Differentiation: gradients, turning points & kinematics – IGCSE Maths exam questions with mark scheme

1Grade 6–7HigherCalculator2 marks

$y = 4x^3 - 5x^2 + 7$

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$. [2]

for one of the two non-zero terms correct: $12x^2$ or $-10x$ (e.g. $12x^2 - 10x + 7$ or $12x^2 + 7$ scores this mark)B1
$12x^2 - 10x$ fully correct, with no constant term (ignore '$+0$')B1
Answer: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12x^2 - 10x$
Examiner tip: The +7 differentiates to 0 — leaving it in costs the second mark.

2Grade 6–7HigherCalculator4 marks

A curve has equation $y = 2x^2 + \dfrac{8}{x}$

Find the gradient of the curve at the point where $x = 2$. [4]

for $\dfrac{8}{x}$ written as $8x^{-1}$ or for one term differentiated correctly ($4x$ or $-8x^{-2}$)M1
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x - 8x^{-2}$ oeA1
(dep on first M1) for substituting $x = 2$ into their $\dfrac{\mathrm{d}y}{\mathrm{d}x}$M1
$6$A1
Answer: $6$
Examiner tip: Rewrite 8/x as 8x⁻¹ before differentiating; the power goes down to −2, not 0.

3Grade 6–7HigherCalculator6 marks

The curve $C$ has equation $y = x^3 - 6x^2 + 9x + 1$

(a) Find the coordinates of the two turning points on $C$. [4]

(b) Determine the nature of each turning point. Give a reason for your answer. [2]

(a) for differentiating with at least two of the three terms correct: $3x^2 - 12x + 9$M1
for setting their $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ and a correct method to solve, e.g. $3(x - 1)(x - 3) = 0$M1
$x = 1$ and $x = 3$A1
$(1, 5)$ and $(3, 1)$A1
(b) for a valid method, e.g. the shape of a positive cubic (maximum then minimum as $x$ increases), or the sign of $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ either side of each point, or $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x - 12$ substitutedM1
$(1, 5)$ is a maximum and $(3, 1)$ is a minimum, with a correct supporting reasonA1
Answer: (a) $(1, 5)$ and $(3, 1)$ (b) $(1, 5)$ maximum, $(3, 1)$ minimum
Examiner tip: Turning points need both coordinates — substitute each x back into the ORIGINAL equation, not into dy/dx.

4Grade 6–7HigherCalculator6 marks

A particle $P$ moves along a straight line. At time $t$ seconds, the displacement of $P$ from a fixed point $O$ is $s$ metres, where

$s = t^3 - 9t^2 + 24t$

(a) Find the velocity of $P$ when $t = 1$ [2]

(b) Find the values of $t$ when $P$ is instantaneously at rest. [2]

(a) for $v = \dfrac{\mathrm{d}s}{\mathrm{d}t} = 3t^2 - 18t + 24$ (at least two terms correct)M1
$9$ m/sA1
(b) for $3t^2 - 18t + 24 = 0$ and a correct method to solve, e.g. $3(t - 2)(t - 4) = 0$M1
$t = 2$ and $t = 4$A1
(c) for $a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 6t - 18$ (ft their $v$)M1
$12$ m/s$^2$A1
Answer: (a) 9 m/s (b) $t = 2$, $t = 4$ (c) 12 m/s²
Examiner tip: Instantaneously at rest means v = 0, not s = 0 — differentiate first, then solve.

5Grade 8–9HigherCalculator6 marks

An open box has a square base of side $x$ cm and a height of $h$ cm. The box has no lid.

The volume of the box is 500 cm$^3$.

(a) Show that the external surface area, $A$ cm$^2$, of the box is given by $A = x^2 + \dfrac{2000}{x}$ [2]

(b) Use calculus to find the minimum value of $A$. [4]

(a) for $x^2h = 500$ or $h = \dfrac{500}{x^2}$, and $A = x^2 + 4xh$M1
for substituting to obtain $A = x^2 + 4x \times \dfrac{500}{x^2} = x^2 + \dfrac{2000}{x}$ with no errors (shown)A1
(b) for $\dfrac{\mathrm{d}A}{\mathrm{d}x} = 2x - 2000x^{-2}$ (at least one term correct)M1
for setting their $\dfrac{\mathrm{d}A}{\mathrm{d}x} = 0$ and rearranging to $x^3 = 1000$ oeM1
$x = 10$A1
$A = 300$ (cm$^2$)A1
Answer: (a) shown (b) minimum $A = 300$ cm²
Examiner tip: The box has no lid, so the surface area is one base plus four sides — x² + 4xh, not 2x² + 4xh.

6Grade 6–7HigherCalculator4 marks

A curve has equation $y = x^3 - 2x^2 - 4x + 5$

Find the $x$-coordinates of the points on the curve where the gradient is 11. [4]

for $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4x - 4$ (at least two terms correct)M1
for $3x^2 - 4x - 4 = 11$ oeM1
for a correct method to solve $3x^2 - 4x - 15 = 0$, e.g. $(3x + 5)(x - 3) = 0$M1
$x = 3$ and $x = -\dfrac{5}{3}$ (accept $-1.67$ or better)A1
Answer: $x = 3$, $x = -\dfrac{5}{3}$
Examiner tip: The gradient is dy/dx, so set the derivative (not y) equal to 11.

7Grade 8–9HigherCalculator5 marks

A particle moves along a straight line. Its displacement, $s$ metres, from a fixed point $O$ at time $t$ seconds is given by

$s = 2t^3 - 15t^2 + 36t \qquad t \geqslant 0$

The particle starts at $O$.

Work out the total distance travelled by the particle in the first 3 seconds. [5]

for $v = 6t^2 - 30t + 36$M1
for $6t^2 - 30t + 36 = 0$ leading to $t = 2$ (and $t = 3$), i.e. identifying that the particle changes direction at $t = 2$M1
for finding $s$ when $t = 2$ $(= 28)$ and $s$ when $t = 3$ $(= 27)$M1
for $28 + (28 - 27)$ oeM1
$29$ (m)A1
Answer: 29 m
Examiner tip: Distance is not the same as displacement: the particle turns back at t = 2, so s(3) = 27 m undercounts by the 1 m it travels backwards.

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