Edexcel International A Level Maths · P1 · Pure Mathematics 1
Differentiation
28 free, high-quality past-paper-style questions on Differentiation for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 256 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P1-6-B-01P1 · ASEasy6 marks
A curve has equation y = 2x5 − 6x2 + 4/x − 9, x ≠ 0.
(a)
Find dy/dx.
(3)
(b)
Find d2y/dx2.
(2)
(c)
Find the gradient of the curve at the point where x = −1.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
4/x written as 4x−1
B1
at least one power reduced by one and multiplied correctly
M1
dy/dx = 10x4 − 12x − 4x−2
A1
Part (b)
differentiates their dy/dx, at least two terms correct ft
M1
d2y/dx2 = 40x3 − 12 + 8x−3
A1
Part (c)
10(−1)4 − 12(−1) − 4(−1)−2 = 10 + 12 − 4 = 18
B1
Common error: Differentiating 4/x as 4. Rewrite it as 4x−1 first: the new power is −2, and the sign changes.
Hence prove, from first principles, that f′(x) = 6x2.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
(x + h)3 = x3 + 3x2h + 3xh2 + h3
M1
2(x3 + 3x2h + 3xh2 + h3) − 2x3 = 6x2h + 6xh2 + 2h3, with no errors seen
A1
Part (b)
[f(x + h) − f(x)]/h = 6x2 + 6xh + 2h2
M1
as h → 0, 6xh + 2h2 → 0
M1
f′(x) = limh→0 (6x2 + 6xh + 2h2) = 6x2, limit notation used correctly and conclusion stated
A1
Exam tip: In a first-principles proof the limit must appear in your working: write limh→0 (or “as h → 0”) on the line where the h terms disappear, and finish with a conclusion.
Figure 1 shows part of the curve C with equation y = x3 − 4x. The curve crosses the x-axis at the point A, at the origin O and at the point B. The point P on C has x-coordinate 1, and the line L is the tangent to C at P.
(a)
Find dy/dx.
(2)
(b)
Show that L has equation y = −x − 2.
(3)
(c)
Find the coordinates of A, and show that L passes through A.
(2)
(d)
Find an equation of the normal to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
at least one term differentiated correctly
M1
dy/dx = 3x2 − 4
A1
Part (b)
y-coordinate of P: 1 − 4 = −3
B1
gradient at P: 3 − 4 = −1, substituting x = 1 into their dy/dx
M1
y + 3 = −(x − 1), so y = −x − 2, as required
A1
Part (c)
x(x2 − 4) = 0, so A is (−2, 0), the root on the negative x-axis in Figure 1
B1
x = −2 in L: y = 2 − 2 = 0, so A lies on L
B1
Part (d)
normal gradient = −1/(−1) = 1 and y + 3 = 1(x − 1)
M1
x − y − 4 = 0 (or any integer multiple)
A1
Common error: Finding the y-coordinate of P by substituting x = 1 into dy/dx. The derivative gives the gradient; the point itself comes from the equation of the curve.
Figure 1 shows the curve C with equation y = x2 + bx + c, where b and c are constants. C crosses the x-axis at the points A(−1, 0) and B(3, 0). The tangents to C at A and at B meet at the point T.
(a)
Use the coordinates of A and B to show that b = −2 and c = −3.
(2)
(b)
Find dy/dx, and hence find the gradient of C at A and at B.
(3)
(c)
Find an equation of each tangent.
(2)
(d)
Find the coordinates of T.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
1 − b + c = 0 and 9 + 3b + c = 0
M1
subtracting: 8 + 4b = 0, so b = −2 and c = −3, as required
A1
Part (b)
dy/dx = 2x − 2
B1
at A: 2(−1) − 2 = −4
A1
at B: 2(3) − 2 = 4
A1
Part (c)
uses y − 0 = m(x − x1) with their gradient at A or B
M1
tangent at A: y = −4x − 4; tangent at B: y = 4x − 12
A1
Part (d)
−4x − 4 = 4x − 12, so x = 1
M1
T is (1, −8)
A1
Exam tip: C is symmetrical about x = 1, halfway between A and B, so the two tangents are mirror images and must meet on that line. T at x = 1 is a quick check on your answer.
