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.