Date | May 2008 | Marks available | 6 | Reference code | 08M.1.hl.TZ2.4 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
Let \(f(x) = \frac{4}{{x + 2}},{\text{ }}x \ne - 2{\text{ and }}g(x) = x - 1\).
If \(h = g \circ f\) , find
(a) h(x) ;
(b) \({h^{ - 1}}(x)\) , where \({h^{ - 1}}\) is the inverse of h.
Markscheme
(a) \(h(x) = g\left( {\frac{4}{{x + 2}}} \right)\) (M1)
\( = \frac{4}{{x + 2}} - 1\,\,\,\,\,\left( { = \frac{{2 - x}}{{2 + x}}} \right)\) A1
(b) METHOD 1
\(x = \frac{4}{{y + 2}} - 1\,\,\,\,\,\)(interchanging x and y) M1
Attempting to solve for y M1
\((y + 2)(x + 1) = 4\,\,\,\,\,\left( {y + 2 = \frac{4}{{x + 1}}} \right)\) (A1)
\({h^{ - 1}}(x) = \frac{4}{{x + 1}} - 2\,\,\,\,\,(x \ne - 1)\) A1 N1
METHOD 2
\(x = \frac{{2 - y}}{{2 + y}}\,\,\,\,\,\)(interchanging x and y) M1
Attempting to solve for y M1
\(xy + y = 2 - 2x\,\,\,\,\,\left( {y(x + 1) = 2(1 - x)} \right)\) (A1)
\({h^{ - 1}}(x) = \frac{{2(1 - x)}}{{x + 1}}\,\,\,\,\,(x \ne - 1)\) A1 N1
Note: In either METHOD 1 or METHOD 2 rearranging first and interchanging afterwards is equally acceptable.
[6 marks]
Examiners report
This question was generally well done, with very few candidates calculating \(f \circ g\) rather than \(g \circ f\).