C Calculus — revision question pack

9 specification points · notes, questions, answers and worked methods

Checked against AQA 8365 section C. Review basis: the qualification registry sourced from the AQA Level 2 Certificate in Further Mathematics (8365) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

C1 · Know that the gradient function dy/dx gives the gradient of the curve and measures the rate of change of y with respect to x

Explanation

For a curve $y=f(x)$ , the derivative $\frac{dy}{dx}$ is its gradient function. It gives the instantaneous rate of change of $y$ with respect to $x$ , rather than an average change across an interval.
Substituting a chosen $x$ -coordinate into $\frac{dy}{dx}$ gives the gradient at that point.
A positive derivative means the curve is increasing locally, a negative derivative means it is decreasing locally, and a zero derivative identifies a stationary point candidate.
Units of a rate are units of $y$ per unit of $x$ .
Examiners expect substitution into the derivative, not into the original function, when a rate of change is requested.

The derivative at a point is the gradient of the tangent to the curve there.

Worked example

For $y=x^2+2x$ , work out the rate of change of $y$ with respect to $x$ when $x=3$ .

1.Differentiate the function: $\frac{dy}{dx}=2x+2$ .
2.Substitute $x=3$ : $\frac{dy}{dx}=2(3)+2$ .
3.Evaluate the derivative.

Answer: The rate of change is $8$ .

Common mistakes

Don't substitute $x=3$ into $y=x^2+2x$ and report the coordinate value $15$ as the rate.
Don't find an average gradient between two nearby points instead of using $\frac{dy}{dx}$ .
Don't interpret a negative derivative as a negative $y$ -coordinate.

Exam tip

Write the gradient function first, then substitute the specified $x$ -value on a separate line.

Tier 1 · Easy

1
At $x=2$ , a curve has $\frac{dy}{dx}=6$ . State its instantaneous rate at this point.
[1 mark]
2
The temperature $T$ in $^\circ\mathrm{C}$ of a liquid is measured $t$ minutes after cooling begins. At $t=5$ , $\frac{dT}{dt}=-1.4$ . Explain what this value tells you.
[1 mark]

Tier 2 · Standard

1
For $y=3x^2-5x$ , work out $\frac{dy}{dx}$ when $x=2$ .
[2 marks]
2
For $y=2x^3-x^2$ , work out the value of $\frac{dy}{dx}$ at $x=-1$ .
[2 marks]
3
The volume $V$ in $\mathrm{cm}^3$ of water in a container is modelled by $V=at^2+4t$ , where $t$ is the time in seconds and $a$ is measured in $\mathrm{cm^3/s^2}$ . At $t=2$ , $\frac{dV}{dt}=20\ \mathrm{cm^3/s}$ . Work out $a$ .
[3 marks]

Tier 3 · Hard

1
A curve has gradient function $\frac{dy}{dx}=3x^2-12x+5$ . Work out the value of $x$ at which the rate of change is $-7$ .
[3 marks]
2
For the curve $y=ax^3+bx^2$ , $\frac{dy}{dx}=7$ at $x=1$ and $\frac{dy}{dx}=-1$ at $x=-1$ . Work out the value of $\frac{dy}{dx}$ at $x=2$ .
[4 marks]
3
For the curve $y=x^4-2x^3$ , the sum of the values of $\frac{dy}{dx}$ at $x=p$ and $x=-p$ is $-48$ . Given that $p>0$ , work out $p$ .
[4 marks]
4
The volume $V$ in $\mathrm{cm}^3$ of liquid in a tank is modelled by $V=t^3-6t^2+15t+40$ , where $t\ge0$ is the time in seconds. At $t=1$ , the instantaneous rate is $r\ \mathrm{cm^3/s}$ . Work out the later time when the instantaneous rate is $4r\ \mathrm{cm^3/s}$ , giving an exact answer.
[4 marks]
5
A curve has gradient function $\frac{dy}{dx}=3x^2-\frac{48}{x^2}$ for $x>0$ . The rate is zero at $x=a$ . Work out the rate at $x=2a$ .
[3 marks]

C2 · Know that the gradient of a function is the gradient of the tangent at that point

Explanation

At a point on a differentiable curve, $f'(x)$ or $\frac{dy}{dx}$ is the gradient of the tangent at that point. The tangent is the straight line that matches the curve’s instantaneous direction locally.
To find its gradient, differentiate the function and substitute the point’s $x$ -coordinate into the derivative.
If a gradient is prescribed, set the derivative equal to that value and solve for every possible $x$ ; the original function then supplies the corresponding coordinates.
Examiners expect a distinction between the point’s $y$ -coordinate and its tangent gradient.
A curve may have the same tangent gradient at more than one point, so all valid solutions must be considered.

Worked example

Work out the gradient of the tangent to $y=x^3-2x$ at the point where $x=1$ .

1.Differentiate: $\frac{dy}{dx}=3x^2-2$ .
2.Substitute the point’s $x$ -coordinate: $3(1)^2-2$ .
3.Evaluate the derivative.

Answer: The tangent gradient is $1$ .

Common mistakes

Don't substitute $x=1$ into the original curve and report $y=-1$ as the gradient.
Don't use the derivative formula but forget to evaluate it at the stated point.
Don't find one $x$ -value when solving $f'(x)=m$ even though the derivative equation has two roots.

Exam tip

When a tangent gradient is given, solve $f'(x)=m$ and then find a coordinate for every resulting $x$ .

Tier 1 · Easy

1
For a function $f$ , $f'(3)=-2$ . State the gradient of the tangent to $y=f(x)$ where $x=3$ .
[1 mark]
2
The tangent to $y=f(x)$ at $x=-1$ has equation $y=5-3x$ . State the value of $f'(-1)$ .
[1 mark]

Tier 2 · Standard

1
Work out the gradient of the tangent to $y=x^2-4x+1$ at the point where $x=1$ .
[2 marks]
2
Work out the gradient of the tangent to $y=x^3+2x^2$ at the point where $x=-2$ .
[2 marks]
3
A tangent to the curve $y=x^3+3x^2$ is parallel to the line $y=24x-5$ . Work out the positive $x$ -coordinate of the point of contact.
[3 marks]

Tier 3 · Hard

1
Work out the coordinates of every point on $y=x^3-3x^2+2$ where the tangent has gradient $9$ .
[4 marks]
2
A curve has equation $y=g(x)$ , where $g(x)=5x^3-15x^2+11$ . Show that its tangents at $x=p$ and $x=2-p$ have equal gradients.
[3 marks]
3
The curve $y=x^3+ax^2+bx$ has horizontal tangents at both $x=-1$ and $x=3$ . Work out $a$ and $b$ .
[4 marks]
4
For the curve $y=x^3+kx^2-5x$ , the tangents at $x=0$ and $x=2$ are perpendicular. Work out $k$ .
[4 marks]
5
The point $P$ lies on $y=x^4-2x^2$ and has a positive $x$ -coordinate. The tangent to the curve at $P$ passes through the origin $O$ . Work out the coordinates of $P$ .
[4 marks]

C3 · Differentiation of kx^n where n is an integer, and the sum of such functions

Explanation

For integer $n$ , the power rule is $\frac{d}{dx}(kx^n)=knx^{n-1}$ . Each term in a sum or difference is differentiated separately, preserving its sign.
A constant differentiates to $0$ . Negative integer powers use the same rule, so a fraction such as $\frac{k}{x^3}$ should first be written $kx^{-3}$ .
Expressions that are products of brackets may need expanding and simplifying before term-by-term differentiation, as required by this specification.
Examiners expect both parts of the power rule: multiply the coefficient by the old power and reduce the power by $1$ .
Final answers may be written with negative powers or converted back to fractions.

Worked example

Differentiate $y=4x^5-\dfrac{3}{x^2}+7$ with respect to $x$ .

1.Rewrite the fraction: $y=4x^5-3x^{-2}+7$ .
2.Apply the power rule term by term: $20x^4+6x^{-3}+0$ .
3.Rewrite the negative power as a fraction if preferred.

Answer: $\dfrac{dy}{dx}=20x^4+\dfrac{6}{x^3}$ .

Common mistakes

Don't differentiate $4x^5$ as $20x^5$ without reducing the power.
Don't differentiate $-3x^{-2}$ as $-6x^{-3}$ instead of $6x^{-3}$ .
Don't leave the constant $7$ in the derivative.

Exam tip

Rewrite denominator powers as negative indices before applying the power rule.

