2016 H2 Maths Paper 1 Q2 — Worked Solution

Part (i)

2 marks

Since $y = 2^{\cos x}$ ,

\frac{dy}{dx} = 2^{\cos x}\ln 2 \cdot \left( - \sin x \right)

By GC,

\left. \ \frac{dy}{dx} \right|_{x = 0} = 2^{1}\ln 2\ \left( - \sin 0 \right) = 0

\left. \ \frac{dy}{dx} \right|_{x = \frac{1}{2}\pi} = 2^{0}\ln 2\ \left( - \sin\frac{1}{2}\pi \right) = - \ln 2 \approx - 0.693

$\therefore$ the gradient at $x = 0$ is $0$ , and the gradient at $x = \frac{1}{2}\pi$ is $- \ln 2 \approx - 0.693$ .

Part (ii)

3 marks

Approach

With both gradients and both points known, each tangent is written down directly from $y - y_{1} = m\left( x - x_{1} \right)$ ; the two lines are then solved simultaneously for their point of intersection.

At $x = 0$ , $y = 2^{\cos 0} = 2^{1} = 2$ , and the gradient there is $0$ , so the tangent is horizontal:

y = 2

At $x = \frac{1}{2}\pi$ , $y = 2^{\cos\frac{1}{2}\pi} = 2^{0} = 1$ , and the gradient there is $- \ln 2$ , so the tangent is

y - 1 = - \ln 2\left( x - \frac{1}{2}\pi \right)

y = 1 - \ln 2\left( x - \frac{1}{2}\pi \right)

To find where the two tangents meet, we substitute $y = 2$ into the second tangent:

2 = 1 - \ln 2\left( x - \frac{1}{2}\pi \right)

1 = - \ln 2\left( x - \frac{1}{2}\pi \right)

x - \frac{1}{2}\pi = - \frac{1}{\ln 2}

x = \frac{1}{2}\pi - \frac{1}{\ln 2}

By GC, $x \approx 0.128$ .

To check your answer: substituting

x = \frac{1}{2}\pi - \frac{1}{\ln 2} \approx 0.128

into

y = 1 - \ln 2\left( x - \frac{1}{2}\pi \right)

gives

y \approx 1 - \ln 2\left( - 1/\ln 2 \right) = 1 + 1 = 2

, matching the first tangent. ✓

$\therefore$ the tangents are $y = 2$ and $y = 1 - \ln 2\left( x - \frac{1}{2}\pi \right)$ , meeting at the point $\left( \frac{1}{2}\pi - \frac{1}{\ln 2},\ 2 \right) \approx (0.128,\ 2)$ .

2016 H2 Maths Paper 1, Question 2

Part (i)

Part (ii)

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