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Date November 2013 Marks available 2 Reference code 13N.1.sl.TZ0.8
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 8 Adapted from N/A

Question

Let \(f(x) = 3x - 2\) and \(g(x) = \frac{5}{{3x}}\), for \(x \ne 0\).

Let \(h(x) = \frac{5}{{x + 2}}\), for \(x \geqslant 0\). 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 » 2.1 » Identity function. Inverse function \({f^{ - 1}}\) .
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