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]