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Date May 2009 Marks available 3 Reference code 09M.1.sl.TZ1.6
Level SL only Paper 1 Time zone TZ1
Command term Show that and Write down Question number 6 Adapted from N/A

Question

Let \(f(x) = {{\rm{e}}^{x + 3}}\) .

(i)     Show that \({f^{ - 1}}(x) = \ln x - 3\) .

(ii)    Write down the domain of \({f^{ - 1}}\) .

[3]
a.

Solve the equation \({f^{ - 1}}(x) = \ln \frac{1}{x}\) .

[4]
b.

Markscheme

(i) interchanging x and y (seen anywhere)     M1

e.g. \(x = {{\rm{e}}^{y + 3}}\)

correct manipulation     A1

e.g. \(\ln x = y + 3\) , \(\ln y = x + 3\)

\({f^{ - 1}}(x) = \ln x - 3\)     AG     N0

(ii) \(x > 0\)     A1     N1 

[3 marks]

a.

collecting like terms; using laws of logs     (A1)(A1)

e.g. \(\ln x - \ln \left( {\frac{1}{x}} \right) = 3\) , \(\ln x + \ln x = 3\) , \(\ln \left( {\frac{x}{{\frac{1}{x}}}} \right) = 3\) , \(\ln {x^2} = 3\)

simplify     (A1)

e.g. \(\ln x = \frac{3}{2}\) ,  \({x^2} = {{\rm{e}}^3}\)

\(x = {{\rm{e}}^{\frac{3}{2}}}\left( { = \sqrt {{{\rm{e}}^3}} } \right)\)     A1     N2

[4 marks]

b.

Examiners report

Many candidates interchanged the \(x\) and \(y\) to find the inverse function, but very few could write down the correct domain of the inverse, often giving \(x \ge 0\) , \(x > 3\) and "all real numbers" as responses.

a.

Where students attempted to solve the equation in (b), most treated \(\ln x - 3\) as \(\ln (x - 3)\) and created an incorrect equation from the outset. The few who applied laws of logarithms often carried the algebra through to completion.

b.

Syllabus sections

Topic 2 - Functions and equations » 2.1 » Domain, range; image (value).
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