Show that \(g \circ f(x) = 3\sin \left( {2x + \frac{\pi }{5}} \right) + 4\).

Find the range of \(g \circ f\).

Given that \(g \circ f\left( {\frac{{3\pi }}{{20}}} \right) = 7\), find the next value of \(x\), greater than \({\frac{{3\pi }}{{20}}}\), for which \(g \circ f(x) = 7\).

The graph of \(y = g \circ f(x)\) can be obtained by applying four transformations to the graph of \(y = \sin x\). State what the four transformations represent geometrically and give the order in which they are applied.

\(g \circ f(x) = g\left( {f(x)} \right)\)     M1

\( = g\left( {2x + \frac{\pi }{5}} \right)\)

\( = 3\sin \left( {2x + \frac{\pi }{5}} \right) + 4\)     AG

[1 mark]

since \( - 1 \le sin \theta  \le  + 1\), range is \([1,{\text{ }}7]\)     (R1)A1

[2 marks]

\(3\sin \left( {2x + \frac{\pi }{5}} \right) + 4 = 7 \Rightarrow 2x + \frac{\pi }{5} = \frac{\pi }{2} + 2n\pi  \Rightarrow x = \frac{{3\pi }}{{20}} + n\pi \)     (M1)

so next biggest value is \(\frac{{23\pi }}{{20}}\)     A1

Note:     Allow use of period.

[2 marks]

Note:     Transformations can be in any order but see notes below.

stretch scale factor \(3\) parallel to \(y\) axis (vertically)     A1

vertical translation of \(4\) up     A1

Note:     Vertical translation is \(\frac{4}{3}\) up if it occurs before stretch parallel to \(y\) axis.

stretch scale factor \(\frac{1}{2}\) parallel to \(x\) axis (horizontally)     A1

horizontal translation of \(\frac{\pi }{{10}}\) to the left     A1

Note:     Horizontal translation is \(\frac{\pi }{{5}}\) to the left if it occurs before stretch parallel to \(x\) axis.

Note:     Award A1 for magnitude and direction in each case.

Accept any correct terminology provided that the meaning is clear eg shift for translation.

[4 marks]

Total [9 marks]

Well done.

Generally well done, some used more complicated methods rather than considering the range of sine.

Fine if they realised the period was \(\pi \), not if they thought it was \(2\pi \).

Typically 3 marks were gained. It was the shift in the axis \(\chi \) of \(\frac{\pi }{{10}}\) that caused the problem.