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Past Papers

## Calculus

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Slopes and Tangents

Maxima and Minima

Curve Sketching

Rates of Change

Differential Equations

## 2022 Paper 1 Question 10(a)

Keith plays hurling.

(a) During a match, Keith hits the ball with his hurl.
The height of the ball could be modelled by the following quadratic function:

\begin{align}h=-2t^2+5t+1.2\end{align}

where $$h$$ is the height of the ball, in metres, $$t$$ seconds after being hit, and $$t\in\mathbb{R}$$.

(i) How high, in metres, was the ball when it was hit (when $$t=0$$)?

(ii) The ball was caught after $$2.4$$ seconds.
How high, in metres, was the ball when it was caught?

(iii) When the ball passed over the halfway line, it was at a height of $$3.2$$ metres and its height was decreasing.

How many seconds after it was hit did the ball pass over the halfway line?
Remember that $$h=-2t^2+5t+1.2$$.

(iv) Find $$\dfrac{dh}{dt}$$ and hence find how long it took the ball to reach its greatest height.

(i) $$1.2\mbox{ m}$$

(ii) $$1.68\mbox{ m}$$

(iii) $$2\mbox{ s}$$

(iv) $$\dfrac{5}{4}\mbox{ s}$$

Solution

(i)

\begin{align}h(0)&=-2(0^2)+5(0)+1.2\\&=1.2\mbox{ m}\end{align}

(ii)

\begin{align}h(2.4)&=-2(2.4^2)+5(2.4)+1.2\\&=1.68\mbox{ m}\end{align}

(iii)

\begin{align}h(t)&=3.2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-2t^2+5t+1.2=3.2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2t^2-5t+2=0\end{align}

\begin{align}\downarrow\end{align}

$$t=\dfrac{1}{2}$$ or $$t=2$$

As the height was decreasing, it must be the second time, i.e. $$2$$ seconds.

(iv)

\begin{align}\frac{dh}{dt}=-4t+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-4t+5=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t=\frac{5}{4}\mbox{ s}\end{align}

Video Walkthrough

## 2021 Paper 1 Question 5(c)

The function, $$f$$ is defined as $$f(x)=3x^2-6x+7$$, where $$x\in\mathbb{R}$$.

(c) Use calculus to find the co-ordinates of the local minimum point of $$f$$.

$$(1,4)$$

Solution

\begin{align}f'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x-6=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=1\end{align}

and

\begin{align}f(1)&=3(1^2)-6(1)+7\\&=4\end{align}

\begin{align}\downarrow\end{align}

Minimum: $$(1,4)$$

Video Walkthrough

## 2020 Paper 1 Question 6

(a)

(i) Differentiate the function $$f(x)=4x^3-3x^2+x-7$$, where $$x\in\mathbb{R}$$, with respect to $$x$$.

(ii) Find the slope of the tangent to the graph of $$f(x)=4x^3-3x^2+x-7$$
at the point $$(1,-5)$$.

(iii) Hence find the equation of the tangent to the graph at this point.

(i) $$12x^2-6x+1$$

(ii) $$7$$

(iii) $$y=7x-12$$

Solution

(i)

\begin{align}f'(x)=12x^2-6x+1\end{align}

(ii)

\begin{align}f'(1)&=12(1^2)-6(1)+1\\&=7\end{align}

(iii)

\begin{align}y-y_1=m(x-x_1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y-(-5)=7(x-1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y+5=7x-7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y=7x-12\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) The function $$g(x)=2x^2+px+q$$, where $$p,q\in\mathbb{Z}$$, and $$x\in\mathbb{R}$$.
Given that $$g(2)=6$$ and $$g'(3)=9$$, find the value of $$p$$ and the value of $$q$$.
Note: $$g'(3)$$ is the value of the derivative of $$g(x)$$ at $$x=3$$.

$$p=-3$$ and $$q=4$$

Solution

\begin{align}g(2)=6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2(2^2)+p(2)+q=6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2p+q=-2\end{align}

and

\begin{align}g'(3)=9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4(3)+p=9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p&=9-12\\&=-3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q&=-2-2p\\&=-2-2(-3)\\&=4\end{align}

Video Walkthrough

## 2020 Paper 1 Question 8(c)

A swimmer is on a starting block at the beginning of a race. When she dives off the block until she resurfaces, the level of the swimmer relative to the level of the water is given by the function:

\begin{align}h(x)=\frac{1}{60}x^2-\frac{1}{4}x+\frac{3}{5}\end{align}

In the function, $$x$$ is the horizontal distance in metres of the swimmer from the block, $$0\leq x\leq 12$$, where $$x\in\mathbb{R}$$ and $$h(x)$$ is measured in metres. (c)

(i) Find $$h'(x)$$, the derivative of $$h(x)=\dfrac{1}{60}x^2-\dfrac{!}{4}x+\dfrac{3}{5}$$.

(ii) Use your answer to Part (c)(i) to find the horizontal distance ($$x$$), in metres, from the starting block to the point at which the swimmer reaches her greatest depth.

