Original questions in the style of Papers 1, 2 and 3, each with a full worked mark scheme. Tick questions to build a printable test. Topic 2 at Standard Level →
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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 14 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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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 \(x=2\): \(1\lt 3\)
26AAHL2-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.
28AAHL2-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.
29AAHL2-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) \(3x+2\)
30AAHL2-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 \(x=1\)
32AAHL2-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 \(=-\dfrac{2}{-\frac{1}{2}}=4\)
33AAHL2-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 \((2,\,0)\)
34AAHL2-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 \(f\).
35AAHL2-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 \(-4\le x\le0\)).
36AAHL2-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
37AAHL2-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) \(x=2\)
38AAHL2-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 \(1\lt x\le1+\sqrt{5}\).
40AAHL2-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 \(0\lt x\lt 10\).
(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) \(10-5\sqrt{2}\le x\le5\)
41AAHL2-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 \(x\gt\tfrac{c}{2}\)
42AAHL2-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 \(-4\lt -\sqrt{k}\lt 0\)
48AAHL2-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 \(L\).
(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
49AAHL2-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) \(x=\ln2\)
51AAHL2-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.
(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 \(k=2\)
54AAHL2-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.
56AAHL2-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 \(\dfrac{1}{\alpha},\ldots,\dfrac{1}{\delta}\)
57AAHL2-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 \(k\).
58AAHL2-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 \((1,1)\)
59AAHL2-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 \(u=\mathrm{e}^{x}\)
60AAHL2-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 \(A(x)=E(x)\).
61AAHL2-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 \([-1,1]\).
62AAHL2-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 \(x=2\)
63AAHL2-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 \(k\gt 3\)
64AAHL2-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 \(x\to\infty\)
Answers: (a) 300 · (b) \(N=6000\) · (c) 12.9 years · (d) 7.36 years
67AAHL2-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.
68AAHL2-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 \(x\lt 0.279\)
69AAHL2-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 \(y=10\)
73AAHL2-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 \(x\gt 4\)
74AAHL2-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.
75AAHL2-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 \(127.417\ldots-(F-150)\)
76AAHL2-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.
77AAHL2-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) \(5.39\le x\le8.73\)
78AAHL2-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) \(x^{2}-10x+50\)
79AAHL2-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 \(0\lt k\lt 1.04\)
80AAHL2-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 \(0.1\).
82AAHL2-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. \(P(12)-96=-4.96\), \(P(13)-104=3.16\).
84AAHL2-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) \(k=1\)
85AAHL2-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
86AAHL2-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 \(t=11.8\)
87AAHL2-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.
(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
89AAHL2-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 \(y=x\)
90AAHL2-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
91AAHL2-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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