The temperature, θ °C, of the water in a solar water heater t hours after sunrise is modelled by θ = 15 + 8t − ½t2, 0 ≤ t ≤ 8. Figure 1 shows the graph of θ against t, together with the tangent to the graph at the point P where t = 2. The points R(1, 23) and S(5, 47) lie on this tangent.
(a)
Use the points R and S to find the gradient of the tangent, and state what this value means in the context of the model.
(2)
(b)
Find dθ/dt, and show that it gives the same value as part (a) when t = 2.
(2)
(c)
Find the temperature of the water at the time when it is increasing at a rate of 2.5 °C per hour.
(3)
(d)
State the time, according to the model, at which the water stops warming up.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
(47 − 23)/(5 − 1) = 6
B1
2 hours after sunrise the temperature of the water is increasing at a rate of 6 °C per hour
B1
Part (b)
dθ/dt = 8 − t
M1
t = 2: 8 − 2 = 6, the same as the gradient of the tangent
A1
Part (c)
8 − t = 2.5
M1
t = 5.5
A1
θ = 15 + 44 − 15.125 = 43.875, so 43.9 °C
A1
Part (d)
dθ/dt = 0 when t = 8: 8 hours after sunrise
B1
Exam tip: A rate of change has units: here °C per hour, because θ is in °C and t is in hours. Include them whenever you interpret a derivative in context.
Air is pumped into a weather balloon. The volume, V cm3, of air in the balloon t seconds after pumping starts is modelled by V = 30t2 − t3, 0 ≤ t ≤ 20.
(a)
Find the rate at which the volume is increasing when t = 5.
(3)
(b)
Find the two times at which the volume is increasing at a rate of 288 cm3 s−1.
(3)
(c)
Find the value of d2V/dt2 when t = 12 and interpret its sign in the context of the question.
negative, so the rate at which the volume is increasing is itself decreasing at t = 12 (the balloon is filling more slowly)
B1
Exam tip: A negative second derivative does not mean the volume is falling. Here dV/dt is still positive at t = 12: the volume is rising, but more and more slowly.
A cubic curve C is defined by y = 2x3 − 9x2 + 7x + 1. At exactly two different points of C the tangent has gradient −5.
(a)
Find dy/dx.
(2)
(b)
Find the coordinates of the two points on C at which the gradient is −5.
(4)
(c)
Find an equation of the tangent to C at the one of these points that lies above the x-axis.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
at least two terms correct
M1
dy/dx = 6x2 − 18x + 7
A1
Part (b)
6x2 − 18x + 7 = −5 ⇒ x2 − 3x + 2 = 0
M1
(x − 1)(x − 2) = 0 ⇒ x = 1, x = 2
A1
substitutes their x values into the equation of C
M1
(1, 1) and (2, −5)
A1
Part (c)
y − 1 = −5(x − 1)
M1
y = −5x + 6
A1
Model sketch
Figure: model sketch: C with the tangent of gradient −5 at (1, 1)
Common error: Substituting the x-values into dy/dx instead of into the equation of C to find the y-coordinates. The derivative gives gradients, not points.
The curve C has equation y = 2x√x − 3x + 5, x > 0. The line L has equation 6x − y + 4 = 0.
(a)
Find dy/dx.
(3)
(b)
The tangent to C at the point P is parallel to L. Find the coordinates of P.
(3)
(c)
Find an equation of the tangent to C at P.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
2x√x = 2x3/2
B1
one term differentiated correctly
M1
dy/dx = 3x1/2 − 3
A1
Part (b)
gradient of L is 6, from y = 6x + 4
B1
3x1/2 − 3 = 6 ⇒ x1/2 = 3 ⇒ x = 9
M1
y = 2 × 27 − 27 + 5: P(9, 32)
A1
Part (c)
y − 32 = 6(x − 9)
M1
y = 6x − 22
A1
Exam tip: Parallel lines have equal gradients. Rearrange the given line into the form y = mx + c to read off m; do not use the 6 in 6x − y + 4 = 0 without checking the coefficient of y.