(iii) Hence find this greatest depth.

(i) $$h'(x)=\frac{1}{30}x-\frac{1}{4}$$

(ii) $$7.5\mbox{ m}$$

(iii) $$\dfrac{27}{80}\mbox{ m}$$

Solution

(i)

\begin{align}h'(x)=\frac{1}{30}x-\frac{1}{4}\end{align}

(ii)

\begin{align}\frac{1}{30}x-\frac{1}{4}=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{30}{4}\\&=7.5\mbox{ m}\end{align}

(iii)

\begin{align}h(7.5)&=\frac{1}{60}(7.5)^2-\frac{1}{4}(7.5)+\frac{3}{5}\\&=\frac{27}{80}\mbox{ m}\end{align}

Video Walkthrough

## 2019 Paper 1 Question 3(b)

The function $$f$$ is defined as $$f(x)=-x^3+4x^2+x-2$$, where $$x\in\mathbb{R}$$.

(b) Find the value of $$x$$ for which $$f^{\prime\prime}(x)=0$$, where $$f^{\prime\prime}(x)$$ is the second derivative of $$f(x)$$.

$$\dfrac{4}{3}$$

Solution

\begin{align}f(x)=-x^3+4x^2+x-2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f'(x)=-3x^2+8x+1-2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f^{\prime\prime}(x)=-6x+8\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f^{\prime\prime}(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-6x+8=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=\frac{4}{3}\end{align}

Video Walkthrough

## 2019 Paper 1 Question 7(d)

A camogie goalkeeper, on a level pitch, hitaball straight up into the air.
The path that the ball travelled can be modelled by the function:

$$f(t)=-4t^2+16t+1$$, $$t\in\mathbb{R}$$

where $$t$$ is the time, in seconds, from when the ball is hit and $$f(t)$$ was the height of the ball, in metres, above the pitch. The ball landed on the ground without being hit again.

(d)

(i) Find $$f'(t)$$, the derivative of $$f(t)=-4t^2+16t+1$$.

(ii) Use your answer from part (d)(i) to find the speed of the ball when it had been in the air for $$4$$ seconds. Give your answer in metres per second.

(iii) Use your answer from part (d)(i) to find the value of $$t$$ for which the ball was descending and travelling at a speed of $$8$$ metres per second.

(i) $$f'(t)=-8t+16$$

(ii) $$16\mbox{ m/s}$$

(iii) $$3\mbox{ s}$$

Solution

(i)

\begin{align}f'(t)&=-8t+16\end{align}

(ii)

\begin{align}f'(4)&=-8(4)+16\\&=-16\mbox{ m/s}\end{align}

Therefore, the speed of the ball is $$16\mbox{ m/s}$$.

(iii)

\begin{align}f'(t)=-8\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-8t+16=-8\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{-8-16}{-8}\\&=3\mbox{ s}\end{align}

Video Walkthrough

## 2018 Paper 1 Question 5(c)

The diagram shows the graph of a quadratic function $$f$$.

It can be shown that this function can be written as $$f(x)=-x^2+x+6$$

(c) Show, using calculus, that the maximum point of $$f(x)$$ is $$(0.5,6.25)$$.

Solution

\begin{align}f'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-2x+1=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=0.5\end{align}

and

\begin{align}f(0.5)&=-(0.5^2)+0.5+6\\&=6.25\end{align}

Video Walkthrough

## 2018 Paper 1 Question 8(e)

The amount, in appropriate units, of a certain medicinal drug in the bloodstream $$t$$ hours after it has been taken can be estimated by the function:

$$C(t)=-t^3+4.5t^2+54t$$, where $$0\leq t\leq9$$, $$t\in\mathbb{R}$$

(e)

(i) Use the drug amount function $$C(t)=-t^3+4.5t^2+54t$$ to find, in terms of $$t$$, the rate at which the drug amount is changing after $$4$$ hours.

(ii) Use your answer to part e(i) to find the rate at which the drug amount is changing after $$4$$ hours.

(iii) Use your answer to part e(i) to find the maximum amount of the drug in the bloodstream over the first $$9$$ hours.

(iv) Use your answer to part e(i) to show that the drug amount in the bloodstream is
decreasing $$7$$ hours after the drug has been taken. Explain your reasoning.

(i) $$C'(t)=-3t^2+9t+54$$

(ii) $$42\mbox{ units/hr}$$

(iii) $$270\mbox{ units}$$

(iv) $$C'(7)=-30\mbox{ units/hr}$$. Therefore, the amount is decreasing as the rate of change is negative.

Solution

(i)

\begin{align}C'(t)&=-3t^2+9t+54\end{align}

(ii)

\begin{align}C'(4)&=-3(4^2)+9(4)+54\\&=42\mbox{ units/hr}\end{align}

(iii)

\begin{align}C'(t)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-3t^2+9t+54=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t^2-3t-18=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(t+3)(t-6)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t=6\mbox{ s}\end{align}
(as $$t>0$$)

\begin{align}\downarrow\end{align}

\begin{align}C(6)&=-6^3+4.5(6^2)+54(6)\\&=270\mbox{ units}\end{align}

(iv)

\begin{align}C'(7)&=-3(7^2)+9(7)+54\\&=-30\mbox{ units/hr}\end{align}

Therefore, the amount is decreasing as the rate of change is negative.

