2022 H2 Maths Paper 1 Q7 — Worked Solution

Part (a)

4 marks

We differentiate $y = x^{- 3}\ln x$ with the product rule, taking $u = x^{- 3}$ , $v = \ln x$ .

\frac{dy}{dx} = - 3x^{- 4}\ln x + x^{- 3} \cdot \frac{1}{x}

= - 3x^{- 4}\ln x + x^{- 4}

= x^{- 4}\left( 1 - 3\ln x \right)

\frac{dy}{dx} = \frac{1 - 3\ln x}{x^{4}}\ \ \left( \text{shown} \right)

To check your answer: at

x = 1

\frac{dy}{dx} = \frac{1 - 0}{1} = 1

, and numerically differentiating

y = x^{- 3}\ln x

near

x = 1

gives a gradient close to

1

✓

At a turning point, $\frac{dy}{dx} = 0$ .

\frac{1 - 3\ln x}{x^{4}} = 0

\Rightarrow 1 - 3\ln x = 0

\Rightarrow \ln x = \frac{1}{3}

\Rightarrow x = e^{1/3}

Substituting into $y = x^{- 3}\ln x$ ,

y = \left( e^{1/3} \right)^{- 3}\ln\left( e^{1/3} \right)

= e^{- 1} \times \frac{1}{3}

= \frac{1}{3e}

To check your answer: at

x = e^{1/3} \approx 1.396

, direct computation gives

x^{- 3}\ln x \approx 0.1226

, matching

\frac{1}{3e} \approx 0.1226

✓

$\therefore$ the turning point is $\left( e^{1/3},\ \frac{1}{3e} \right)$ .

Part (b)

5 marks

Approach

the region is bounded by the curve, the $x$ -axis and $x = 3$ ; we first need the $x$ -intercept to know where the enclosed region starts. Since $x^{- 3} > 0$ for $x > 0$ , the sign of $y$ matches the sign of $\ln x$ , so the curve crosses the $x$ -axis only at $x = 1$ , and the enclosed region runs from $x = 1$ to $x = 3$ .

Since $x^{- 3} > 0$ for $x > 0$ , $y = 0$ only when $\ln x = 0$ , i.e. $x = 1$ ; for $0 < x < 1$ , $\ln x < 0$ so $y < 0$ , and for $x > 1$ , $\ln x > 0$ so $y > 0$ . Hence the curve crosses the $x$ -axis at $x = 1$ , and the region enclosed by $C$ , the $x$ -axis and $x = 3$ lies between $x = 1$ and $x = 3$ .

\text{Area} = \int_{1}^{3}x^{- 3}\ln\text{x dx}

Using integration by parts with $u = \ln x$ , $dv = x^{- 3}dx$ , so $du = \frac{1}{x}dx$ , $v = - \frac{1}{2x^{2}}$ ,

\int x^{- 3}\ln\text{x dx} = - \frac{\ln x}{2x^{2}} - \int\left( - \frac{1}{2x^{2}} \right)\left( \frac{1}{x} \right)dx

= - \frac{\ln x}{2x^{2}} + \frac{1}{2}\int x^{- 3}dx

= - \frac{\ln x}{2x^{2}} + \frac{1}{2}\left( - \frac{1}{2x^{2}} \right) + c

= - \frac{\ln x}{2x^{2}} - \frac{1}{4x^{2}} + c

Let $F(x) = - \frac{\ln x}{2x^{2}} - \frac{1}{4x^{2}}$ . At $x = 3$ , $F(3) = - \frac{\ln 3}{18} - \frac{1}{36}$ ; at $x = 1$ , $F(1) = - \frac{\ln 1}{2} - \frac{1}{4} = 0 - \frac{1}{4} = - \frac{1}{4}$ .

\text{Area} = F(3) - F(1)

= \left( - \frac{\ln 3}{18} - \frac{1}{36} \right) - \left( - \frac{1}{4} \right)

= - \frac{\ln 3}{18} - \frac{1}{36} + \frac{1}{4}

Using a common denominator of $36$ : $- \frac{1}{36} + \frac{9}{36} = \frac{8}{36} = \frac{2}{9}$ , so

\text{Area} = \frac{2}{9} - \frac{\ln 3}{18}

= \frac{4}{18} - \frac{\ln 3}{18}

= \frac{4 - \ln 3}{18}

To check your answer:

\frac{4 - \ln 3}{18} \approx \frac{4 - 1.0986}{18} \approx 0.1612

, matching a numerical estimate of

\int_{1}^{3}x^{- 3}\ln\text{x dx} \approx 0.1612

✓

$\therefore$ the area is $\frac{4 - \ln 3}{18}$ (exact), $\text{≈}0.161\ \text{units}^{2}$ .

2022 H2 Maths Paper 1, Question 7

Part (a)

Part (b)

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