The function f is defined by f(x) = px3 + qx2 + rx + 5, where p, q and r are constants. Figure 1 shows the graph of the gradient function, y = f′(x). It is a parabola that crosses the y-axis at (0, 10) and the x-axis at (1, 0) and (5, 0).
(a)
Use Figure 1 to show that f′(x) = 2x2 − 12x + 10.
(3)
(b)
Find the values of p, q and r.
(3)
(c)
Find f″(x).
(1)
(d)
Find the exact coordinates of the two points on the curve y = f(x) at which the gradient is −6.
Common error: Reading Figure 1 as the graph of y = f(x). It shows the gradient function, so its x-intercepts are where the gradient of f is zero, not where f(x) = 0.
Figure 1 shows a closed storage box in the shape of a cuboid. Its length, width and height, in centimetres, are marked on Figure 1 in terms of x, where 0 < x < 9. The volume of the box is V cm3.
(a)
Show that V = 36x2 − 4x3.
(2)
(b)
Find dV/dx.
(2)
(c)
Find the value of dV/dx when x = 2 and when x = 7, and explain what the sign of each value tells you about the box.
(3)
(d)
Find the value of x for which dV/dx = 0, and the volume of the box for this value of x.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
V = 2x × x × (18 − 2x), using the three dimensions in Figure 1
M1
= 2x2(18 − 2x) = 36x2 − 4x3, as required
A1
Part (b)
at least one term correct
M1
dV/dx = 72x − 12x2
A1
Part (c)
substitutes x = 2 and x = 7 into their dV/dx
M1
96 and −84
A1
at x = 2 the volume increases as x increases; at x = 7 the volume decreases as x increases
B1
Part (d)
12x(6 − x) = 0
M1
x = 6, since x > 0
A1
V = 36 × 36 − 4 × 216 = 432 cm3
A1
Exam tip: dV/dx measures how fast the volume changes as the width x changes, not as time passes. Its sign tells you whether a slightly wider box holds more or less.
Figure 1 shows the curve C1 with equation y = ax2 and the curve C2 with equation y = b/x2, x > 0, where a and b are constants. The curves meet at the point P(4, 2).
(a)
Find the value of a and the value of b.
(2)
(b)
Show that the tangents to C1 and to C2 at P are perpendicular.
(4)
(c)
Explain why the normal to C1 at P is the same line as the tangent to C2 at P, and find its equation.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
16a = 2, so a = 1/8
B1
b/16 = 2, so b = 32
B1
Part (b)
for C1: dy/dx = x/4
M1
gradient of C1 at P = 1
A1
for C2: y = 32x−2, so dy/dx = −64x−3 and the gradient at P = −64/64 = −1
M1
1 × (−1) = −1, so the tangents are perpendicular
A1
Part (c)
normal to C1 at P has gradient −1/1 = −1
M1
this equals the gradient of the tangent to C2 at P, and both lines pass through P, so they are the same line
B1
y − 2 = −(x − 4): y = 6 − x
A1
Exam tip: Write b/x² as bx⁻² before differentiating. The power goes down to −3, so the derivative is −2bx⁻³, with a change of sign.
Figure 1 shows the curve C with equation y = ½x2 and the point K(0, 5). The normals to C at the points P and Q both pass through K.
(a)
Show that the normal to C at the point with x-coordinate a, where a ≠ 0, has equation x + ay = a + ½a3.
(4)
(b)
Find the exact coordinates of P and Q.
(3)
(c)
Show that the distance KP is exactly 3.
(2)
(d)
There is a third normal to C that passes through K. Write down its equation.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
dy/dx = x, so the gradient of the tangent is a
B1
gradient of the normal = −1/a
M1
y − ½a2 = −(1/a)(x − a)
M1
× a: ay − ½a3 = −x + a, so x + ay = a + ½a3, as required
A1
Part (b)
substitutes (0, 5): 5a = a + ½a3
M1
a ≠ 0, so a2 = 8 and a = ±2√2
A1
P(2√2, 4) and Q(−2√2, 4)
A1
Part (c)
KP2 = (2√2 − 0)2 + (4 − 5)2 = 8 + 1
M1
KP = √9 = 3, as required
A1
Part (d)
the normal at the origin, where the tangent is horizontal: x = 0 (the y-axis)
B1
Exam tip: Working with a general point (a, ½a²) gives one normal equation for every point at once. Substituting the coordinates of K then turns the geometry into an equation in a.
