2018 H2 Maths Paper 1 Q1 — Worked Solution

Part (i)

2 marks

We use the quotient rule with $u = \ln x$ and $v = x$ , so $u'= \frac{1}{x}$ and $v'= 1$ .

\frac{dy}{dx} = \frac{vu'- uv'}{v^{2}}

= \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^{2}}

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

$\therefore\frac{dy}{dx} = \frac{1 - \ln x}{x^{2}}$ .

Part (ii)

4 marks

Approach

The integrand $\frac{\ln x}{x^{2}}$ is not itself $\frac{dy}{dx}$ , but comparing the two shows how they relate: split $\frac{dy}{dx}$ from (i) into two fractions and isolate the term we want.

Since

\frac{dy}{dx} = \frac{1}{x^{2}} - \frac{\ln x}{x^{2}},

rearranging gives

\frac{\ln x}{x^{2}} = \frac{1}{x^{2}} - \frac{dy}{dx}.

Integrating both sides from $1$ to $e$ :

\int_{1}^{e}\frac{\ln x}{x^{2}}\, dx

= \int_{1}^{e}\frac{1}{x^{2}}\, dx - \int_{1}^{e}\frac{dy}{dx}\, dx

= \left\lbrack - \frac{1}{x} \right\rbrack_{1}^{e} - \lbrack y\rbrack_{1}^{e}

Since $y = \frac{\ln x}{x}$ , we have $y(e) = \frac{1}{e}$ and $y(1) = 0$ .

\int_{1}^{e}\frac{\ln x}{x^{2}}\, dx

= \left( - \frac{1}{e} + 1 \right) - \left( \frac{1}{e} - 0 \right)

= 1 - \frac{2}{e}

To check your answer: By GC,

\int_{1}^{e}\frac{\ln x}{x^{2}}\, dx \approx 0.264

, and

1 - \frac{2}{e} \approx 0.264

. ✓

$\therefore\int_{1}^{e}\frac{\ln x}{x^{2}}\, dx = 1 - \frac{2}{e}$ .

2018 H2 Maths Paper 1, Question 1

Part (i)

Part (ii)

This is how we mark, and how we teach.