \(p\gt 12\), joins the group.
Write down the set of meeting points that minimize the total distance travelled by the six friends.
(1)
(f)(ii)
Given that this minimum total distance is 30 km, find \(p\).
(2)
Show mark scheme
Working
Mark
Notes
(a)(i)
\(x\lt 1\): \(S_{2}(x)=6-2x\)
A1
\(1\le x\le5\): \(S_{2}(x)=4\)
A1
\(x\gt 5\): \(S_{2}(x)=2x-6\)
A1
(a)(ii)
flat-bottomed V: horizontal segment \(y=4\) from \((1,\,4)\) to \((5,\,4)\)
A1
lines of gradient \(-2\) and \(2\) on either side, \(y\)-intercept \((0,\,6)\)
A1
(a)(iii)
\(6-2x=6\) or \(2x-6=6\)
M1
\(x=0\) or \(x=6\)
A1
(b)(i)
minimum value 5
A1
at \(x=4\)
A1
(b)(ii)
on each interval between consecutive values 1, 4, 6, every modulus term equals \(\pm(x-a_{i})\), so \(S_{3}\) is linear there
R1
(c)
for \(a_{k}\lt x\lt a_{k+1}\): \(|x-a_{i}|=x-a_{i}\) for \(i\le k\) and \(|x-a_{i}|=a_{i}-x\) for \(i\gt k\)
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