Date | May 2011 | Marks available | 4 | Reference code | 11M.1.hl.TZ1.8 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Consider the functions given below.
\[f(x) = 2x + 3\]\[g(x) = \frac{1}{x},x \ne 0\]
(i) Find \(\left( {g \circ f} \right)\left( x \right)\) and write down the domain of the function.
(ii) Find \(\left( {f \circ g} \right)\left( x \right)\) and write down the domain of the function.
Find the coordinates of the point where the graph of \(y = f(x)\) and the graph of \(y = \left( {{g^{ - 1}} \circ f \circ g} \right)(x)\) intersect.
Markscheme
(i) \(\left( {g \circ f} \right)\left( x \right) = \frac{1}{{2x + 3}}\), \(x \ne - \frac{3}{2}\) (or equivalent) A1
(ii) \(\left( {f \circ g} \right)\left( x \right) = \frac{2}{x} + 3\), \(x \ne 0\) (or equivalent) A1
[2 marks]
EITHER
\(f(x) = \left( {{g^{ - 1}} \circ f \circ g} \right)(x) \Rightarrow \left( {f \circ g} \right)\left( x \right)\) (M1)
\(\frac{1}{{2x + 3}} = \frac{2}{x} + 3\) A1
OR
\(\left( {{g^{ - 1}} \circ f \circ g} \right)(x) = \frac{1}{{\frac{2}{x} + 3}}\) A1
\(2x + 3 = \frac{1}{{\frac{2}{x} + 3}}\) M1
THEN
\(6{x^2} + 12x + 6 = 0\) (or equivalent) A1
\(x = - 1\), \(y = 1\) (coordinates are (−1, 1) ) A1
[4 marks]
Examiners report
Part (a) was in general well answered and part (b) well attempted. Some candidates had difficulties with the order of composition and in using correct notation to represent the domains of the functions.
Part (a) was in general well answered and part (b) well attempted. Some candidates had difficulties with the order of composition and in using correct notation to represent the domains of the functions.