2019 H2 Maths Paper 1 Q7 — Worked Solution

Part (i)

6 marks

Since $y = xe^{- x}$ , we use the product rule to differentiate.

\frac{dy}{dx} = (1)e^{- x} + x\left( - e^{- x} \right)

= e^{- x}(1 - x).

At $x = 1$ :

y = (1)e^{- 1} = \frac{1}{e},\quad\quad\frac{dy}{dx} = e^{- 1}(1 - 1) = 0.

The tangent has gradient $0$ and passes through $\left( 1,\ \frac{1}{e} \right)$ , so it is the horizontal line

y = \frac{1}{e}.

At $x = - 1$ :

y = ( - 1)e^{1} = - e,\quad\quad\frac{dy}{dx} = e^{1}\left( 1 - ( - 1) \right) = 2e.

Using $y - y_{1} = m\left( x - x_{1} \right)$ with $\left( x_{1},y_{1} \right) = ( - 1, - e)$ and $m = 2e$ ,

y - ( - e) = 2e\left( x - ( - 1) \right)

y + e = 2e(x + 1)

y = 2ex + 2e - e

y = 2ex + e.

To check your answer: at

x = - 1

2e( - 1) + e = - e

, matching

y = - e

found above. ✓

$\therefore$ the tangent at $x = 1$ is $y = \frac{1}{e}$ , and the tangent at $x = - 1$ is $y = 2ex + e$ .

Part (ii)

2 marks

Approach

The acute angle between two lines of gradients $m_{1}$ and $m_{2}$ satisfies $\tan\theta = \left| \frac{m_{2} - m_{1}}{1 + m_{1}m_{2}} \right|$ . Here one gradient is $0$ , so this simplifies a great deal.

Using the gradients found in (i), $m_{1} = 0$ and $m_{2} = 2e$ ,

\tan\theta = \left| \frac{2e - 0}{1 + (0)(2e)} \right|

= 2e.

By GC, $\theta = \arctan(2e) = {79.6}^{\circ}$ (1 d.p.), which is already acute.

$\therefore$ the acute angle between the tangents is ${79.6}^{\circ}$ (1 d.p.).

2019 H2 Maths Paper 1, Question 7

Part (i)

Part (ii)

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