Tier 1 · Easy

1
Differentiate $5x^4$ with respect to $x$ .
[1 mark]
2
Differentiate $7x^{-1}$ with respect to $x$ .
[1 mark]

Tier 2 · Standard

1
Given $y=4x^3-3x^2+7x-9$ , work out $\frac{dy}{dx}$ .
[3 marks]
2
Expand the brackets and hence work out $\frac{dy}{dx}$ when $y=(2x-1)(x^2+3)$ .
[3 marks]
3
The curve $y=ax^4+\frac{6}{x}$ has gradient $10$ at $x=1$ . Work out $a$ .
[3 marks]

Tier 3 · Hard

1
For $y=2x^5-\frac{3}{x^2}+4x$ , work out $\frac{dy}{dx}$ and hence its value when $x=1$ .
[4 marks]
2
Expand and simplify, then work out $\frac{dy}{dx}$ when $y=(x^2+1)(3x^2-2x)+\frac{4}{x^2}$ .
[4 marks]
3
The function $f(x)=kx^n$ , where $k>0$ and $n$ is an integer greater than $1$ , satisfies $f'(1)=8$ and $f'(2)=64$ . Work out $k$ and $n$ .
[4 marks]
4
The function $f(x)=(x^2+px+q)(x^2-2x)+\frac{8}{x}$ has a derivative containing no $x^2$ term. Given that $f'(2)=20$ , work out $p$ and $q$ .
[4 marks]
5
The function $f(x)=x^m(ax^n+b)$ , where $m$ and $n$ are positive integers, has derivative $f'(x)=15x^4-6x^2$ . Work out $a$ , $b$ , $m$ and $n$ .
[4 marks]

C4 · The equation of a tangent and normal at any point on a curve

Explanation

To find a tangent equation, first determine the point on the curve and evaluate $\frac{dy}{dx}$ there to obtain the tangent gradient. The normal is perpendicular to the tangent, so when both gradients are defined its gradient is the negative reciprocal.
Use the common point of contact in point-gradient form $y-y_1=m(x-x_1)$ .
If only an $x$ -coordinate is supplied, substitute into the original curve to find $y$ before forming either line.
Horizontal tangents have vertical normals, which require an equation $x=\text{constant}$ rather than a finite gradient.
Examiners expect the curve point, gradient calculation and line equation as distinct method stages.

The tangent and normal meet at right angles at the point of contact.

Worked example

Work out the equation of the normal to $y=x^2+1$ at the point where $x=2$ .

1.The point is $(2,5)$ because $2^2+1=5$ .
2. $\frac{dy}{dx}=2x$ , so the tangent gradient is $4$ and the normal gradient is $-\frac14$ .
3.Use point-gradient form through $(2,5)$ .

Answer: $y-5=-\dfrac14(x-2)$ .

Common mistakes

Don't use tangent gradient $4$ for the normal instead of the negative reciprocal $-\frac14$ .
Don't substitute $x=2$ into the derivative but never finds the point’s $y$ -coordinate.
Don't use the curve’s intercept rather than the point of contact in the line equation.

Exam tip

For a normal, show the tangent gradient and its negative reciprocal before applying point-gradient form.

Tier 1 · Easy

1
The curve $y=x^2$ passes through $(2,4)$ . State the gradients of the tangent and the normal at this point.
[2 marks]
2
At $(2,1)$ , a curve has tangent gradient $-3$ . Work out an equation of the normal at this point.
[2 marks]

Tier 2 · Standard

1
Work out the equation of the tangent to $y=x^2+3x-1$ at the point where $x=1$ .
[3 marks]
2
Work out the equation of the normal to $y=x^2-6x$ at the point where $x=3$ .
[2 marks]
3
A tangent to the curve $y=x^2+1$ is parallel to the line $6x-y=4$ . Work out the equation of this tangent.
[3 marks]

Tier 3 · Hard

1
The normal to $y=x^3-x$ at the point where $x=2$ meets the $x$ -axis at $R$ . Work out the coordinates of $R$ .
[4 marks]
2
The curve $y=x^2+k$ has a tangent at $x=1$ which passes through $(0,2)$ . Work out $k$ and hence work out an equation of the normal to the curve at $x=1$ .
[4 marks]
3
At each point on the curve $y=x^2-\frac{9}{2}$ with a non-zero $x$ -coordinate, a normal is drawn. Two of these normals pass through the origin. Work out the equations of these two normals.
[4 marks]
4
The tangent to the curve $y=x^3-3x$ at the point $(2,2)$ meets the $y$ -axis at $A=(0,-16)$ . The normal to the curve at the same point meets the $y$ -axis at $B$ . Work out the length $AB$ .
[4 marks]
5
The tangents to the curve $y=x^2$ at the points $P=(p,p^2)$ and $Q=(q,q^2)$ , where $p\neq q$ , are perpendicular. Show that $pq=-\frac14$ , and work out the $y$ -coordinate of the point where the two tangents intersect.
[4 marks]

C5 · Increasing and decreasing functions

Explanation

A differentiable function is increasing where $\frac{dy}{dx}>0$ and decreasing where $\frac{dy}{dx}<0$ . Solve $\frac{dy}{dx}=0$ to find stationary boundary values, then determine the derivative’s sign in every interval created by those values.
This can be done by factor signs or test values.
The sign of $y$ itself is irrelevant: a curve below the $x$ -axis may still be increasing.
At a stationary point the derivative is zero, so strict increasing and decreasing intervals normally use open endpoints.
Examiners expect interval notation or inequalities covering every region and a derivative sign argument, not a visual judgment from an unscaled sketch.

Derivative signs divide a curve into increasing and decreasing intervals.

Worked example

Work out where $f(x)=x^3-3x^2$ is increasing and where it is decreasing.

1. $f'(x)=3x^2-6x=3x(x-2)$ , so stationary values are $x=0$ and $x=2$ .
2. $f'(x)>0$ for $x<0$ and $x>2$ .
3. $f'(x)<0$ for $0<x<2$ .

Answer: Increasing for $x<0$ and $x>2$ ; decreasing for $0<x<2$ .

Common mistakes

Don't use where $f(x)>0$ instead of where $f'(x)>0$ .
Don't find stationary values $0$ and $2$ and fail to test the derivative signs between and outside them.
Don't state that the function is increasing at a stationary point where the derivative equals zero.

Exam tip

Factor the derivative, mark its zeros on a sign line and test every resulting interval.

Tier 1 · Easy

1
A function has derivative $f'(x)=2x-6$ . State the values of $x$ for which the function is increasing.
[2 marks]
2
At $x=2$ , a function has $f'(2)=-7$ . State whether the function is increasing or decreasing at this point.
[1 mark]

Tier 2 · Standard

1
Work out where $y=x^2-8x+1$ is decreasing and where it is increasing.
[3 marks]
2
A function has derivative $f'(x)=(x+2)(x-5)$ . Work out the interval on which the function is decreasing.
[2 marks]
3
Show that $f(x)=x^3+3x^2+6x-4$ is increasing for every real value of $x$ .
[3 marks]

Tier 3 · Hard

1
Work out the intervals on which $f(x)=x^3-3x^2-9x+4$ is increasing and the interval on which it is decreasing.
[4 marks]
2
Work out the values of $x$ for which $f(x)=x^4-2x^2$ is increasing and the values of $x$ for which it is decreasing.
[4 marks]
3
The function $f(x)=x^3+3kx^2+3x$ , where $k$ is a real constant, is never decreasing. Work out all possible values of $k$ .
[4 marks]
4
For $x\ne0$ , work out the intervals on which $f(x)=x^2+\frac{16}{x}$ is increasing and the intervals on which it is decreasing.
[4 marks]
5
The function $f(x)=x^3+ax^2+bx+7$ is decreasing exactly when $-2<x<5$ . Work out $a$ and $b$ .
[4 marks]

C6 · Understand and use the notation d2y/dx2; know that it measures the rate of change of the gradient function

Explanation

The second derivative $\frac{d^2y}{dx^2}$ is the derivative of the gradient function $\frac{dy}{dx}$ , so it measures how the gradient changes as $x$ changes. It is found by differentiating the original function twice.
At a point, $\frac{d^2y}{dx^2}>0$ means the gradient is increasing and $\frac{d^2y}{dx^2}<0$ means the gradient is decreasing.
This sign can help classify stationary points.
The notation does not mean $\left(\frac{dy}{dx}\right)^2$ ; the superscript records a second differentiation.
Examiners expect both derivative stages when the original function is supplied, with negative powers simplified consistently.