Figure 1 shows the curve C with equation y = √x(x − k), x ≥ 0, where k is a positive constant. C passes through the origin O and the point B, and crosses the x-axis again at A(9, 0).
(a)
Write down the value of k.
(1)
(b)
Show that y = x3/2 − 9x1/2, and hence find dy/dx.
(3)
(c)
Find an equation of the tangent to C at A.
(3)
(d)
At B the tangent to C is parallel to the x-axis. Find the exact coordinates of B.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
√9(9 − k) = 0, so k = 9
B1
Part (b)
√x × x = x3/2 and √x × 9 = 9x1/2
B1
at least one power reduced by one
M1
dy/dx = (3/2)x1/2 − (9/2)x−1/2
A1
Part (c)
substitutes x = 9: (3/2)(3) − (9/2)(1/3)
M1
gradient = 9/2 − 3/2 = 3
A1
y = 3x − 27
A1
Part (d)
(3/2)x1/2 = (9/2)x−1/2, setting their dy/dx = 0
M1
multiplying by 2x1/2: 3x = 9, so x = 3
A1
y = √3(3 − 9): B is (3, −6√3)
A1
Common error: Differentiating √x and (x − 9) separately and multiplying the results, giving (1/2)x^(−1/2) × 1. Multiply out first, then differentiate each power of x.
Figure 1 shows part of the curve C with equation y = x3. The point P(2, 8) lies on C, and Q is the point on C with x-coordinate 2 + h, where h > 0. The chord PQ is drawn, together with the horizontal and vertical distances from P to Q. In Figure 1, h = 1.
(a)
Show that the gradient of the chord PQ is 12 + 6h + h2.
(3)
(b)
Find the gradient of the chord PQ shown in Figure 1.
(1)
(c)
Explain how your answer to part (a) gives the gradient of the tangent to C at P, and state this gradient.
(2)
(d)
The tangent to C at P meets C again at the point S. Find the coordinates of S.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
(2 + h)3 = 8 + 12h + 6h2 + h3
M1
gradient = [(2 + h)3 − 8]/h = (12h + 6h2 + h3)/h
M1
= 12 + 6h + h2, as required
A1
Part (b)
h = 1: 12 + 6 + 1 = 19 (or (27 − 8)/1 = 19)
B1
Part (c)
as h → 0, Q moves towards P and 6h + h2 → 0
M1
gradient of tangent = limh→0 (12 + 6h + h2) = 12
A1
Part (d)
tangent: y − 8 = 12(x − 2), so y = 12x − 16
M1
x3 = 12x − 16, so x3 − 12x + 16 = 0
M1
(x − 2)2(x + 4) = 0, using x = 2 as a repeated root
M1
S is (−4, −64)
A1
Exam tip: A tangent touches the curve, so x = 2 is a repeated root of the cubic: (x − 2)² is a factor. Divide it out to find the other intersection quickly.
The curve C has equation y = 12/x − x, x ≠ 0. The point P(2, 4) lies on C.
Figure: the curve C and the normal at P meeting C again at Q
(a)
Find dy/dx.
(2)
(b)
Show that the normal to C at P has equation x − 4y + 14 = 0.
(4)
(c)
The normal at P meets C again at the point Q. Find the exact coordinates of Q.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
12/x written as 12x−1 and differentiated to kx−2
M1
dy/dx = −12x−2 − 1
A1
Part (b)
gradient of tangent at x = 2: −12/4 − 1 = −4
M1
gradient of normal = ¼
M1
y − 4 = ¼(x − 2)
M1
4y − 16 = x − 2 ⇒ x − 4y + 14 = 0, as required
A1
Part (c)
(x + 14)/4 = 12/x − x, eliminating y
M1
5x2 + 14x − 48 = 0
A1
(x − 2)(5x + 24) = 0, using x = 2 as a known root or solving
M1
Q(−24/5, 23/10)
A1
Exam tip: When a line meets a curve “again”, you already know one solution: here x = 2 at P. Use it as a factor to solve the quadratic quickly, and as a check on your algebra.
