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Date November 2013 Marks available 1 Reference code 13N.1.hl.TZ0.3
Level HL only Paper 1 Time zone TZ0
Command term State Question number 3 Adapted from N/A

Question

The diagram below shows a sketch of the graph of \(y = f(x)\).


Sketch the graph of \(y = {f^{ - 1}}(x)\) on the same axes.

[2]
a.

State the range of \({f^{ - 1}}\).

[1]
b.

Given that \(f(x) = \ln (ax + b),{\text{ }}x > 1\), find the value of \(a\) and the value of \(b\).

[4]
c.

Markscheme

(a)     

 

shape with y-axis intercept (0, 4)     A1

 

Note:     Accept curve with an asymptote at \(x = 1\) suggested.

 

correct asymptote \(y = 1\)     A1

[2 marks]

a.

range is \({f^{ - 1}}(x) > 1{\text{ (or }}\left] {1,{\text{ }}\infty } \right[)\)     A1

 

Note:     Also accept \(\left] {1,{\text{ 10}}} \right]\) or \(\left] {1,{\text{ 10}}} \right[\).

 

Note:     Do not allow follow through from incorrect asymptote in (a).

 

[1 mark]

b.

\((4,{\text{ }}0) \Rightarrow \ln (4a + b) = 0\)     M1

\( \Rightarrow 4a + b = 1\)     A1

asymptote at \(x = 1 \Rightarrow a + b = 0\)     M1

\( \Rightarrow a = \frac{1}{3},{\text{ }}b =  - \frac{1}{3}\)     A1

[4 marks]

c.

Examiners report

A number of candidates were able to answer a) and b) correctly but found part c) more challenging. Correct sketches for the inverse were seen, but with a few missing a horizontal asymptote. The range in part b) was usually seen correctly. In part c), only a small number of very good candidates were able to gain full marks. A large number used the point \((4,{\text{ 0)}}\) to form the equation \(4a + b = 1\) but were unable (or did not recognise the need) to use the asymptote to form a second equation.

a.

A number of candidates were able to answer a) and b) correctly but found part c) more challenging. Correct sketches for the inverse were seen, but with a few missing a horizontal asymptote. The range in part b) was usually seen correctly. In part c), only a small number of very good candidates were able to gain full marks. A large number used the point \((4,{\text{ 0)}}\) to form the equation \(4a + b = 1\) but were unable (or did not recognise the need) to use the asymptote to form a second equation.

b.

A number of candidates were able to answer a) and b) correctly but found part c) more challenging. Correct sketches for the inverse were seen, but with a few missing a horizontal asymptote. The range in part b) was usually seen correctly. In part c), only a small number of very good candidates were able to gain full marks. A large number used the point \((4,{\text{ 0)}}\) to form the equation \(4a + b = 1\) but were unable (or did not recognise the need) to use the asymptote to form a second equation.

c.

Syllabus sections

Topic 2 - Core: Functions and equations » 2.2 » Investigation of key features of graphs, such as maximum and minimum values, intercepts, horizontal and vertical asymptotes and symmetry, and consideration of domain and range.
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