A particle P moves in a straight line with displacement relative to origin given by

\[s = 2\sin (\pi t) + \sin (2\pi t),{\text{ }}t \geqslant 0,\]

where t is the time in seconds and the displacement is measured in centimetres.

(i)     Write down the period of the function s.

(ii)     Find expressions for the velocity, v, and the acceleration, a, of P.

(iii)     Determine all the solutions of the equation v = 0 for \(0 \leqslant t \leqslant 4\).

Consider the function

\[f(x) = A\sin (ax) + B\sin (bx),{\text{ }}A,{\text{ }}a,{\text{ }}B,{\text{ }}b,{\text{ }}x \in \mathbb{R}.\]

Use mathematical induction to prove that the\({(2n)^{{\text{th}}}}\) derivative of f is given by \(({f^{(2n)}}(x) = {( - 1)^n}\left( {A{a^{2n}}\sin (ax) + B{b^{2n}}\sin (bx)} \right)\), for all \(n \in {\mathbb{Z}^ + }\).

(i)     the period is 2     A1

(ii)     \(v = \frac{{{\text{d}}s}}{{{\text{d}}t}} = 2\pi \cos (\pi t) + 2\pi \cos (2\pi t)\)     (M1)A1

\(a = \frac{{{\text{d}}v}}{{{\text{d}}t}} = - 2{\pi ^2}\sin (\pi t) - 4{\pi ^2}\sin (2\pi t)\)     (M1)A1

(iii)     \(v = 0\)

\(2\pi \left( {\cos (\pi t) + \cos (2\pi t)} \right) = 0\)

EITHER

\(\cos (\pi t) + 2{\cos ^2}(\pi t) - 1 = 0\)     M1

\(\left( {2\cos (\pi t) - 1} \right)\left( {\cos (\pi t) + 1} \right) = 0\)     (A1)

\(\cos (\pi t) = \frac{1}{2}{\text{ or }}\cos (\pi t) = - 1\)     A1

\(t = \frac{1}{3},{\text{ }}t = 1\)     A1

\(t = \frac{5}{3},{\text{ }}t = \frac{7}{3},{\text{ }}t = \frac{{11}}{3},{\text{ }}t = 3\)     A1

OR

\(2\cos \left( {\frac{{\pi t}}{2}} \right)\cos \left( {\frac{{3\pi t}}{2}} \right) = 0\)     M1

\(\cos \left( {\frac{{\pi t}}{2}} \right) = 0{\text{ or }}\cos \left( {\frac{{3\pi t}}{2}} \right) = 0\)     A1A1

\(t = \frac{1}{3},{\text{ 1}}\)     A1

\(t = \frac{5}{3},{\text{ }}\frac{7}{3},{\text{ }}3,{\text{ }}\frac{{11}}{3}\)     A1

[10 marks]

\(P(n):{f^{(2n)}}(x) = {( - 1)^n}\left( {A{a^{2n}}\sin (ax) + B{b^{2n}}\sin (bx)} \right)\)

\(P(1):f''(x) = {\left( {Aa\cos (ax) + Bb\cos (bx)} \right)^\prime }\)     M1

\( = - A{a^2}\sin (ax) - B{b^2}\sin (bx)\)

\( = - 1\left( {A{a^2}\sin (ax) + B{b^2}\sin (bx)} \right)\)     A1

\(\therefore P(1)\) true

assume that

\(P(k):{f^{(2k)}}(x) = {( - 1)^k}\left( {A{a^{2k}}\sin (ax) + B{b^{2k}}\sin (bx)} \right)\) is true     M1

consider \(P(k + 1)\)

\({f^{(2k + 1)}}(x) = {( - 1)^k}\left( {A{a^{2k + 1}}\cos (ax) + B{b^{2k + 1}}\cos (bx)} \right)\)     M1A1

\({f^{(2k + 2)}}(x) = {( - 1)^k}\left( { - A{a^{2k + 2}}\sin (ax) - B{b^{2k + 2}}\sin (bx)} \right)\)     A1

\( = {( - 1)^{k + 1}}\left( {A{a^{2k + 2}}\sin (ax) + B{b^{2k + 2}}\sin (bx)} \right)\)     A1

\(P(k)\) true implies \(P(k + 1)\) true, \(P(1)\) true so \(P(n)\) true \(\forall n \in {\mathbb{Z}^ + }\)     R1

Note: Award the final R1 only if the previous three M marks have been awarded.

[8 marks]

In (a), only a few candidates gave the correct period but the expressions for velocity and acceleration were correctly obtained by most candidates. In (a)(iii), many candidates manipulated the equation v = 0 correctly to give the two possible values for \(\cos (\pi t)\) but then failed to find all the possible values of t.

Solutions to (b) were disappointing in general with few candidates giving a correct solution.