| Date | May 2008 | Marks available | 4 | 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 \({f^{ - 1}}(x)\) .
Let \(g(x) = {{\rm{e}}^x}\) .
Find \((g \circ f)(x)\) , giving your answer in the form \(ax + b\) , where \(a,b \in \mathbb{Z}\) .
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]
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]
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.
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.
