Show that the $x$-coordinate of the minimum point on the curve $y = f(x)$ satisfies the equation $\tan x = 2x$.

Determine the values of $x$ for which $f(x)$ is a decreasing function.

Sketch the graph of $y = f(x)$ showing clearly the minimum point and any asymptotic behaviour.

Find the coordinates of the point on the graph of $f$ where the normal to the graph is parallel to the line $y =  - x$.

This region is now rotated through $2\pi$ radians about the $x$-axis. Find the volume of revolution.

attempt to use quotient rule or product rule     M1

$f’(x) = \frac{{\sin x\left( {\frac{1}{2}{x^{ - \frac{1}{2}}}} \right) - \sqrt x \cos x}}{{{{\sin }^2}x}}{\text{ }}\left( { = \frac{1}{{2\sqrt x \sin x}} - \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}} \right)$     A1A1

Note:     Award A1 for $\frac{1}{{2\sqrt x \sin x}}$ or equivalent and A1 for $- \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}$ or equivalent.

setting $f’(x) = 0$     M1

$\frac{{\sin x}}{{2\sqrt x }} - \sqrt x \cos x = 0$

$\frac{{\sin x}}{{2\sqrt x }} = \sqrt x \cos x$ or equivalent     A1

$\tan x = 2x$     AG

[5 marks]

$x = 1.17$

$0 < x \leqslant 1.17$     A1A1

Note:     Award A1 for $0 < x$ and A1 for $x \leqslant 1.17$. Accept $x < 1.17$.

[2 marks]

concave up curve over correct domain with one minimum point above the $x$-axis.     A1

approaches $x = 0$ asymptotically     A1

approaches $x = \pi$ asymptotically     A1

Note:     For the final A1 an asymptote must be seen, and $\pi$ must be seen on the $x$-axis or in an equation.

[3 marks]

$f’(x){\text{ }}\left( { = \frac{{\sin x\left( {\frac{1}{2}{x^{ - \frac{1}{2}}}} \right) - \sqrt x \cos x}}{{{{\sin }^2}x}}} \right) = 1$     (A1)

attempt to solve for $x$     (M1)

$x = 1.96$     A1

$y = f(1.96 \ldots )$

$= 1.51$     A1

[4 marks]

$V = \pi \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{x{\text{d}}x}}{{{{\sin }^2}x}}}$     (M1)(A1)

Note:     M1 is for an integral of the correct squared function (with or without limits and/or $\pi$).

$= 2.68{\text{ }}( = 0.852\pi )$     A1

[3 marks]