Video Walkthrough

## 2017 Paper 1 Question 3(b)

(b) Find the co-ordinates of the minimum point of the function $$f(x)=3x^2-6x-8$$, where $$x\in\mathbb{R}$$.

$$(1,-11)$$

Solution

\begin{align}f'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x-6=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f(1)&=3(1^2)-6(1)-8\\&=-11\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(1,-11)\end{align}

Video Walkthrough

## 2017 Paper 1 Question 8(d)

A company makes and sells fibre optic cable. It can sell, at most, $$200$$ kilometres of cable in a week. For a certain range of its production the company has found that profit can be modelled using the function:

$$P(x)=275x-x^2-2000$$, where $$x\leq 200$$

In the function, $$x$$ is the number of kilometres of fibre optic cable sold and $$P(x)$$ is the profit in euro.

(d) Use calculus to find the number of kilometres of cable sold when the profit is increasing at a rate of €$$105$$ per $$\mbox{km}$$.

$$85\mbox{ km}$$

Solution

\begin{align}P'(x)=105\end{align}

\begin{align}\downarrow\end{align}

\begin{align}275-2x=105\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{275-105}{2}\\&=85\mbox{ km}\end{align}

Video Walkthrough

## 2016 Paper 1 Question 4(c)

The function $$f:x \rightarrow x^3+x^2-2x+7$$ is defined for $$x\in\mathbb{R}$$.

(c)

(i) Find $$f'(x)$$, the derivative of $$f(x)$$.
Hence find the slope of the tangent to the graph of $$f$$ when $$x=1$$.

(ii) Hence, find the equation of the tangent to the graph of f at the point $$A(1,7)$$.

(i) $$f'(x)=3x^2+2x-2$$ and $$m=3$$

(ii) $$y=3x+4$$

Solution

(i)

\begin{align}f'(x)=3x^2+2x-2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f'(1)&=3(1^2)+2(1)-2\\&=3\end{align}

(ii)

\begin{align}y-7=3(x-1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y-7=3x-3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y=3x+4\end{align}

Video Walkthrough

## 2016 Paper 1 Question 8(d)

Kieran has $$21$$ metres of fencing. He wants to enclose a vegetable garden in a rectangular shape as shown.

It can be shown that $$y=10.5-x$$.

(d)

(i) Show that the area of the rectangle can be written as $$A=10.5x-x^2$$.

(ii) Find $$\dfrac{dA}{dx}$$.

(iii) Hence, find the value of $$x$$ which will give the maximum area.

(iv) Find this maximum area.

(ii) $$\dfrac{dA}{dx}=10.5-2x$$

(iii) $$5.25\mbox{ m}$$

(iv) $$27.5625\mbox{ m}^2$$

Solution

(i)

\begin{align}A&=xy\\&=x(10.5-x)\\&=10.5x-x^2\end{align}

as required.

(ii)

\begin{align}\frac{dA}{dx}=10.5-2x\end{align}

(iii)

\begin{align}A'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}10.5-2x=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{10.5}{2}\\&=5.25\mbox{ m}\end{align}

(iv)

\begin{align}A(5.25)&=10.5(5.25)-5.25^2\\&=27.5625\mbox{ m}^2\end{align}

Video Walkthrough

## 2015 Paper 1 Question 4(b) & (c)

(b) Find the co-ordinates of the turning point of the function $$f(x)=-x^2+6x-4$$, $$x\in\mathbb{R}$$.

$$(3,5)$$

Solution

\begin{align}f'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-2x+6=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=3\end{align}

and

\begin{align}f(3)&=-3^2+6(3)-4\\&=5\end{align}

Therefore, the turning point is $$(3,5)$$.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c) The roots of this function, correct to one decimal place, can be found to be $$x=0.8$$ and $$x=5.2$$.

Use this, together your answer to parts (b) above, to sketch the curve $$y=f(x)$$.
Show your scale on both axes.

Solution
Video Walkthrough

## 2015 Paper 1 Question 8(d)

The daily profit of an oil trader is given by the profit function $$p=96x-0.03x^2$$, where $$p$$ is the daily profit, in euro, and $$x$$ is the number of barrels of oil traded in a day.

(d)

(i) Use calculus to find the number of barrels of oil traded that will earn the maximum daily profit.

(ii) Find this maximum profit.

(i) $$1{,}600\mbox{ barrels}$$

(ii) $$76{,}800\mbox{ euro}$$

Solution

(i)

\begin{align}p'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}96-0.03x=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{96}{0.06}\\&=1{,}600\mbox{ barrels}\end{align}

(ii)

\begin{align}p(1{,}600)&=96(1{,}600)-0.03(1{,}600)^2\\&=76{,}800\mbox{ euro}\end{align}

Video Walkthrough