Show that $f''(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)$ .

Obtain a similar expression for ${f^{(4)}}(x)$ .

Suggest an expression for ${f^{(2n)}}(x)$, $n \in {\mathbb{Z}^ + }$, and prove your conjecture using mathematical induction.

$f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x$     A1

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

$= 2{{\text{e}}^x}\cos x$     A1

$= 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)$     AG

[3 marks]

$f'''(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)$     A1

${f^{(4)}}(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) - 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)$     A1

$= 4{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)$     A1

$= 4{{\text{e}}^x}\sin (x + \pi )$     A1

[4 marks]

the conjecture is that

${f^{(2n)}}(x) = {2^n}{{\text{e}}^x}\sin \left( {x + \frac{{n\pi }}{2}} \right)$     A1

for n  = 1, this formula gives

$f''(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)$ which is correct     A1

let the result be true for n = k , $\left( {i.e.{\text{ }}{f^{(2k)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)} \right)$     M1

consider ${f^{(2k + 1)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)$     M1

${f^{\left( {2(k + 1)} \right)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) - {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)$     A1

$= {2^{k + 1}}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)$     A1

$= {2^{k + 1}}{{\text{e}}^x}\sin \left( {x + \frac{{(k + 1)\pi }}{2}} \right)$     A1

therefore true for $n = k \Rightarrow$ true for $n = k + 1$ and since true for $n = 1$

the result is proved by induction.     R1

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

[8 marks]