Determine an expression for \(f’(x)\) in terms of \(x\).

Sketch a graph of \(y = f’(x)\) for \(0 \leqslant x < \frac{\pi }{2}\).

Find the \(x\)-coordinate(s) of the point(s) of inflexion of the graph of \(y = f(x)\), labelling these clearly on the graph of \(y = f’(x)\).

Express \(\sin x\) in terms of \(\mu \).

Express \(\sin 2x\) in terms of \(u\).

Hence show that \(f(x) = 0\) can be expressed as \({u^3} - 7{u^2} + 15u - 9 = 0\).

Solve the equation \(f(x) = 0\), giving your answers in the form \(\arctan k\) where \(k \in \mathbb{Z}\).

\(f’(x) = 4\sin x\cos x + 14\cos 2x + {\sec ^2}x\) (or equivalent)     (M1)A1

[2 marks]

     A1A1A1A1

Note:     Award A1 for correct behaviour at \(x = 0\), A1 for correct domain and correct behaviour for \(x \to \frac{\pi }{2}\), A1 for two clear intersections with \(x\)-axis and minimum point, A1 for clear maximum point.

[4 marks]

\(x = 0.0736\)     A1

\(x = 1.13\)     A1

[2 marks]

attempt to write \(\sin x\) in terms of \(u\) only     (M1)

\(\sin x = \frac{u}{{\sqrt {1 + {u^2}} }}\)     A1

[2 marks]

\(\cos x = \frac{1}{{\sqrt {1 + {u^2}} }}\)     (A1)

attempt to use \(\sin 2x = 2\sin x\cos x{\text{ }}\left( { = 2\frac{u}{{\sqrt {1 + {u^2}} }}\frac{1}{{\sqrt {1 + {u^2}} }}} \right)\)     (M1)

\(\sin 2x = \frac{{2u}}{{1 + {u^2}}}\)     A1

[3 marks]

\(2{\sin ^2}x + 7\sin 2x + \tan x - 9 = 0\)

\(\frac{{2{u^2}}}{{1 + {u^2}}} + \frac{{14u}}{{1 + {u^2}}} + u - 9{\text{ }}( = 0)\)     M1

\(\frac{{2{u^2} + 14u + u(1 + {u^2}) - 9(1 + {u^2})}}{{1 + {u^2}}} = 0\) (or equivalent)     A1

\({u^3} - 7{u^2} + 15u - 9 = 0\)     AG

[2 marks]

\(u = 1\) or \(u = 3\)     (M1)

\(x = \arctan (1)\)     A1

\(x = \arctan (3)\)     A1

Note:     Only accept answers given the required form.

[3 marks]