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Date May 2008 Marks available 3 Reference code 08M.1.sl.TZ1.7
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 7 Adapted from N/A

Question

Let \(f(x) = \ln (x + 5) + \ln 2\) , for \(x > - 5\) .

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

[4]
a.

Let \(g(x) = {{\rm{e}}^x}\) .

Find \((g \circ f)(x)\) , giving your answer in the form \(ax + b\) , where \(a,b \in \mathbb{Z}\) .

 

[3]
b.

Markscheme

METHOD 1

\(\ln (x + 5) + \ln 2 = \ln (2(x + 5))\) \(( = \ln (2x + 10))\)     (A1)

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = \ln (2y + 10)\)

evidence of correct manipulation     (A1)

e.g. \({{\rm{e}}^x} = 2y + 10\)

\({f^{ - 1}}(x) = \frac{{{{\rm{e}}^x} - 10}}{2}\)    A1     N2

METHOD 2

\(y = \ln (x + 5) + \ln 2\)

\(y - \ln 2 = ln(x + 5)\)     (A1)

evidence of correct manipulation     (A1)

e.g. \({{\rm{e}}^{y - \ln 2}} = x + 5\)

interchanging x and y (seen anywhere)     (M1)

e.g. \({{\rm{e}}^{x - \ln 2}} = y + 5\)

\({f^{ - 1}}(x) = {{\rm{e}}^{x - \ln 2}} - 5\)     A1     N2

[4 marks]

a.

METHOD 1

evidence of composition in correct order     (M1)

e.g. \((g \circ f)(x) = g(\ln (x + 5) + \ln 2)\)

\( = {{\rm{e}}^{\ln (2(x + 5))}} = 2(x + 5)\)

\((g \circ f)(x) = 2x + 10\)     A1A1     N2

METHOD 2

evidence of composition in correct order     (M1)

e.g. \((g \circ f)(x) = {{\rm{e}}^{\ln (x + 5) + \ln 2}}\)

\( = {{\rm{e}}^{\ln (x + 5)}} \times {{\rm{e}}^{\ln 2}} = (x + 5)2\)

\((g \circ f)(x) = 2x + 10\)     A1A1     N2

[3 marks]

b.

Examiners report

This was one of the more difficult problems for the candidates. Knowledge of the laws of logarithms appeared weak as did the inverse nature of the exponential and logarithmic functions. There were a number of candidates who mistook the notation for the inverse to mean either the derivative or the reciprocal. The order of composition seemed well understood by most candidates but they were unable to simplify by the rules of indices to obtain the correct final answer.

a.

This was one of the more difficult problems for the candidates. Knowledge of the laws of logarithms appeared weak as did the inverse nature of the exponential and logarithmic functions. There were a number of candidates who mistook the notation for the inverse to mean either the derivative or the reciprocal. The order of composition seemed well understood by most candidates but they were unable to simplify by the rules of indices to obtain the correct final answer.

b.

Syllabus sections

Topic 2 - Functions and equations » 2.1 » Composite functions.
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