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IG.M1 Differentiation: Gradients and Turning Points

EDEXCEL 4MA1 · Calculator allowed · about 80 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

y = 9. Write down the value of dy/dx.

(Total for Question 1 is 1 mark)

y = x⁵. Find dy/dx.

(Total for Question 2 is 1 mark)

y = -3x². Find dy/dx.

(Total for Question 3 is 1 mark)

y = 6x³ - 4x + 1. Find dy/dx.

(Total for Question 4 is 2 marks)

y = 5x² - 7x - 2. Find dy/dx.

(Total for Question 5 is 2 marks)

y = 2x³ + 5x². Find d²y/dx², the second derivative of y with respect to x.

(Total for Question 6 is 2 marks)

Find the gradient of the curve y = x³ - 4x at the point where x = 2.

(Total for Question 7 is 2 marks)

The curve y = 2x² - 8x + 3 has gradient 4 at the point P. Find the x-coordinate of P.

(Total for Question 8 is 3 marks)

Find the coordinates of the point on the curve y = x² - 6x + 5 at which the gradient is zero.

(Total for Question 9 is 3 marks)

y = x³ - 3x² - 9x + 5

(a)Find dy/dx.(2)

(b)Find the x-coordinates of the turning points of the curve.(3)

(Total for Question 10 is 5 marks)

This question continues from Question 10, for the curve y = x³ - 3x² - 9x + 5, which has turning points at x = -1 and x = 3.

(a)Find the y-coordinate of each turning point.(2)

(b)By finding d²y/dx², determine whether each turning point is a maximum or a minimum.(3)

(Total for Question 11 is 5 marks)

A particle moves in a straight line so that its displacement, s metres, from a fixed point O at time t seconds is given by s = t³ - 6t² + 9t, for t ≥ 0.

(a)Find an expression for the velocity, v m/s, of the particle at time t.(2)

(b)Find the velocity of the particle when t = 4.(2)

(c)Find the values of t at which the particle is instantaneously at rest.(3)

(Total for Question 12 is 7 marks)

For the particle described in Question 12, with displacement s = t³ - 6t² + 9t, find the acceleration of the particle when t = 1, and state whether the particle is speeding up or slowing down at this instant.

(Total for Question 13 is 3 marks)

A farmer has 40 m of fencing. She wants to fence off a rectangular area for sheep, using a straight wall as one side of the rectangle, so fencing is only needed for the other three sides. She uses x metres of fencing for each of the two sides perpendicular to the wall, and the rest of the fencing for the side parallel to the wall. Let A m² be the area enclosed.

Figure (to be drawn): A rectangle representing the sheep enclosure. The top side lies along a straight wall (drawn as a thick line, not fenced). The two vertical sides, each labelled x metres, are perpendicular to the wall. The bottom side, parallel to the wall, is labelled (40 - 2x) metres.

(a)Show that A = 40x - 2x².(2)

(b)Find dA/dx.(2)

(c)Find the value of x that gives the maximum possible enclosed area, and find this maximum area.(3)

(d)Explain how you know that this value of x gives a maximum area, not a minimum area.(1)

(Total for Question 14 is 8 marks)

y = x⁴ - 8x² + 3

(a)Find dy/dx.(2)

(b)Find the x-coordinates of all the stationary points of the curve.(4)

(Total for Question 15 is 6 marks)

Find the value(s) of x for which the tangent to the curve y = x³ - 2x² + 1 is parallel to the line y = 4x - 5.

(Total for Question 16 is 4 marks)

Show that the curve y = x³ - 3x² + 3x + 5 has exactly one stationary point, and determine its nature.

(Total for Question 17 is 4 marks)

The number of bacteria, N, in a laboratory culture, t hours after the start of an experiment, is modelled by N = 200 + 30t² - 2t³, for 0 ≤ t ≤ 10.

(a)Find dN/dt.(2)

(b)Find the rate at which the number of bacteria is increasing when t = 3.(2)

(c)Find the value of t, for 0 ≤ t ≤ 10, at which the rate of increase of the number of bacteria is at its greatest.(3)

(Total for Question 18 is 7 marks)

Mark scheme · IG.M1 Differentiation: Gradients and Turning Points

Question 1

B1 dy/dx = 0 cao
Answer: dy/dx = 0

Question 2

B1 dy/dx = 5x⁴ cao
Answer: dy/dx = 5x⁴

Question 3

B1 dy/dx = -6x cao
Answer: dy/dx = -6x

Question 4

M1 differentiates at least two terms correctly, e.g. 18x² or -4 seen
A1 dy/dx = 18x² - 4 cao (the constant +1 differentiates to 0)
Answer: dy/dx = 18x² - 4

Question 5

M1 differentiates at least two terms correctly, e.g. 10x or -7 seen
A1 dy/dx = 10x - 7 cao
Answer: dy/dx = 10x - 7

Question 6

M1 finds dy/dx = 6x² + 10x (differentiating once)
A1 d²y/dx² = 12x + 10 cao (differentiating a second time)
Answer: d²y/dx² = 12x + 10

Question 7

M1 dy/dx = 3x² - 4 found
A1 gradient = 8 cao, from correctly substituting x = 2
Answer: Gradient = 8

Question 8

M1 dy/dx = 4x - 8 found
M1 sets 4x - 8 = 4 and rearranges, e.g. 4x = 12
A1 x = 3 cao
Answer: x = 3

Question 9

M1 dy/dx = 2x - 6 found and set equal to 0
A1 x = 3 cao
A1 y = -4 cao, so the point is (3, -4), ft from their x
Answer: (3, -4)