Worked example

For $y=2x^4-3x^3+x$ , work out $\frac{d^2y}{dx^2}$ .

1.Differentiate once: $\frac{dy}{dx}=8x^3-9x^2+1$ .
2.Differentiate the gradient function term by term.
3.The constant $1$ differentiates to $0$ .

Answer: $\dfrac{d^2y}{dx^2}=24x^2-18x$ .

Common mistakes

Don't square the first derivative instead of differentiating it again.
Don't stop after finding $\frac{dy}{dx}$ when the second derivative is requested.
Don't keep the constant $1$ in the second derivative.

Exam tip

Write the first derivative on its own line before differentiating again, so the two stages are visible.

Tier 1 · Easy

1
For $y=3x^4-5x^2+7$ , work out $\frac{d^2y}{dx^2}$ .
[2 marks]
2
The gradient function of a curve is $5x^3-6x+4$ . Work out the rate of change of the gradient when $x=-1$ .
[2 marks]

Tier 2 · Standard

1
A curve has $\frac{dy}{dx}=6x^2-4x$ . Work out the value of $x$ at which the gradient is changing at a rate of $20$ , given $x>0$ .
[3 marks]
2
For the curve $y=ax^3+3x^2-x$ , the gradient is changing at a rate of $30$ when $x=2$ . Work out the value of $a$ .
[3 marks]
3
For the curve $y=x^4-8x^3+18x^2+5$ , work out the values of $x$ for which the rate of change of the gradient is zero.
[3 marks]

Tier 3 · Hard

1
For the curve $y=x^4-4x^3+2x^2$ , work out the ranges of $x$ for which the gradient is increasing. Give exact values.
[4 marks]
2
The gradient function of a curve is $g(x)=ax^2+bx+c$ . The rate of change of the gradient is $-6$ when $x=-1$ and $14$ when $x=4$ . Work out the constants $a$ and $b$ , and hence the values of $x$ for which the gradient is increasing.
[4 marks]
3
For the curve $y=x^4+ax^3+bx^2$ , the gradient is changing at a rate of zero when $x=-1$ and when $x=3$ . Work out the least value of $\frac{d^2y}{dx^2}$ .
[4 marks]
4
The gradient function of a curve is $g(x)=2x^3-3px^2+12x$ . The gradient is changing at the same rate when $x=1$ and when $x=4$ . Work out $p$ and this common rate.
[3 marks]
5
For the curve $y=ax^4+bx^{-1}$ , where $x\ne0$ , the gradient is changing at a rate of $30$ when $x=1$ and at a rate of $18$ when $x=-1$ . Work out $a$ , $b$ and the rate at which the gradient is changing when $x=2$ .
[4 marks]

C7 · Use of differentiation to find maxima and minima points on a curve

Explanation

At a stationary point, $\frac{dy}{dx}=0$ . Solve this equation and substitute each resulting $x$ -value into the original curve to obtain full coordinates.
The nature can be determined using the second derivative: a negative value gives a local maximum and a positive value gives a local minimum.
Alternatively, a derivative sign change from positive to negative proves a maximum, while negative to positive proves a minimum.
If the second derivative is zero, this test is inconclusive and the sign-change method is needed.
Examiners expect every stationary point, its coordinates and its nature; reporting only the $x$ -coordinates or assuming that every stationary point is a turning point is incomplete.

At local maxima and minima, the tangent is horizontal and the first derivative is zero.

Worked example

Work out the coordinates and nature of the stationary points of $y=x^3-3x^2-9x+5$ .

1. $\frac{dy}{dx}=3x^2-6x-9=3(x-3)(x+1)$ , so $x=-1$ or $x=3$ .
2.Substitution gives $y(-1)=10$ and $y(3)=-22$ .
3. $\frac{d^2y}{dx^2}=6x-6$ , which is negative at $x=-1$ and positive at $x=3$ .

Answer: $(-1,10)$ is a local maximum and $(3,-22)$ is a local minimum.

Common mistakes

Don't solve $\frac{dy}{dx}=0$ but report $x=-1$ and $x=3$ without the corresponding $y$ -coordinates.
Don't substitute stationary values into the derivative instead of the original curve to find $y$ .
Don't reverse the second-derivative test and call a negative value a minimum.

Exam tip

Present each stationary point as a coordinate followed by ‘maximum’ or ‘minimum’, supported by a second-derivative or sign-change line.

Tier 1 · Easy

1
The curve $y=x^3-3x^2+2$ has a stationary point where $x=2$ . Using the second derivative $\frac{d^2y}{dx^2}$ , work out its nature.
[2 marks]
2
The curve $y=x^2-6x+11$ has a stationary point where $x=3$ . Work out its coordinates and state its nature.
[2 marks]

Tier 2 · Standard

1
The curve $y=x^3-6x^2+9x+1$ has stationary points where $x=1$ and $x=3$ . Work out their coordinates and determine their nature.
[3 marks]
2
The curve $y=x^3-12x$ has a stationary point with a positive $x$ -coordinate. Work out its coordinates and determine its nature.
[3 marks]
3
The curve $y=x^3-3x^2+k$ passes through the stationary point $(0,5)$ . Work out the coordinates and nature of its other stationary point.
[3 marks]

Tier 3 · Hard

1
Work out the coordinates and nature of every stationary point of $y=2x^4-16x^2+3$ .
[4 marks]
2
The curve $y=x^3+ax^2+bx$ has stationary points at $x=1$ and $x=3$ . Work out $a$ and $b$ , and hence work out the coordinates and nature of both stationary points.
[4 marks]
3
For $a>0$ , the curve $y=x^3-3a^2x$ has two stationary points whose vertical distance apart is $32$ units. Work out the coordinates and nature of both stationary points.
[4 marks]
4
For the curve $y=x^4-4x^3$ , work out the coordinates of every stationary point. State which point is a maximum or minimum, and explain why the other point is neither.
[4 marks]
5
The curve $y=x^3-3x^2-9x+k$ has a local maximum on the line $y=2x+15$ . Work out $k$ and the coordinates and nature of both stationary points.
[4 marks]

C8 · Using calculus to find maxima and minima in simple problems

Explanation

In an optimisation problem, express the quantity to be maximised or minimised as a function of one variable. Use any perimeter, length or other constraint to remove additional variables.
Differentiate, solve the derivative equation $\frac{dQ}{dx}=0$ , and verify the required nature with the second derivative, a derivative sign change or the known shape of a quadratic.
Reject values outside the contextual domain, such as negative lengths.
Finally substitute the stationary input back into the requested quantity: the value of $x$ is not the final answer when the question asks for a maximum area, minimum cost or another output.
Examiners expect the model, stationary calculation, nature check and contextual conclusion.

A rectangle constrained by side lengths x and 20-x has area as a one-variable function.

Worked example

A rectangle has perimeter $40$ cm. Use calculus to work out its maximum area.

1.If one side is $x$ , the other is $20-x$ , so $A=x(20-x)=20x-x^2$ .
2. $\frac{dA}{dx}=20-2x=0$ gives $x=10$ .
3. $\frac{d^2A}{dx^2}=-2<0$ , so the area is maximal; $A(10)=100$ .

Answer: The maximum area is $100\text{ cm}^2$ .

Common mistakes

Don't use the full perimeter and write the other side as $40-x$ instead of $20-x$ .
Don't report the stationary side length $x=10$ as though it were the maximum area.
Don't accept a stationary value outside the allowed length domain.

Exam tip

End an optimisation solution by substituting back into the requested quantity and attaching its units.

