Use the quotient rule to show that $f'(x) = \frac{{ - 1}}{{{{\sin }^2}x}}$ .

Find $f''(x)$ .

Find the value of p and of q.

Use information from the table to explain why there is a point of inflexion on the graph of f where $x = \frac{\pi }{2}$ .

$\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x$ , $\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = - \sin x$ (seen anywhere)     (A1)(A1)

evidence of using the quotient rule     M1

correct substitution     A1

e.g. $\frac{{\sin x( - \sin x) - \cos x(\cos x)}}{{{{\sin }^2}x}}$ , $\frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}}$

$f'(x) = \frac{{ - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}$     A1

$f'(x) = \frac{{ - 1}}{{{{\sin }^2}x}}$     AG      N0

[5 marks]

METHOD 1

appropriate approach     (M1)

e.g. $f'(x) = - {(\sin x)^{ - 2}}$

$f''(x) = 2({\sin ^{ - 3}}x)(\cos x)$ $\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)$     A1A1     N3

Note: Award A1 for $2{\sin ^{ - 3}}x$ , A1 for $\cos x$ .

METHOD 2

derivative of ${\sin ^2}x = 2\sin x\cos x$ (seen anywhere)     A1

evidence of choosing quotient rule     (M1)

e.g. $u = - 1$ ,  $v = {\sin ^2}x$ , $f'' = \frac{{{{\sin }^2}x \times 0 - ( - 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}$

$f''(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}$ $\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)$     A1     N3

[3 marks]

evidence of substituting $\frac{\pi }{2}$     M1

e.g. $\frac{{ - 1}}{{{{\sin }^2}\frac{\pi }{2}}}$ , $\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}$

$p = - 1$ ,  $q = 0$    A1A1     N1N1

[3 marks]

second derivative is zero, second derivative changes sign     R1R1     N2

[2 marks]

Many candidates comfortably applied the quotient rule, although some did not completely show that the Pythagorean identity achieves the numerator of the answer given. Whether changing to $- {(\sin x)^{ - 2}}$ , or applying the quotient rule a second time, most candidates neglected the chain rule in finding the second derivative.

Whether changing to $- {(\sin x)^{ - 2}}$ , or applying the quotient rule a second time, most candidates neglected the chain rule in finding the second derivative.

Those who knew the trigonometric ratios at $\frac{\pi }{2}$typically found the values of p and of q, sometimes in follow-through from an incorrect $f''(x)$ .

Few candidates gave two reasons from the table that supported the existence of a point of inflexion. Most stated that the second derivative is zero and neglected to consider the sign change to the left and right of q. Some discussed a change of concavity, but without supporting this statement by referencing the change of sign in $f''(x)$ , so no marks were earned.