Find all values of x for $0.1 \leqslant x \leqslant 1$ such that $\sin (\pi {x^{ - 1}}) = 0$.

Find $\int_{\frac{1}{{n + 1}}}^{\frac{1}{n}} {\pi {x^{ - 2}}\sin (\pi {x^{ - 1}}){\text{d}}x}$, showing that it takes different integer values when n is even and when n is odd.

Evaluate $\int_{0.1}^1 {\left| {\pi {x^{ - 2}}\sin (\pi {x^{ - 1}})} \right|{\text{d}}x}$.

$\sin (\pi {x^{ - 1}}) = 0{\text{ }}\frac{\pi }{x} = \pi ,{\text{ }}2\pi ( \ldots )$     (A1) 

$x = 1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8},\frac{1}{9},\frac{1}{{10}}$     A1

[2 marks]

$\left[ {\cos (\pi {x^{ - 1}})} \right]_{\frac{1}{{n + 1}}}^{\frac{1}{n}}$     M1

$= \cos (\pi n) - \cos \left( {\pi (n + 1)} \right)$     A1

= 2 when n is even and = –2 when n is odd     A1

[3 marks]

$\int_{0.1}^1 {\left| {\pi {x^{ - 2}}\sin (\pi {x^{ - 1}})} \right|{\text{d}}x}  = 2 + 2 +  \ldots  + 2 = 18$     (M1)A1

[2 marks]

There were a pleasing number of candidates who answered part (a) correctly. Fewer were successful with part (b). It was expected by this stage of the paper that candidates would be able to just write down the value of the integral rather than use substitution to evaluate it.

There were a pleasing number of candidates who answered part (a) correctly. Fewer were successful with part (b). It was expected by this stage of the paper that candidates would be able to just write down the value of the integral rather than use substitution to evaluate it.

There were disappointingly few correct answers to part (c) with candidates not realising that it was necessary to combine the previous two parts in order to write down the answer.