Tier 1 · Easy

1
A quantity is modelled by $P=x(14-x)$ for $0<x<14$ . Work out the greatest possible value of $P$ by differentiating.
[2 marks]
2
The height, $h$ metres, of a model rocket $t$ seconds after launch is $h=20+12t-3t^2$ , for $t>0$ . Use differentiation to work out the time at which the height is greatest.
[2 marks]

Tier 2 · Standard

1
A rectangle has side lengths $(x+2)$ cm and $(10-x)$ cm, where $0<x<10$ . Use calculus to work out its maximum area.
[3 marks]
2
A rectangular enclosure is built against a straight wall using $24$ metres of fencing for the other three sides. Use calculus to work out the maximum possible area of the enclosure.
[3 marks]
3
Positive lengths $x$ cm and $y$ cm satisfy $xy=144$ . Use calculus to work out the least possible value of $x+y$ .
[3 marks]

Tier 3 · Hard

1
The upper corners of a rectangle lie on $y=12-x^2$ at $(x,y)$ and $(-x,y)$ , where $x>0$ . The lower corners lie on the $x$ -axis. Use calculus to work out the maximum area of the rectangle.
[4 marks]
2
A cuboid has a square base of side $x$ cm and height $h$ cm, where $2x+h=18$ . Use calculus to work out the maximum possible volume of the cuboid.
[4 marks]
3
An open-topped box has a square base of side $x$ cm and height $h$ cm, where $x>0$ and $h>0$ . Its volume is $256\text{ cm}^3$ . Use calculus to work out the least possible surface area of the box.
[4 marks]
4
The height $h$ metres of a ball $t$ seconds after launch is $h=2+pt-4t^2$ , where $t\ge0$ . The ball is at the same height when $t=1$ and when $t=3$ . Use calculus to work out its greatest height.
[4 marks]
5
The point $P=(x,x^2)$ lies on the curve $y=x^2$ , where $x>0$ , and $Q=(0,3)$ . Show that $PQ^2=x^4-5x^2+9$ . Hence use calculus to work out the least value of $PQ^2$ , and the shortest distance $PQ$ .
[4 marks]

C9 · Sketch/interpret a curve with known maximum and minimum points

Explanation

A local maximum is a turning point where the curve changes from increasing to decreasing; a local minimum changes from decreasing to increasing. To sketch from known information, plot and label the turning points and any intercepts in increasing $x$ -order, then join them with one smooth curve.
Horizontal tangents should be visible at smooth maxima and minima.
The derivative sign determines which sections rise or fall.
A sketch need not be to scale, but relative positions, crossings and the number of turning points must match the information.
Examiners expect no invented roots or stationary points, and an interpretation question should use intervals between the stated turning-point coordinates.

A smooth curve rises to a local maximum, decreases to a local minimum, then rises again.

Worked example

A smooth curve has a local maximum at $(-1,4)$ and a local minimum at $(2,-3)$ , with no other stationary points. State its increasing and decreasing intervals.

1.Before the local maximum, the curve rises as $x$ increases.
2.Between the maximum and minimum, it falls.
3.After the local minimum, it rises again.

Answer: Increasing for $x<-1$ and $x>2$ ; decreasing for $-1<x<2$ .

Common mistakes

Don't state that the curve is decreasing to the left of its local maximum.
Don't include extra turning points not present in the supplied information.
Don't join labelled points with straight segments instead of a smooth curve with horizontal tangents.

Exam tip

Order every labelled feature by its $x$ -coordinate before drawing the curve through them.

Tier 1 · Easy

1
A smooth curve has a local maximum at $x=-2$ and a local minimum at $x=1$ , with no other stationary points. State where the curve is decreasing.
[1 mark]
2
A smooth graph rises until $x=-1$ , falls from $x=-1$ to $x=4$ , and then rises again. Write down the $x$ -value at which a local minimum occurs.
[1 mark]

Tier 2 · Standard

1
Sketch the cubic $y=2x^3-3x^2-12x+2$ , which has local maximum $L=(-1,9)$ and local minimum $M=(2,-18)$ . Label $L$ , $M$ and the $y$ -intercept $N$ .
[3 marks]
2
A smooth curve $y=f(x)$ has a local maximum at $(-3,7)$ , a local minimum at $(2,-1)$ and no other turning points. Work out the values of $x$ for which the curve is decreasing, and explain how the two turning points show that the curve crosses the $x$ -axis between $x=-3$ and $x=2$ .
[3 marks]
3
A smooth curve $y=f(x)$ has a local maximum at $P=(-1,6)$ and a local minimum at $Q=(3,-2)$ , and no other turning points. Work out the values of $x$ for which the curve is increasing, and sketch a possible curve.
[3 marks]

Tier 3 · Hard

1
A continuous smooth curve $y=f(x)$ has roots $x=-4$ , $x=1$ and $x=5$ . It has a local maximum at $(-2,5)$ and a local minimum at $(3,-4)$ , with no other turning points. Sketch a possible curve, labelling all five given points.
[4 marks]
2
A smooth curve has local minima at $(-3,-2)$ and $(2,-5)$ , and a local maximum at $(0,4)$ . It has no other turning points, and the curve rises without bound as $x$ tends to either positive or negative infinity. Sketch a possible curve, labelling the three turning points, and state where the curve is increasing.
[4 marks]
3
A continuous smooth curve $y=f(x)$ has a local maximum at $P=(-2,4)$ and a local minimum at $Q=(1,0)$ , with no other turning points. It meets the $x$ -axis only at $R=(-4,0)$ and at $Q$ . The curve falls without bound as $x$ tends to negative infinity and rises without bound as $x$ tends to positive infinity. State which root the curve crosses and which it touches, and use this information to sketch a possible curve.
[4 marks]
4
A smooth curve $y=f(x)$ has a local maximum at $(-2,7)$ and a local minimum at $(3,-1)$ , with no other turning points. Also, $f(x)$ falls without bound as $x$ tends to negative infinity and rises without bound as $x$ tends to positive infinity. Work out the values of $k$ for which $f(x)=k$ has exactly three distinct solutions.
[3 marks]
5
On the interval $-5\le x\le6$ , a smooth curve $y=f(x)$ starts at $A=(-5,-4)$ , has a local maximum at $P=(-2,7)$ , then a local minimum at $Q=(3,-2)$ , and ends at $B=(6,9)$ . It has no other turning points. Work out the range of $f(x)$ on this interval, using a sketch that labels $A$ , $P$ , $Q$ and $B$ .
[4 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

C1 · Know that the gradient function dy/dx gives the gradient of the curve and measures the rate of change of y with respect to x

Tier 1 · Easy

Mark scheme for C1 Tier 1 · Easy
Q	Answer	Mark	Comments
1	Rate of change $=6$	1	$\frac{dy}{dx}$ directly gives the rate of change of $y$ with respect to $x$ . Its value at the point is $6$ .
2	At $t=5$ minutes, the temperature is decreasing at $1.4^\circ\mathrm{C}$ per minute	1	The derivative is the instantaneous rate of change of temperature. Its negative sign means the temperature is decreasing, and its magnitude gives a rate of $1.4^\circ\mathrm{C}$ per minute.

Tier 2 · Standard

Mark scheme for C1 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\dfrac{dy}{dx}=7$	2	Differentiate: $\frac{dy}{dx}=6x-5$ . At $x=2$ , $\frac{dy}{dx}=6(2)-5=7$ .
2	$\frac{dy}{dx}=8$	2	Differentiate: $\frac{dy}{dx}=6x^2-2x$ . At $x=-1$ , $\frac{dy}{dx}=6(-1)^2-2(-1)=6+2=8$ .
3	$a=4$ (accept $4\ \mathrm{cm^3/s^2}$ )	3	Differentiate to obtain $\frac{dV}{dt}=2at+4$ . At $t=2$ , this gives $4a+4=20$ , so $4a=16$ and $a=4\ \mathrm{cm^3/s^2}$ .

Tier 3 · Hard

Mark scheme for C1 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$x=2$	3	Set the gradient function equal to the required rate: $3x^2-12x+5=-7$ . Then $3x^2-12x+12=0$ , so $x^2-4x+4=0$ . Since $(x-2)^2=0$ , $x=2$ .
2	The rate of change at $x=2$ is $20$	4	$\frac{dy}{dx}=3ax^2+2bx$ . The two given rates give $3a+2b=7$ and $3a-2b=-1$ . Adding gives $6a=6$ , so $a=1$ ; then $3+2b=7$ , so $b=2$ . At $x=2$ , $\frac{dy}{dx}=3(1)(2^2)+2(2)(2)=12+8=20$ .
3	$p=2$	4	$\frac{dy}{dx}=4x^3-6x^2$ . The two derivative values are $4p^3-6p^2$ and $-4p^3-6p^2$ , so their sum is $-12p^2$ . Hence $-12p^2=-48$ , giving $p^2=4$ . Since $p>0$ , $p=2$ .
4	$t=2+\sqrt{7}$ seconds	4	$\frac{dV}{dt}=3t^2-12t+15$ , so $r=3-12+15=6$ . The later rate is therefore $24\ \mathrm{cm^3/s}$ . Solving $3t^2-12t+15=24$ gives $t^2-4t-3=0$ , so $t=2\pm\sqrt{7}$ . The value $2-\sqrt{7}$ is negative, so the later time is $t=2+\sqrt{7}$ seconds.
5	$45$	3	At $x=a$ , $3a^2-\frac{48}{a^2}=0$ . Multiplying by $a^2$ gives $3a^4=48$ , so $a^4=16$ . As $a>0$ , $a=2$ , and hence $2a=4$ . The rate there is $3(4)^2-\frac{48}{4^2}=48-3=45$ .

