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Date	May 2018	Marks available	3	Reference code	18M.1.AHL.TZ1.H_9
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Find	Question number	H_9	Adapted from	N/A

Question

Let $f\left( x \right) = \frac{{2 - 3{x^5}}}{{2{x^3}}},\,\,x \in \mathbb{R},\,\,x \ne 0$ .

The graph of $y = f\left( x \right)$ has a local maximum at A. Find the coordinates of A.

[5]

a.

Show that there is exactly one point of inflexion, B, on the graph of $y = f\left( x \right)$ .

[5]

b.i.

The coordinates of B can be expressed in the form B $\left( {{2^a},\,b \times {2^{ - 3a}}} \right)$ where a, b $\in \mathbb{Q}$ . Find the value of a and the value of b.

[3]

b.ii.

Sketch the graph of $y = f\left( x \right)$ showing clearly the position of the points A and B.

[4]

c.

Markscheme

attempt to differentiate (M1)

$f'\left( x \right) = - 3{x^{ - 4}} - 3x$ A1

Note: Award M1 for using quotient or product rule award A1 if correct derivative seen even in unsimplified form, for example $f'\left( x \right) = \frac{{ - 15{x^4} \times 2{x^3} - 6{x^2}\left( {2 - 3{x^5}} \right)}}{{{{\left( {2{x^3}} \right)}^2}}}$ .

$- \frac{3}{{{x^4}}} - 3x = 0$ M1

$\Rightarrow {x^5} = - 1 \Rightarrow x = - 1$ A1

${\text{A}}\left( { - 1,\, - \frac{5}{2}} \right)$ A1

[5 marks]

a.

$f''\left( x \right) = 0$ M1

$f''\left( x \right) = 12{x^{ - 5}} - 3\left( { = 0} \right)$ A1

Note: Award A1 for correct derivative seen even if not simplified.

$\Rightarrow x = \sqrt[5]{4}\left( { = {2^{\frac{2}{5}}}} \right)$ A1

hence (at most) one point of inflexion R1

Note: This mark is independent of the two A1 marks above. If they have shown or stated their equation has only one solution this mark can be awarded.

$f''\left( x \right)$ changes sign at $x = \sqrt[5]{4}\left( { = {2^{\frac{2}{5}}}} \right)$ R1

so exactly one point of inflexion

[5 marks]

b.i.

$x = \sqrt[5]{4} = {2^{\frac{2}{5}}}\left( { \Rightarrow a = \frac{2}{5}} \right)$ A1

$f\left( {{2^{\frac{2}{5}}}} \right) = \frac{{2 - 3 \times {2^2}}}{{2 \times {2^{\frac{6}{5}}}}} = - 5 \times {2^{ - \frac{6}{5}}}\left( { \Rightarrow b = - 5} \right)$ (M1)A1

Note: Award M1 for the substitution of their value for $x$ into $f\left( x \right)$ .

[3 marks]

b.ii.

A1A1A1A1

A1 for shape for x < 0
A1 for shape for x > 0
A1 for maximum at A
A1 for POI at B.

Note: Only award last two A1s if A and B are placed in the correct quadrants, allowing for follow through.

[4 marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.

Syllabus sections

Topic 2—Functions » SL 2.3—Graphing

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