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Date May 2018 Marks available 3 Reference code 18M.1.hl.TZ1.9
Level HL only Paper 1 Time zone TZ1
Command term Find Question number 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 6 - Core: Calculus » 6.3 » Points of inflexion with zero and non-zero gradients.
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