C2 · Know that the gradient of a function is the gradient of the tangent at that point

Tier 1 · Easy

Mark scheme for C2 Tier 1 · Easy
Q	Answer	Mark	Comments
1	Tangent gradient $=-2$	1	$f'(3)$ is the gradient of the tangent at $x=3$ , so the required gradient is $-2$ .
2	$f'(-1)=-3$	1	The coefficient of $x$ in the tangent equation is its gradient. Since the tangent gradient equals the derivative at the point, $f'(-1)=-3$ .

Tier 2 · Standard

Mark scheme for C2 Tier 2 · Standard
Q	Answer	Mark	Comments
1	Tangent gradient $=-2$	2	Differentiate to get $\frac{dy}{dx}=2x-4$ . At $x=1$ , the tangent gradient is $2(1)-4=-2$ .
2	Tangent gradient $=4$	2	$\frac{dy}{dx}=3x^2+4x$ . At $x=-2$ , the tangent gradient is $3(-2)^2+4(-2)=12-8=4$ .
3	$x=2$	3	The tangent gradient is $24$ . Since $\frac{dy}{dx}=3x^2+6x$ , set $3x^2+6x=24$ . This gives $x^2+2x-8=0$ , so $(x+4)(x-2)=0$ . The positive solution is $x=2$ .

Tier 3 · Hard

Mark scheme for C2 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$(-1,-2)$ and $(3,2)$	4	$\frac{dy}{dx}=3x^2-6x$ . Set this equal to $9$ : $3x^2-6x-9=0$ , so $x^2-2x-3=0$ and $(x-3)(x+1)=0$ . Thus $x=3$ or $x=-1$ . Substitution into the curve gives $y=2$ at $x=3$ and $y=-2$ at $x=-1$ .
2	Both tangent gradients are $15p^2-30p$	3	$g'(x)=15x^2-30x$ . At $x=p$ , the gradient is $15p^2-30p$ . At $x=2-p$ , it is $15(2-p)^2-30(2-p)=15(4-4p+p^2)-60+30p=15p^2-30p$ . Therefore the two gradients are equal.
3	$a=-3$ and $b=-9$	4	$\frac{dy}{dx}=3x^2+2ax+b$ . The two horizontal tangents give $3-2a+b=0$ and $27+6a+b=0$ . Subtracting the first equation from the second gives $24+8a=0$ , so $a=-3$ . Substitution into $3-2a+b=0$ then gives $3+6+b=0$ , so $b=-9$ .
4	$k=-\frac{17}{10}$	4	$\frac{dy}{dx}=3x^2+2kx-5$ . The tangent gradients at $x=0$ and $x=2$ are $-5$ and $7+4k$ respectively. Perpendicular gradients have product $-1$ , so $-5(7+4k)=-1$ . Hence $35+20k=1$ , giving $k=-\frac{34}{20}=-\frac{17}{10}$ .
5	$P=\left(\sqrt{\frac{2}{3}},-\frac{8}{9}\right)$ (or $P=\left(\frac{\sqrt{6}}{3},-\frac{8}{9}\right)$ )	4	Let the $x$ -coordinate of $P$ be $a>0$ . A tangent through the origin has gradient $\frac{a^4-2a^2}{a}=a^3-2a$ , and the tangent gradient at $P$ is $4a^3-4a$ . Equating them gives $3a^3-2a=0$ , so $a(3a^2-2)=0$ . Since $a>0$ , $a=\sqrt{\frac{2}{3}}$ . Then $y=a^4-2a^2=\frac{4}{9}-\frac{4}{3}=-\frac{8}{9}$ .

C3 · Differentiation of kx^n where n is an integer, and the sum of such functions

Tier 1 · Easy

Mark scheme for C3 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$20x^3$	1	Multiply the coefficient by the power and reduce the power by $1$ : $\frac{d}{dx}(5x^4)=5\times4x^3=20x^3$ .
2	$-7x^{-2}$ (or $-\frac{7}{x^2}$ )	1	The power rule gives $7(-1)x^{-2}=-7x^{-2}=-\frac{7}{x^2}$ .

Tier 2 · Standard

Mark scheme for C3 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\frac{dy}{dx}=12x^2-6x+7$	3	Differentiate term by term: $4x^3$ gives $12x^2$ , $-3x^2$ gives $-6x$ , $7x$ gives $7$ , and the constant $-9$ gives $0$ . Therefore $\frac{dy}{dx}=12x^2-6x+7$ .
2	$\frac{dy}{dx}=6x^2-2x+6$	3	First expand: $y=2x^3-x^2+6x-3$ . Differentiating term by term gives $\frac{dy}{dx}=6x^2-2x+6$ .
3	$a=4$	3	Write $y=ax^4+6x^{-1}$ , so $\frac{dy}{dx}=4ax^3-6x^{-2}$ . At $x=1$ , $4a-6=10$ . Therefore $4a=16$ and $a=4$ .

Tier 3 · Hard

Mark scheme for C3 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\frac{dy}{dx}=10x^4+\frac{6}{x^3}+4$ $\dfrac{dy}{dx}=20$ when $x=1$	4	Write $-\frac{3}{x^2}$ as $-3x^{-2}$ . Differentiating gives $10x^4+6x^{-3}+4=10x^4+\frac{6}{x^3}+4$ . At $x=1$ , the value is $10+6+4=20$ .
2	$\frac{dy}{dx}=12x^3-6x^2+6x-2-\frac{8}{x^3}$ (or $12x^3-6x^2+6x-2-8x^{-3}$ )	4	Expand and rewrite the fractional term: $y=3x^4-2x^3+3x^2-2x+4x^{-2}$ . Differentiate term by term to obtain $\frac{dy}{dx}=12x^3-6x^2+6x-2-8x^{-3}=12x^3-6x^2+6x-2-\frac{8}{x^3}$ .
3	$k=2$ and $n=4$	4	$f'(x)=knx^{n-1}$ . From $f'(1)=8$ , $kn=8$ . Dividing $f'(2)=kn2^{n-1}=64$ by this equation gives $2^{n-1}=8=2^3$ , so $n=4$ . Then $4k=8$ , giving $k=2$ .
4	$p=2$ and $q=3$	4	Expanding gives $f(x)=x^4+(p-2)x^3+(q-2p)x^2-2qx+8x^{-1}$ . Therefore $f'(x)=4x^3+3(p-2)x^2+2(q-2p)x-2q-8x^{-2}$ . There is no $x^2$ term, so $3(p-2)=0$ and $p=2$ . Then $f'(2)=32+4(q-4)-2q-2=14+2q$ . Since this equals $20$ , $q=3$ .
5	$a=3$ , $b=-2$ , $m=3$ and $n=2$	4	Expanding first gives $f(x)=ax^{m+n}+bx^m$ , so $f'(x)=a(m+n)x^{m+n-1}+bmx^{m-1}$ . Matching the $x^2$ term gives $m-1=2$ , so $m=3$ , and $bm=-6$ then gives $b=-2$ . Matching the $x^4$ term gives $m+n-1=4$ , so $n=2$ , and $a(m+n)=15$ gives $5a=15$ , hence $a=3$ .

C4 · The equation of a tangent and normal at any point on a curve

Tier 1 · Easy

Mark scheme for C4 Tier 1 · Easy
Q	Answer	Mark	Comments
1	Tangent gradient $=4$ Normal gradient $=-\frac{1}{4}$	2	$\frac{dy}{dx}=2x$ , so at $x=2$ the tangent gradient is $4$ . The normal gradient is the negative reciprocal, $-\frac{1}{4}$ .
2	$y-1=\frac{1}{3}(x-2)$ (or any algebraically equivalent equation, e.g. $y=\frac{1}{3}x+\frac{1}{3}$ )	2	The normal gradient is the negative reciprocal of $-3$ , which is $\frac{1}{3}$ . Through $(2,1)$ , its equation is $y-1=\frac{1}{3}(x-2)$ .

