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## Calculus

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

**(BLUE = Paper 1, RED = Paper 2)**

## 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.

Give your answer in seconds.

**(i) **\(1.2\mbox{ m}\)

**(ii) **\(1.68\mbox{ m}\)

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

**(iv) **\(\dfrac{5}{4}\mbox{ s}\)

**(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}

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## 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 \(f\).

\((1,4)\)

\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:** \((1,4)\)

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## 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) **\(y=7x-12\)

**(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}

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**(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 \(q=4\)

\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}

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## 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) **\(\dfrac{27}{80}\mbox{ m}\)

**(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}

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## 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 \(f(x)\).

\(\dfrac{4}{3}\)

\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}

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## 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) **\(3\mbox{ s}\)

**(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}

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## 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 \((0.5,6.25)\).

The answer is already in the question!

\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}

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## 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.

**(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.

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## 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 \(x\in\mathbb{R}\).

\((1,-11)\)

\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}

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## 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 \(\mbox{km}\).

\(85\mbox{ km}\)

\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}

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## 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) **\(y=3x+4\)

**(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}

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## 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.

**(i) **The answer is already in the question!

**(ii) **\(\dfrac{dA}{dx}=10.5-2x\)

**(iii) **\(5.25\mbox{ m}\)

**(iv) **\(27.5625\mbox{ m}^2\)

**(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}

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

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

\((3,5)\)

\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 \((3,5)\).

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**(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.

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## 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) **\(76{,}800\mbox{ euro}\)

**(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}