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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-x3 and g(x)=lnx, for x>0.

Find (fg)(x).

[2]
a.

Solve the equation (fg)(x)=x.

[2]
b.i.

Hence or otherwise, given that g(2a)=f-1(2a), 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(lnx) , g(4-x3)

(fg)(x)=4-(lnx)3      A1  N2

[2 marks]

a.

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(2a))=f(f-1(2a))  (=2a)

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

eg     2a=2.85056

1.42528

a=1.43       A1  N2

 

METHOD 2 – (finding inverse)

interchanging x and y (seen anywhere)       (M1)

eg     x=4-y3 , f-1(x)=34-x

correct working       (A1)

eg     34-2a=ln(2a), sketch showing intersection of f-1(2x) and g(2x)

1.42528

a=1.43       A1  N2

 

[3 marks]

b.ii.

Examiners report

[N/A]
a.
[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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