Tier 2 · Standard

Mark scheme for C4 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$y=5x-2$	3	At $x=1$ , $y=1+3-1=3$ , so the point is $(1,3)$ . Also $\frac{dy}{dx}=2x+3$ , giving tangent gradient $5$ . Thus $y-3=5(x-1)$ , so $y=5x-2$ .
2	$x=3$	2	At $x=3$ , $y=3^2-6(3)=-9$ , so the point is $(3,-9)$ . Also $\frac{dy}{dx}=2x-6$ , which is $0$ at $x=3$ . The tangent is horizontal, so the normal is the vertical line through the point. Therefore its equation is $x=3$ .
3	$y=6x-8$	3	The line $6x-y=4$ has gradient $6$ . Since $\frac{dy}{dx}=2x$ , the point of contact has $2x=6$ , so $x=3$ and $y=3^2+1=10$ . The tangent through $(3,10)$ is $y-10=6(x-3)$ , which simplifies to $y=6x-8$ .

Tier 3 · Hard

Mark scheme for C4 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$R=(68,0)$	4	At $x=2$ , $y=2^3-2=6$ , so the point is $(2,6)$ . Since $\frac{dy}{dx}=3x^2-1$ , the tangent gradient is $11$ and the normal gradient is $-\frac{1}{11}$ . The normal is $y-6=-\frac{1}{11}(x-2)$ . At the $x$ -axis $y=0$ , so $-6=-\frac{1}{11}(x-2)$ , giving $66=x-2$ and $x=68$ .
2	$k=3$ $y-4=-\frac{1}{2}(x-1)$ (or any equivalent form, e.g. $y=-\frac{1}{2}x+\frac{9}{2}$ or $x+2y=9$ )	4	The tangent gradient at $x=1$ is $2$ , and the point of contact is $(1,1+k)$ . Its equation is $y-(1+k)=2(x-1)$ . Since $(0,2)$ lies on it, $2-(1+k)=-2$ , so $k=3$ . The point of contact is then $(1,4)$ . The normal gradient is $-\frac{1}{2}$ , giving $y-4=-\frac{1}{2}(x-1)$ .
3	$y=-\frac{1}{4}x$ and $y=\frac{1}{4}x$	4	Let a point of contact have $x$ -coordinate $a$ , where $a\ne0$ . It is $(a,a^2-\frac{9}{2})$ . The tangent gradient is $2a$ , so the normal gradient is $-\frac{1}{2a}$ . As the normal also passes through the origin, its equation is $y=-\frac{1}{2a}x$ . Substituting the point of contact gives $a^2-\frac{9}{2}=-\frac{1}{2}$ , so $a^2=4$ and $a=\pm2$ . The corresponding equations are $y=-\frac{1}{4}x$ and $y=\frac{1}{4}x$ .
4	$AB=\frac{164}{9}$ units	4	$\frac{dy}{dx}=3x^2-3$ , so the tangent gradient at $(2,2)$ is $9$ and the normal gradient is $-\frac{1}{9}$ . The normal is $y-2=-\frac{1}{9}(x-2)$ , which meets the $y$ -axis at $B=(0,\frac{20}{9})$ . Hence $AB=\frac{20}{9}-(-16)=\frac{164}{9}$ units.
5	$2p\times2q=-1$ , so $pq=-\frac14$ $y=pq=-\frac14$	4	The tangent gradients are $2p$ and $2q$ . Perpendicular gradients multiply to $-1$ , so $4pq=-1$ and $pq=-\frac14$ . The tangent at $P$ is $y=2px-p^2$ and the tangent at $Q$ is $y=2qx-q^2$ . Equating them, $2(p-q)x=p^2-q^2$ , so $x=\frac{p+q}{2}$ . Substituting into $y=2px-p^2$ gives $y=p(p+q)-p^2=pq$ . Hence the intersection has $y$ -coordinate $-\frac14$ .

C5 · Increasing and decreasing functions

Tier 1 · Easy

Mark scheme for C5 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x>3$	2	The function is increasing when $f'(x)>0$ . Solve $2x-6>0$ : $2x>6$ , so $x>3$ .
2	The function is decreasing at $x=2$	1	Since $f'(2)=-7<0$ , the function is decreasing at this point.

Tier 2 · Standard

Mark scheme for C5 Tier 2 · Standard
Q	Answer	Mark	Comments
1	Decreasing for $x<4$ Increasing for $x>4$	3	$\frac{dy}{dx}=2x-8=2(x-4)$ . This derivative is negative when $x<4$ and positive when $x>4$ . Therefore the curve decreases before $x=4$ and increases after $x=4$ .
2	Decreasing for $-2<x<5$ (accept $-2\le x\le 5$ )	2	The derivative is zero at $x=-2$ and $x=5$ . Between these values, $x+2$ is positive and $x-5$ is negative, so $f'(x)<0$ . Outside them the two factors have the same sign. Therefore the function is decreasing for $-2<x<5$ .
3	$f'(x)=3(x+1)^2+3>0$ for every real $x$ , so $f$ is increasing everywhere	3	$f'(x)=3x^2+6x+6$ . Completing the square gives $f'(x)=3(x+1)^2+3$ . Since $(x+1)^2\ge0$ , $f'(x)\ge3>0$ for every real $x$ , so the function is increasing everywhere.

Tier 3 · Hard

Mark scheme for C5 Tier 3 · Hard
Q	Answer	Mark	Comments
1	Increasing for $x<-1$ and $x>3$ Decreasing for $-1<x<3$	4	$f'(x)=3x^2-6x-9=3(x-3)(x+1)$ . The derivative is $0$ at $x=-1$ and $x=3$ . It is positive outside these roots and negative between them, so $f$ increases for $x<-1$ and $x>3$ , and decreases for $-1<x<3$ .
2	Increasing for $-1<x<0$ and $x>1$ Decreasing for $x<-1$ and $0<x<1$ (accept $\le$ and $\ge$ at the stationary values)	4	$f'(x)=4x^3-4x=4x(x-1)(x+1)$ , so the stationary values are $-1$ , $0$ and $1$ . Testing the factor signs gives $f'(x)<0$ for $x<-1$ , $f'(x)>0$ for $-1<x<0$ , $f'(x)<0$ for $0<x<1$ , and $f'(x)>0$ for $x>1$ . These signs give the stated decreasing and increasing intervals.
3	$-1\le k\le1$	4	$f'(x)=3x^2+6kx+3=3\big((x+k)^2+1-k^2\big)$ . This derivative is non-negative for every real $x$ precisely when its minimum value, $3(1-k^2)$ , is non-negative. Hence $1-k^2\ge0$ , so $k^2\le1$ and $-1\le k\le1$ .
4	Increasing for $x>2$ Decreasing for $x<0$ and $0<x<2$	4	$f'(x)=2x-16x^{-2}=\frac{2(x^3-8)}{x^2}=\frac{2(x-2)(x^2+2x+4)}{x^2}$ . For $x\ne0$ , both $x^2$ and $x^2+2x+4=(x+1)^2+3$ are positive, so the sign of $f'(x)$ is the sign of $x-2$ . Therefore $f'(x)<0$ on $x<0$ and $0<x<2$ , while $f'(x)>0$ on $x>2$ .
5	$a=-\frac{9}{2}$ and $b=-30$	4	$f'(x)=3x^2+2ax+b$ . Since the function is decreasing exactly between $-2$ and $5$ , these are the two zeros of the derivative. Its leading coefficient is $3$ , so $f'(x)=3(x+2)(x-5)=3x^2-9x-30$ . Comparing coefficients gives $2a=-9$ and $b=-30$ , hence $a=-\frac{9}{2}$ .

C6 · Understand and use the notation d2y/dx2; know that it measures the rate of change of the gradient function

Tier 1 · Easy

Mark scheme for C6 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$\frac{d^2y}{dx^2}=36x^2-10$	2	Differentiate once: $\frac{dy}{dx}=12x^3-10x$ . Differentiate again: $\frac{d^2y}{dx^2}=36x^2-10$ .
2	$9$	2	Differentiate the gradient function: $\frac{d^2y}{dx^2}=15x^2-6$ . At $x=-1$ , this is $15(-1)^2-6=9$ .

Tier 2 · Standard

Mark scheme for C6 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$x=2$	3	The rate of change of the gradient is $\frac{d^2y}{dx^2}$ . Differentiate again: $\frac{d^2y}{dx^2}=12x-4$ . Set $12x-4=20$ , so $12x=24$ and $x=2$ (consistent with $x>0$ ).
2	$a=2$	3	$\frac{dy}{dx}=3ax^2+6x-1$ , so $\frac{d^2y}{dx^2}=6ax+6$ . At $x=2$ , $12a+6=30$ . Hence $12a=24$ and $a=2$ .
3	$x=1$ or $x=3$	3	$\frac{dy}{dx}=4x^3-24x^2+36x$ , so $\frac{d^2y}{dx^2}=12x^2-48x+36$ . A stationary gradient has $\frac{d^2y}{dx^2}=0$ , so $12(x^2-4x+3)=12(x-1)(x-3)=0$ . Hence $x=1$ or $x=3$ .