The curve C has equation y = ax2 + b/x, x ≠ 0, where a and b are constants. The point P(2, 5) lies on C, and the gradient of C at P is 7/2.
(a)
Show that 8a + b = 10.
(2)
(b)
Find the value of a and the value of b.
(4)
(c)
Find an equation of the tangent to C at P, in the form px + qy + r = 0, where p, q and r are integers.
(2)
(d)
Show that P is the only point on C at which the gradient is 7/2.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
substitutes x = 2, y = 5: 5 = 4a + b/2
M1
× 2: 8a + b = 10, as required
A1
Part (b)
dy/dx = 2ax − bx−2
M1
at x = 2: 4a − b/4 = 7/2, i.e. 16a − b = 14
A1
solves with 8a + b = 10, e.g. adding: 24a = 24
M1
a = 1, b = 2
A1
Part (c)
y − 5 = (7/2)(x − 2)
M1
7x − 2y − 4 = 0 (or any integer multiple)
A1
Part (d)
2x − 2/x2 = 7/2, using their a and b
M1
4x3 − 7x2 − 4 = 0
A1
(x − 2)(4x2 + x + 2) = 0, using x = 2 as a factor
M1
discriminant of 4x2 + x + 2 is 1 − 32 = −31 < 0, so no other real roots: x = 2 is the only solution and P is the only such point
A1
Exam tip: To show that a quadratic factor gives no more solutions, quote its discriminant b2 − 4ac, show it is negative, and write the conclusion in words.
Figure: model sketch: the two tangents to C that pass through the origin
Exam tip: “Hence” means use the general tangent from the previous part. Substituting the origin into it gives an equation in k directly; no new differentiation is needed.
Figure 1 shows part of the curve C with equation y = 8√x − x, x ≥ 0. The point P on C has x-coordinate 4. The tangent to C at P meets the x-axis at T, and the normal to C at P meets the x-axis at N. The triangle TPN is shaded.
(a)
Find dy/dx.
(2)
(b)
Show that the tangent to C at P has equation y = x + 8.
(3)
(c)
Find an equation of the normal to C at P, and the coordinates of N.
(3)
(d)
Find the area of the shaded triangle TPN.
(2)
(e)
Show that the normal to C at P does not meet C at any other point.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
8√x = 8x1/2, differentiated to 4x−1/2
M1
dy/dx = 4x−1/2 − 1
A1
Part (b)
y-coordinate of P: 8 × 2 − 4 = 12
B1
gradient at P: 4/2 − 1 = 1
M1
y − 12 = 1(x − 4), so y = x + 8, as required
A1
Part (c)
normal gradient = −1
M1
y − 12 = −(x − 4): y = 16 − x
A1
y = 0: N is (16, 0)
A1
Part (d)
T is (−8, 0), from x + 8 = 0
B1
area = ½ × (16 − (−8)) × 12 = 144
A1
Part (e)
8√x − x = 16 − x, so 8√x = 16
M1
√x = 2 gives only x = 4, which is P, so there is no other point
A1
Exam tip: In the triangle, the base TN lies along the x-axis, so the height is simply the y-coordinate of P. No distance formula is needed.
A BMX track includes a mound. The cross-section of the mound is modelled by the curve with equation y = px2 − qx3, 0 ≤ x ≤ 6, where x m is the horizontal distance from the start O of the mound, y m is the height above level ground, and p and q are positive constants. Figure 1 shows the cross-section. The top of the mound is the point T(4, 4), where the gradient of the curve is zero.
(a)
Show that p = 3/4 and q = 1/8.
(4)
(b)
Find the angle between the surface of the mound and the horizontal at the point where x = 1.
