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Date November 2013 Marks available 3 Reference code 13N.1.sl.TZ0.8
Level SL only Paper 1 Time zone TZ0
Command term Hence and Sketch Question number 8 Adapted from N/A

Question

Let f(x)=3x2 and g(x)=53x, for x0.

Let h(x)=5x+2, for x. The graph of h has a horizontal asymptote at y = 0.

Find {f^{ - 1}}(x).

[2]
a.

Show that \left( {g \circ {f^{ - 1}}} \right)(x) = \frac{5}{{x + 2}}.

[2]
b.

Find the y-intercept of the graph of h.

[2]
c(i).

Hence, sketch the graph of h.

[3]
c(ii).

For the graph of {h^{ - 1}}, write down the x-intercept;

[1]
d(i).

For the graph of {h^{ - 1}}, write down the equation of the vertical asymptote.

[1]
d(ii).

Given that {h^{ - 1}}(a) = 3, find the value of a.

[3]
e.

Markscheme

interchanging x and y     (M1)

eg     x = 3y - 2

{f^{ - 1}}(x) = \frac{{x + 2}}{3}{\text{   }}\left( {{\text{accept }}y = \frac{{x + 2}}{3},{\text{ }}\frac{{x + 2}}{3}} \right)     A1     N2

[2 marks]

a.

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

eg     g\left( {\frac{{x + 2}}{3}} \right),{\text{ }}\frac{{\frac{5}{{3x}} + 2}}{3}

correct substitution     A1

eg     \frac{5}{{3\left( {\frac{{x + 2}}{3}} \right)}}

\left( {g \circ {f^{ - 1}}} \right)(x) = \frac{5}{{x + 2}}     AG     N0

[2 marks]

b.

valid approach     (M1)

eg     h(0),{\text{ }}\frac{5}{{0 + 2}}

y = \frac{5}{2}{\text{   }}\left( {{\text{accept (0, 2.5)}}} \right)     A1     N2

[2 marks]

c(i).

     A1A2     N3

Notes:     Award A1 for approximately correct shape (reciprocal, decreasing, concave up).

     Only if this A1 is awarded, award A2 for all the following approximately correct features: y-intercept at (0, 2.5), asymptotic to x-axis, correct domain x \geqslant 0.

     If only two of these features are correct, award A1.

 

[3 marks]

c(ii).

x = \frac{5}{2}{\text{   }}\left( {{\text{accept (2.5, 0)}}} \right)     A1     N1

[1 mark]

d(i).

x = 0   (must be an equation)     A1     N1

[1 mark]

d(ii).

METHOD 1

attempt to substitute 3 into h (seen anywhere)     (M1)

eg     h(3),{\text{ }}\frac{5}{{3 + 2}}

correct equation     (A1)

eg     a = \frac{5}{{3 + 2}},{\text{ }}h(3) = a

a = 1     A1     N2

[3 marks]

METHOD 2

attempt to find inverse (may be seen in (d))     (M1)

eg     x = \frac{5}{{y + 2}},{\text{ }}{h^{ - 1}} = \frac{5}{x} - 2,{\text{ }}\frac{5}{x} + 2

correct equation, \frac{5}{x} - 2 = 3     (A1)

a = 1     A1     N2

[3 marks]

e.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c(i).
[N/A]
c(ii).
[N/A]
d(i).
[N/A]
d(ii).
[N/A]
e.

Syllabus sections

Topic 2 - Functions and equations » 2.5 » The rational function x \mapsto \frac{{ax + b}}{{cx + d}} and its graph.

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