Tier 3 · Hard

Mark scheme for C6 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$x<1-\frac{\sqrt{6}}{3}$ or $x>1+\frac{\sqrt{6}}{3}$	4	First $\frac{dy}{dx}=4x^3-12x^2+4x$ , so $\frac{d^2y}{dx^2}=12x^2-24x+4=4(3x^2-6x+1)$ . The gradient is increasing when $3x^2-6x+1>0$ . The boundary values are $x=\frac{6\pm\sqrt{36-12}}{6}=1\pm\frac{\sqrt{6}}{3}$ . The quadratic opens upwards, so it is positive outside the two roots.
2	$a=2$ and $b=-2$ The gradient is increasing for $x>\frac12$ .	4	Since $g'(x)=2ax+b$ , the two conditions give $-2a+b=-6$ and $8a+b=14$ . Subtracting gives $10a=20$ , so $a=2$ ; then $b=-2$ . Thus $g'(x)=4x-2$ . The gradient is increasing when $4x-2>0$ , so $x>\frac12$ .
3	The least value is $-48$ , occurring when $x=1$ .	4	$\frac{d^2y}{dx^2}=12x^2+6ax+2b$ . Using $x=-1$ and $x=3$ gives $12-6a+2b=0$ and $108+18a+2b=0$ . Subtracting gives $24a=-96$ , so $a=-4$ and then $b=-18$ . Therefore $\frac{d^2y}{dx^2}=12x^2-24x-36=12(x-1)^2-48$ , whose least value is $-48$ at $x=1$ .
4	$p=5$ The common rate is $-12$	3	Differentiate the gradient function: $g'(x)=6x^2-6px+12$ . The equal rates give $18-6p=108-24p$ , so $18p=90$ and $p=5$ . Substitution into either rate gives $g'(1)=18-30=-12$ .
5	$a=2$ and $b=3$ The rate when $x=2$ is $\frac{387}{4}$ (or $96.75$ )	4	$\frac{dy}{dx}=4ax^3-bx^{-2}$ , so $\frac{d^2y}{dx^2}=12ax^2+2bx^{-3}$ . The two given rates produce $12a+2b=30$ and $12a-2b=18$ . Adding gives $24a=48$ , so $a=2$ ; then $b=3$ . At $x=2$ , the rate is $12(2)(2^2)+2(3)(2^{-3})=96+\frac34=\frac{387}{4}$ .

C7 · Use of differentiation to find maxima and minima points on a curve

Tier 1 · Easy

Mark scheme for C7 Tier 1 · Easy
Q	Answer	Mark	Comments
1	The stationary point is a minimum.	2	$\frac{dy}{dx}=3x^2-6x$ , so $\frac{d^2y}{dx^2}=6x-6$ . At $x=2$ , $\frac{d^2y}{dx^2}=6>0$ , so the stationary point is a minimum.
2	$(3,2)$ is a minimum.	2	$\frac{dy}{dx}=2x-6=0$ gives $x=3$ . Then $y(3)=9-18+11=2$ . Since $\frac{d^2y}{dx^2}=2>0$ , $(3,2)$ is a minimum.

Tier 2 · Standard

Mark scheme for C7 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$(1,5)$ is a maximum. $(3,1)$ is a minimum.	3	Substitution gives $y(1)=1-6+9+1=5$ and $y(3)=27-54+27+1=1$ . Also $\frac{d^2y}{dx^2}=6x-12$ . This is $-6$ at $x=1$ , giving a maximum, and $6$ at $x=3$ , giving a minimum.
2	$(2,-16)$ is a minimum.	3	$\frac{dy}{dx}=3x^2-12=3(x-2)(x+2)$ . The positive stationary value is $x=2$ , giving $y(2)=8-24=-16$ . Since $\frac{d^2y}{dx^2}=6x$ is positive at $x=2$ , $(2,-16)$ is a minimum.
3	$(2,1)$ is a minimum.	3	Since $(0,5)$ lies on the curve, $k=5$ . Also $\frac{dy}{dx}=3x^2-6x=3x(x-2)$ , so the other stationary value is $x=2$ . Then $y(2)=8-12+5=1$ . As $\frac{d^2y}{dx^2}=6x-6$ is positive at $x=2$ , $(2,1)$ is a minimum.

Tier 3 · Hard

Mark scheme for C7 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$(-2,-29)$ is a minimum. $(0,3)$ is a maximum. $(2,-29)$ is a minimum.	4	$\frac{dy}{dx}=8x^3-32x=8x(x-2)(x+2)$ , so $x=-2,0,2$ . Substitution gives $y=-29,3,-29$ respectively. Since $\frac{d^2y}{dx^2}=24x^2-32$ , its values are $64,-32,64$ . Positive values give minima at $(-2,-29)$ and $(2,-29)$ ; the negative value gives a maximum at $(0,3)$ .
2	$a=-6$ and $b=9$ $(1,4)$ is a maximum. $(3,0)$ is a minimum.	4	The derivative is $3x^2+2ax+b$ . Its roots are $1$ and $3$ , so $3x^2+2ax+b=3(x-1)(x-3)=3x^2-12x+9$ . Hence $a=-6$ and $b=9$ . The curve is $y=x^3-6x^2+9x$ , giving $(1,4)$ and $(3,0)$ . Since $\frac{d^2y}{dx^2}=6x-12$ , its values are $-6$ and $6$ , so the points are respectively a maximum and a minimum.
3	$(-2,16)$ is a maximum. $(2,-16)$ is a minimum.	4	$\frac{dy}{dx}=3(x^2-a^2)$ , so the stationary values are $x=-a$ and $x=a$ . Their $y$ -coordinates are $2a^3$ and $-2a^3$ , so their vertical distance is $4a^3$ . Thus $4a^3=32$ , giving $a=2$ . The points are $(-2,16)$ and $(2,-16)$ . Since $\frac{d^2y}{dx^2}=6x$ , the first is a maximum and the second is a minimum.
4	$(3,-27)$ is a local minimum $(0,0)$ is neither a maximum nor a minimum, because the curve is decreasing on both sides of $x=0$	4	$\frac{dy}{dx}=4x^3-12x^2=4x^2(x-3)$ , so the stationary values are $x=0$ and $x=3$ , giving $(0,0)$ and $(3,-27)$ . Also $\frac{d^2y}{dx^2}=12x(x-2)$ , which is positive at $x=3$ , so $(3,-27)$ is a local minimum. At $x=0$ the second derivative is zero, but $4x^2(x-3)<0$ on both sides of $0$ , so the curve is decreasing on both sides and $(0,0)$ is neither a maximum nor a minimum.
5	$k=8$ $(-1,13)$ is a local maximum and $(3,-19)$ is a local minimum	4	$\frac{dy}{dx}=3x^2-6x-9=3(x+1)(x-3)$ , so the stationary values are $x=-1$ and $x=3$ . Since $\frac{d^2y}{dx^2}=6x-6$ , the point with $x=-1$ is the local maximum and the point with $x=3$ is the local minimum. The line gives the maximum's ordinate as $2(-1)+15=13$ . On the curve this ordinate is $5+k$ , so $k=8$ . The other ordinate is $3^3-3(3^2)-9(3)+8=-19$ .

C8 · Using calculus to find maxima and minima in simple problems

Tier 1 · Easy

Mark scheme for C8 Tier 1 · Easy
Q	Answer	Mark	Comments
1	The maximum value of $P$ is $49$ .	2	$P=14x-x^2$ , so $\frac{dP}{dx}=14-2x$ . Setting this to zero gives $x=7$ . Then $P(7)=7(14-7)=49$ ; the negative $x^2$ coefficient confirms a maximum.
2	$t=2$ seconds	2	$\frac{dh}{dt}=12-6t$ . Setting this to zero gives $t=2$ . Since $\frac{d^2h}{dt^2}=-6<0$ , this gives the greatest height.

