Date | May 2016 | Marks available | 2 | Reference code | 16M.1.sl.TZ2.6 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Hence | Question number | 6 | Adapted from | N/A |
Question
Let \(f(x) = 6x\sqrt {1 - {x^2}} \), for \( - 1 \leqslant x \leqslant 1\), and \(g(x) = \cos (x)\), for \(0 \leqslant x \leqslant \pi \).
Let \(h(x) = (f \circ g)(x)\).
Write \(h(x)\) in the form \(a\sin (bx)\), where \(a,{\text{ }}b \in \mathbb{Z}\).
Hence find the range of \(h\).
Markscheme
attempt to form composite in any order (M1)
eg\(\,\,\,\,\,\)\(f\left( {g(x)} \right),{\text{ }}\cos \left( {6x\sqrt {1 - {x^2}} } \right)\)
correct working (A1)
eg\(\,\,\,\,\,\)\(6\cos x\sqrt {1 - {{\cos }^2}x} \)
correct application of Pythagorean identity (do not accept \({\sin ^2}x + {\cos ^2}x = 1\)) (A1)
eg\(\,\,\,\,\,\)\({\sin ^2}x = 1 - {\cos ^2}x,{\text{ }}6\cos x\sin x,{\text{ }}6\cos x \left| \sin x\right|\)
valid approach (do not accept \(2\sin x\cos x = \sin 2x\)) (M1)
eg\(\,\,\,\,\,\)\(3(2\cos x\sin x)\)
\(h(x) = 3\sin 2x\) A1 N3
[5 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)amplitude \( = 3\), sketch with max and min \(y\)-values labelled, \( - 3 < y < 3\)
correct range A1 N2
eg\(\,\,\,\,\,\)\( - 3 \leqslant y \leqslant 3\), \([ - 3,{\text{ }}3]\) from \( - 3\) to 3
Note: Do not award A1 for \( - 3 < y < 3\) or for “between \( - 3\) and 3”.
[2 marks]
Examiners report
In part (a), nearly all candidates found the correct composite function in terms of \(\cos x\), though many did not get any further than this first step in their solution to the question. While some candidates seemed to recognize the need to use trigonometric identities, most were unsuccessful in finding the correct expression in the required form.
In part (b), very few candidates were able to provide the correct range of the function.