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Date	November 2020	Marks available	3	Reference code	20N.2.SL.TZ0.S_4
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find and Hence or otherwise	Question number	S_4	Adapted from	N/A

Question

Let $f (x) = 4 - x^{3}$ and $g (x) = \ln x$ , for $x > 0$ .

Find $(f \circ g) (x)$ .

[2]

Solve the equation $(f \circ g) (x) = x$ .

[2]

b.i.

Hence or otherwise, given that $g (2 a) = f^{- 1} (2 a)$ , find the value of $a$ .

[3]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to form composite (in any order) (M1)

eg $f (\ln x), g (4 - x^{3})$

$(f \circ g) (x) = 4 - {(\ln x)}^{3}$ A1 N2

[2 marks]

valid approach using GDC (M1)

eg , $(2.85, 2.85)$

$2.85056$

$2.85$ A1 N2

[2 marks]

b.i.

METHOD 1 – (using properties of functions)

recognizing inverse relationship (M1)

eg $f (g (2 a)) = f (f^{- 1} (2 a)) (= 2 a)$

equating $2 a$ to their $x$ from (i) (A1)

eg $2 a = 2.85056$

$1.42528$

$a = 1.43$ A1 N2

METHOD 2 – (finding inverse)

interchanging $x$ and $y$ (seen anywhere) (M1)

eg $x = 4 - y^{3}, f^{- 1} (x) = \sqrt[3]{4 - x}$

correct working (A1)

eg $\sqrt[3]{4 - 2 a} = \ln (2 a)$ , sketch showing intersection of $f^{- 1} (2 x)$ and $g (2 x)$

$1.42528$

$a = 1.43$ A1 N2

[3 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

Syllabus sections

Topic 2—Functions » SL 2.5—Composite functions, identity, finding inverse

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