(2)
(c)
There is one other point on the mound at which the surface makes the same angle with the horizontal as at x = 1. Find the value of x at this point and the height of the mound there.
(3)
(d)
Find the exact value of x at which the gradient of the mound is −3/2.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
T on the curve: 16p − 64q = 4
M1
dy/dx = 2px − 3qx2
M1
gradient zero at x = 4: 8p − 48q = 0, so p = 6q
M1
96q − 64q = 4, so q = 1/8 and p = 3/4, as required
A1
Part (b)
gradient at x = 1: 3/2 − 3/8 = 9/8
M1
angle = tan−1(9/8) = 48.4°
A1
Part (c)
(3/2)x − (3/8)x2 = 9/8, so x2 − 4x + 3 = 0
M1
(x − 1)(x − 3) = 0, so x = 3
A1
y = 27/4 − 27/8 = 27/8: height 3.375 m (accept 3.38 m)
A1
Part (d)
(3/2)x − (3/8)x2 = −3/2, so x2 − 4x − 4 = 0
M1
x = 2 ± 2√2
M1
2 − 2√2 < 0, so x = 2 + 2√2 (≈ 4.83, within 0 ≤ x ≤ 6)
A1
Exam tip: A point where the gradient is zero gives two equations: the point lies on the curve, and dy/dx = 0 there. Use both to find two unknown constants.
Figure 1 shows the curve C1 with equation y = x2 and the curve C2 with equation y = −x2 + px + q, where p and q are constants. C2 passes through the points V(4, 6) and W(6, 2). The straight lines L1 and L2 are each a tangent to both C1 and C2.
(a)
Show that p = 8 and q = −10.
(2)
(b)
Show that the tangent to C1 at the point where x = a has equation y = 2ax − a2.
(2)
(c)
Show that the tangent to C2 at the point where x = b has equation y = (8 − 2b)x + b2 − 10.
(3)
(d)
Hence find the equations of L1 and L2.
(4)
(e)
Find the coordinates of the point where L1 and L2 meet.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
−16 + 4p + q = 6 and −36 + 6p + q = 2
M1
subtracting: 2p − 20 = −4, so p = 8 and q = −10, as required
Exam tip: Two lines y = mx + c are the same line only if both the gradients and the intercepts are equal. Matching both gives two equations in the two unknown points of contact.
Figure 1 shows part of the curve C with equation y = x4 − 4x3 − 2x2 + 20x + 1. The points A and B on C have x-coordinates −1 and 3 respectively. The straight line L is the tangent to C at A.
(a)
Find dy/dx.
(2)
(b)
Find an equation of L.
(3)
(c)
Show that L is also the tangent to C at B.
(3)
(d)
Show that x4 − 4x3 − 2x2 + 20x + 1 − (8x − 8) = (x + 1)2(x − 3)2, and hence explain why no part of C lies below L.
(3)
(e)
Find d2y/dx2, and show that it takes the same value at A and at B.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
at least two terms correct
M1
dy/dx = 4x3 − 12x2 − 4x + 20
A1
Part (b)
y-coordinate of A: 1 + 4 − 2 − 20 + 1 = −16
B1
gradient at A: −4 − 12 + 4 + 20 = 8
M1
y + 16 = 8(x + 1): y = 8x − 8
A1
Part (c)
y-coordinate of B: 81 − 108 − 18 + 60 + 1 = 16, and 8(3) − 8 = 16, so B lies on L
B1
gradient at B: 108 − 108 − 12 + 20 = 8
M1
B is on L and C has the same gradient as L at B, so L is the tangent at B
left-hand side = x4 − 4x3 − 2x2 + 12x + 9, the same, as required
A1
a square is never negative, so the curve's y is never less than the line's y: C never goes below L (it touches L only at A and B)
B1
Part (e)
d2y/dx2 = 12x2 − 24x − 4
B1
at A: 12 + 24 − 4 = 32; at B: 108 − 72 − 4 = 32, the same
B1
Exam tip: A line is a tangent at a point when two things hold there: the point lies on both, and the gradients are equal. Check both to show that one line touches a curve twice.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.