Tier 2 · Standard

Mark scheme for C8 Tier 2 · Standard
Q	Answer	Mark	Comments
1	The maximum area is $36\text{ cm}^2$ .	3	$A=(x+2)(10-x)=-x^2+8x+20$ . Hence $\frac{dA}{dx}=-2x+8$ , so the stationary value is at $x=4$ . The second derivative is $-2<0$ , so this is a maximum. $A(4)=6\times6=36\text{ cm}^2$ .
2	The maximum area is $72\text{ m}^2$ .	3	Let each side perpendicular to the wall be $x$ metres. The remaining side is $24-2x$ , so $A=x(24-2x)=24x-2x^2$ . Then $\frac{dA}{dx}=24-4x=0$ gives $x=6$ . Since $\frac{d^2A}{dx^2}=-4<0$ , this is a maximum, and $A(6)=6\times12=72\text{ m}^2$ .
3	The least value of $x+y$ is $24\text{ cm}$ .	3	Since $y=\frac{144}{x}$ , let $S=x+y=x+144x^{-1}$ for $x>0$ . Then $\frac{dS}{dx}=1-144x^{-2}$ . Setting this to zero gives $x^2=144$ , so $x=12$ and $y=12$ . Also $\frac{d^2S}{dx^2}=288x^{-3}>0$ , confirming a minimum, and $x+y=24\text{ cm}$ .

Tier 3 · Hard

Mark scheme for C8 Tier 3 · Hard
Q	Answer	Mark	Comments
1	The maximum area is $32$ square units.	4	The width is $2x$ and the height is $12-x^2$ , so $A=2x(12-x^2)=24x-2x^3$ . Then $\frac{dA}{dx}=24-6x^2$ . Setting this to zero and using $x>0$ gives $x=2$ . Also $\frac{d^2A}{dx^2}=-12x=-24<0$ , so this is a maximum. $A(2)=4(12-4)=32$ .
2	The maximum volume is $216\text{ cm}^3$ .	4	The constraint gives $h=18-2x$ , so $V=x^2h=x^2(18-2x)=18x^2-2x^3$ , with $0<x<9$ . Then $\frac{dV}{dx}=36x-6x^2=6x(6-x)$ , so the interior stationary value is $x=6$ . Also $\frac{d^2V}{dx^2}=36-12x=-36<0$ at $x=6$ , so this is a maximum. Here $h=6$ and $V=6^3=216\text{ cm}^3$ .
3	The least surface area is $192\text{ cm}^2$ .	4	From $x^2h=256$ , $h=256x^{-2}$ . The base and four sides have total area $S=x^2+4xh=x^2+1024x^{-1}$ . Hence $\frac{dS}{dx}=2x-1024x^{-2}$ . Setting this to zero gives $2x^3=1024$ , so $x=8$ and $h=4$ . Since $\frac{d^2S}{dx^2}=2+2048x^{-3}=6>0$ at $x=8$ , this is a minimum, and $S=8^2+4(8)(4)=192\text{ cm}^2$ .
4	The greatest height is $18\text{ m}$ .	4	The equal heights give $2+p-4=2+3p-36$ , so $p=16$ . Hence $h=2+16t-4t^2$ and $\frac{dh}{dt}=16-8t$ . The stationary time is $t=2$ , and $\frac{d^2h}{dt^2}=-8<0$ , so it gives the greatest height. Therefore $h(2)=2+32-16=18\text{ m}$ .
5	$PQ^2=x^2+(x^2-3)^2=x^4-5x^2+9$ Least $PQ^2=\frac{11}{4}$ , so the shortest distance is $PQ=\frac{\sqrt{11}}{2}$ units (or $\sqrt{\frac{11}{4}}$ units)	4	The squared distance is $D^2=x^2+(x^2-3)^2=x^4-5x^2+9$ . Minimising $D^2$ also minimises the positive distance $D$ . Differentiate: $\frac{d(D^2)}{dx}=4x^3-10x=2x(2x^2-5)$ . Since $x>0$ , the stationary value is $x=\sqrt{\frac52}$ . The second derivative is $12x^2-10=20>0$ there, so this gives a minimum. Substitution gives $D^2=\frac{11}{4}$ , hence $D=\frac{\sqrt{11}}{2}$ units.

C9 · Sketch/interpret a curve with known maximum and minimum points

Tier 1 · Easy

Mark scheme for C9 Tier 1 · Easy
Q	Answer	Mark	Comments
1	The curve is decreasing for $-2<x<1$ .	1	After a local maximum the curve falls, and it continues to fall until the local minimum. Therefore it is decreasing between the two stationary $x$ -values.
2	$x=4$	1	At $x=4$ the curve changes from decreasing to increasing, so this is the local minimum.

Tier 2 · Standard

Mark scheme for C9 Tier 2 · Standard
Q	Answer	Mark	Comments
1	A smooth cubic rising to $L=(-1,9)$ , falling through $N=(0,2)$ to $M=(2,-18)$ , then rising again.	3	At $x=0$ , $y=2$ , so $N=(0,2)$ . Plot $L$ , $M$ and $N$ . The positive-leading cubic comes from the lower left and rises to $L$ , decreases through $N$ to $M$ , then rises towards the upper right. Join the points smoothly with horizontal tangents at $L$ and $M$ .
2	The curve is decreasing for $-3<x<2$ . Between the turning points the curve falls continuously from $y=7$ to $y=-1$ ; since it passes from a positive value to a negative value, it must cross the $x$ -axis between $x=-3$ and $x=2$ .	3	A curve decreases between a local maximum and the next local minimum, so it is decreasing for $-3<x<2$ . On that interval the curve falls from $f(-3)=7$ to $f(2)=-1$ ; a smooth curve passing from a positive value to a negative value must cross the $x$ -axis, so there is a root between $x=-3$ and $x=2$ .
3	The curve is increasing for $x<-1$ and for $x>3$ . A smooth curve rising to $P$ , falling to $Q$ , then rising again, with horizontal tangents at both labelled points.	3	A curve rises before a local maximum and after the following local minimum, so it is increasing for $x<-1$ and for $x>3$ . Draw a smooth curve rising to $P(-1,6)$ , falling to $Q(3,-2)$ and then rising, with a horizontal tangent at each turning point.

Tier 3 · Hard

Mark scheme for C9 Tier 3 · Hard
Q	Answer	Mark	Comments
1	A smooth curve crossing at $(-4,0)$ , rising to $(-2,5)$ , crossing at $(1,0)$ , falling to $(3,-4)$ , then crossing at $(5,0)$ and rising.	4	Place the three roots on the $x$ -axis and the maximum and minimum at their stated coordinates. The order in $x$ is $-4,-2,1,3,5$ . Draw one smooth curve that crosses each root, turns at the maximum and minimum only, and has horizontal tangents at those turning points.
2	A smooth W-shaped curve descending from the upper left to $(-3,-2)$ , rising to $(0,4)$ , falling to $(2,-5)$ , then rising to the upper right, with horizontal tangents at all three labelled points. The curve is increasing for $-3<x<0$ and for $x>2$ .	4	Order the turning points by their $x$ -coordinates: $-3,0,2$ . Starting high on the left, draw the curve down to the first minimum, up to the maximum, down to the second minimum and then up to the right, using horizontal tangents at each turning point. The rising sections are therefore $-3<x<0$ and $x>2$ .
3	The curve crosses the $x$ -axis at $R=(-4,0)$ and touches it at $Q=(1,0)$ . Sketch: the curve rises from the lower left through $R$ , reaches the maximum $P$ , falls to touch the axis at $Q$ , then rises to the upper right, with horizontal tangents at $P$ and $Q$ .	4	The curve begins below the $x$ -axis and rises through $R=(-4,0)$ , so $R$ is a crossing. It continues to the maximum $P$ , then falls to the minimum $Q=(1,0)$ . Since $Q$ is a minimum on the axis, the curve touches the axis there and turns upwards. Complete the sketch by making it rise without bound to the right, with horizontal tangents at $P$ and $Q$ .
4	$-1<k<7$	3	A horizontal line $y=k$ must meet each of the three sections of the curve: before the maximum, between the two turning points and after the minimum. This happens when its height is strictly above the minimum value $-1$ and strictly below the maximum value $7$ . At either boundary it meets a turning point and gives only two distinct intersections. Therefore $-1<k<7$ .
5	A smooth curve rising from $A$ to $P$ , falling to $Q$ , then rising to $B$ , with horizontal tangents at $P$ and $Q$ . $-4\le f(x)\le9$	4	Order the points by increasing $x$ : $A$ , $P$ , $Q$ , $B$ . Draw the curve rising from $A$ to the local maximum $P$ , falling to the local minimum $Q$ , then rising to $B$ , with no extra turns and with horizontal tangents at $P$ and $Q$ . The smallest ordinate is $-4$ at the left endpoint and the greatest is $9$ at the right endpoint, so the range is $-4\le f(x)\le9$ .