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Date	May 2018	Marks available	2	Reference code	18M.2.sl.TZ1.1
Level	SL only	Paper	2	Time zone	TZ1
Command term	Find	Question number	1	Adapted from	N/A

Question

Let f(x) = ln x − 5x , for x > 0 .

Find f '(x).

[2]

Find f "(x).

[1]

Solve f '(x) = f "(x).

[2]

Markscheme

$f'\left( x \right) = \frac{1}{x} - 5$ A1A1 N2

[2 marks]

f "(x) = −x⁻² A1 N1

[1 mark]

METHOD 1 (using GDC)

valid approach (M1)

0.558257

x = 0.558 A1 N2

Note: Do not award A1 if additional answers given.

METHOD 2 (analytical)

attempt to solve their equation f '(x) = f "(x) (do not accept $\frac{1}{x} - 5 = - \frac{1}{{{x^2}}}$ ) (M1)

eg $5{x^2} - x - 1 = 0,\,\,\frac{{1 \pm \sqrt {21} }}{{10}},\,\,\frac{1}{x} = \frac{{ - 1 \pm \sqrt {21} }}{2},\,\, - 0.358$

0.558257

x = 0.558 A1 N2

Note: Do not award A1 if additional answers given.

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 6 - Calculus » 6.2 » Derivative of

${x^n}\left( {n \in \mathbb{Q}} \right)$ ,

$\sin x$ ,

$\cos x$ ,

$\tan x$ ,

${{\text{e}}^x}$ and

$\ln x$ .

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