21M.1.SL.TZ1.8

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Let $y = \frac{\ln x}{x^{4}}$ for $x > 0$ .

Consider the function defined by $f (x) \frac{\ln x}{x^{4}}$ for $x > 0$ and its graph $y = f (x)$ .

Show that $\frac{d y}{d x} = \frac{1 - 4 \ln x}{x^{5}}$ .

[3]

The graph of $f$ has a horizontal tangent at point $P$ . Find the coordinates of $P$ .

[5]

Given that $f'' (x) = \frac{20 \ln x - 9}{x^{6}}$ , show that $P$ is a local maximum point.

[3]

Solve $f (x) > 0$ for $x > 0$ .

[2]

Sketch the graph of $f$ , showing clearly the value of the $x$ -intercept and the approximate position of point $P$ .

[3]

Markscheme / solution

attempt to use quotient or product rule (M1)

$\frac{d y}{d x} = \frac{x^{4} (\frac{1}{x}) - (\ln x) (4 x^{3})}{{(x^{4})}^{2}}$ OR $(\ln x) (- 4 x^{- 5}) + (x^{- 4}) (\frac{1}{x})$ A1

correct working A1

$= \frac{x^{3} (1 - 4 \ln x)}{x^{8}}$ OR cancelling $x^{3}$ OR $\frac{- 4 \ln x}{x^{5}} + \frac{1}{x^{5}}$

$= \frac{1 - 4 \ln x}{x^{5}}$ AG

[3 marks]

$f' (x) = \frac{d y}{d x} = 0$ (M1)

$\frac{1 - 4 \ln x}{x^{5}} = 0$

$\ln x = \frac{1}{4}$ (A1)

$x = e^{\frac{1}{4}}$ A1

substitution of their $x$ to find $y$ (M1)

$y = \frac{\ln e^{\frac{1}{4}}}{{(e^{\frac{1}{4}})}^{4}}$

$= \frac{1}{4 e} (= \frac{1}{4} e^{- 1})$ A1

$P (e^{\frac{1}{4}}, \frac{1}{4 e})$

[5 marks]

$f'' (e^{\frac{1}{4}}) = \frac{20 \ln e^{\frac{1}{4}} - 9}{{(e^{\frac{1}{4}})}^{6}}$ (M1)

$= \frac{5 - 9}{e^{1.5}} (= - \frac{4}{e^{1.5}})$ A1

which is negative R1

hence $P$ is a local maximum AG

Note: The R1 is dependent on the previous A1 being awarded.

[3 marks]

$\ln x > 0$ (A1)

$x > 1$ A1

[2 marks]

A1A1A1

Note: Award A1 for one $x$ -intercept only, located at $1$

A1 for local maximum, $P$ , in approximately correct position
A1 for curve approaching $x$ -axis as $x \to \infty$ (including change in concavity).

[3 marks]

Examiners’ report

[N/A]