Question 10

(a) M1 differentiates at least two terms correctly
(a) A1 dy/dx = 3x² - 6x - 9 cao
(a) Answer: dy/dx = 3x² - 6x - 9
(b) M1 sets their dy/dx = 0 and simplifies to a 3-term quadratic equal to zero, e.g. x² - 2x - 3 = 0, ft from part (a)
(b) M1 factorises or solves the quadratic correctly, e.g. (x - 3)(x + 1) = 0
(b) A1 x = 3 and x = -1 cao (both values)
(b) Answer: x = -1 and x = 3

Question 11

(a) M1 substitutes x = -1 and x = 3 into y = x³ - 3x² - 9x + 5
(a) A1 y = 10 at x = -1 and y = -22 at x = 3, both correct, cao
(a) Answer: (-1, 10) and (3, -22)
(b) M1 finds d²y/dx² = 6x - 6
(b) A1 at x = -1, d²y/dx² = -12 < 0, so (-1, 10) is a maximum
(b) A1 at x = 3, d²y/dx² = 12 > 0, so (3, -22) is a minimum
(b) Answer: (-1, 10) is a maximum; (3, -22) is a minimum

Question 12

(a) M1 differentiates s with respect to t
(a) A1 v = 3t² - 12t + 9 cao
(a) Answer: v = 3t² - 12t + 9
(b) M1 substitutes t = 4 into their expression for v, ft from part (a)
(b) A1 v = 9 m/s cao
(b) Answer: v = 9 m/s
(c) M1 sets their v = 0, ft from part (a), e.g. 3t² - 12t + 9 = 0, and simplifies to t² - 4t + 3 = 0
(c) M1 factorises or solves the quadratic correctly, e.g. (t - 1)(t - 3) = 0
(c) A1 t = 1 and t = 3 cao (both values)
(c) Answer: t = 1 second and t = 3 seconds

Question 13

M1 finds acceleration a = dv/dt = 6t - 12
M1 substitutes t = 1 into their expression for a
A1 a = -6 m/s² cao, with a correct statement that the particle is speeding up (accelerating) in the negative direction: velocity at t = 1 is 0 and acceleration is negative, so the particle is momentarily at rest and then moves in the negative direction with increasing speed, oe sensible interpretation accepted
Answer: a = -6 m/s²; the particle is momentarily at rest and then speeds up in the negative direction

Question 14

(a) M1 correctly finds the length of the side parallel to the wall as (40 - 2x) metres
(a) A1 forms A = x(40 - 2x) and expands fully to A = 40x - 2x², cso
(a) Answer: A = 40x - 2x² (shown)
(b) M1 differentiates A with respect to x
(b) A1 dA/dx = 40 - 4x cao
(b) Answer: dA/dx = 40 - 4x
(c) M1 sets their dA/dx = 0 and solves, ft from part (b)
(c) A1 x = 10 cao
(c) A1 maximum area = 200 m² cao, from substituting x = 10 into A = 40x - 2x² (or into x times the parallel side, length 20 m)
(c) Answer: x = 10 m, maximum area = 200 m²
(d) B1 correct justification, e.g. d²A/dx² = -4, which is negative, so it is a maximum; oe (e.g. A = 40x - 2x² is a negative quadratic in x, so its graph is an upside-down parabola with a single maximum turning point)
(d) Answer: d²A/dx² = -4 < 0, so it is a maximum

Question 15

(a) M1 differentiates at least one term correctly
(a) A1 dy/dx = 4x³ - 16x cao
(a) Answer: dy/dx = 4x³ - 16x
(b) M1 sets their dy/dx = 0, ft from part (a)
(b) M1 takes out a common factor correctly, e.g. 4x(x² - 4) = 0
(b) A1 x = 0 cao
(b) A1 x = 2 and x = -2 cao (both values), from correctly factorising x² - 4 = (x - 2)(x + 2)
(b) Answer: x = -2, x = 0, x = 2

Question 16

M1 finds dy/dx = 3x² - 4x
M1 sets their dy/dx equal to 4 (the gradient of the given line) and rearranges to 3x² - 4x - 4 = 0, oe
A1 x = 2 cao
A1 x = -2/3 oe cao
Answer: x = 2 or x = -2/3

Question 17

M1 finds dy/dx = 3x² - 6x + 3
M1 factorises fully, e.g. dy/dx = 3(x - 1)²
A1 correct conclusion that 3(x - 1)² ≥ 0 for all x, equal to 0 only when x = 1, so there is exactly one stationary point (at x = 1), cso
B1 correctly identifies the nature: since dy/dx does not change sign around x = 1 (it is ≥ 0 on both sides), this is a stationary point of inflection, not a maximum or minimum; oe supported by d²y/dx² = 6x - 6 = 0 at x = 1
Answer: Exactly one stationary point, at x = 1; it is a stationary point of inflection

Question 18

(a) M1 differentiates N with respect to t
(a) A1 dN/dt = 60t - 6t² cao
(a) Answer: dN/dt = 60t - 6t²
(b) M1 substitutes t = 3 into their dN/dt, ft from part (a)
(b) A1 126 (bacteria per hour) cao
(b) Answer: 126 bacteria per hour
(c) M1 differentiates dN/dt to find d²N/dt² = 60 - 12t
(c) M1 sets their d²N/dt² = 0 and solves
(c) A1 t = 5 cao, with 0 ≤ 5 ≤ 10 confirmed to be in the given range
(c) Answer: t = 5 hours

Differentiation: Gradients and Turning Points - Worksheets, Questions and Revision