2023 H2 Maths Paper 2 Q2 — Worked Solution

Part (a)

3 marks

y = \ln\left( \sec x \right)

Differentiating using the chain rule,

\frac{dy}{dx} = \frac{\sec x\tan x}{\sec x}

= \tan x

Differentiating again,

\frac{d^{2}y}{dx^{2}} = \sec^{2}x

Differentiating once more,

\frac{d^{3}y}{dx^{3}} = 2\sec x \cdot \sec x\tan x

= 2\sec^{2}x\tan x

Since $\frac{d^{2}y}{dx^{2}} = \sec^{2}x$ and $\frac{dy}{dx} = \tan x$ ,

\frac{d^{3}y}{dx^{3}} = 2\sec^{2}x\tan x = 2\left( \frac{d^{2}y}{dx^{2}} \right)\left( \frac{dy}{dx} \right)\ \ \left( \text{shown} \right)

Part (b)

4 marks

Approach

Part (a) gives $y'''$ in terms of $y''$ and $y'$ , so rather than differentiating $2\sec^{2}x\tan x$ from scratch, we differentiate the relation in (a) once more (product rule) to get $y⁗$ , then evaluate every derivative at $x = 0$ .

Evaluating the derivatives at $x = 0$ ,

y(0) = \ln\left( \sec 0 \right) = \ln 1 = 0,\quad y'(0) = \tan 0 = 0,\quad y''(0) = \sec^{2}(0) = 1

Using the result of (a), $y'''(0) = 2\, y''(0)\, y'(0) = 2(1)(0) = 0$ .

Differentiating the relation in (a) once more,

\frac{d^{4}y}{dx^{4}} = 2\left\lbrack \frac{d^{3}y}{dx^{3}} \cdot \frac{dy}{dx} + \frac{d^{2}y}{dx^{2}} \cdot \frac{d^{2}y}{dx^{2}} \right\rbrack

= 2\left\lbrack \frac{d^{3}y}{dx^{3}} \cdot \frac{dy}{dx} + \left( \frac{d^{2}y}{dx^{2}} \right)^{2} \right\rbrack

At $x = 0$ ,

y⁗(0) = 2\left\lbrack (0)(0) + (1)^{2} \right\rbrack = 2

Using the Maclaurin series $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^{2} + \frac{y'''(0)}{3!}x^{3} + \frac{y⁗(0)}{4!}x^{4} + \cdots$ ,

y = 0 + 0 + \frac{1}{2}x^{2} + 0 + \frac{2}{24}x^{4} + \cdots

= \frac{x^{2}}{2} + \frac{x^{4}}{12} + \cdots

To check your answer: at

x = 0.1

, the series gives

\frac{(0.1)^{2}}{2} + \frac{(0.1)^{4}}{12} \approx 0.005008

, close to the direct value

\ln\left( \sec 0.1 \right) \approx 0.005005

, as expected for small

x

✓

$\therefore$ $\ln\left( \sec x \right) \approx \frac{x^{2}}{2} + \frac{x^{4}}{12}$ .

Part (c)

2 marks

Putting $x = \frac{\pi}{4}$ ,

y = \ln\left( \sec\frac{\pi}{4} \right) = \ln\sqrt{2} = \frac{1}{2}\ln 2

Using the expansion from (b),

\frac{1}{2}\ln 2 \approx \frac{(\pi/4)^{2}}{2} + \frac{(\pi/4)^{4}}{12}

= \frac{\pi^{2}}{32} + \frac{\pi^{4}}{3072}

\ln 2 \approx \frac{\pi^{2}}{16} + \frac{\pi^{4}}{1536}

To check your answer:

\frac{\pi^{2}}{16} + \frac{\pi^{4}}{1536} \approx 0.6803

, reasonably close to the true value

\ln 2 \approx 0.6931

, given that

x = \pi/4

is not especially small ✓

$\therefore$ $\ln 2 \approx \frac{\pi^{2}}{16} + \frac{\pi^{4}}{1536}$ .

Part (d)

Using the expansion from (b),

\int_{0}^{\pi/10}\ln\left( \sec x \right)\, dx \approx \int_{0}^{\pi/10}\left( \frac{x^{2}}{2} + \frac{x^{4}}{12} \right)dx

= \left\lbrack \frac{x^{3}}{6} + \frac{x^{5}}{60} \right\rbrack_{0}^{\pi/10}

= \frac{(\pi/10)^{3}}{6} + \frac{(\pi/10)^{5}}{60}

= \frac{\pi^{3}}{6000} + \frac{\pi^{5}}{6\, 000\, 000}

$\therefore$ $\int_{0}^{\pi/10}\ln\left( \sec x \right)\, dx \approx 0.005219$ (4 s.f.).

2023 H2 Maths Paper 2, Question 2

Part (a)

Part (b)

Part (c)

Part (d)

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