Question

If f(x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x) ...... (cos nx + i sin nx) and f(1) = 1, then f'' (1) is equal to

(a) $\frac{n\left(n+1\right)}{2}$
(b) ${\left\{\frac{n\left(n+1\right)}{2}\right\}}^2$
(c) $-{\left\{\frac{n\left(n+1\right)}{2}\right\}}^2$
(d) none of these

Answer 1

(c) $-{\left\{\frac{n\left(n+1\right)}{2}\right\}}^2$

Here,
$$\begin{gathered} f\left(x\right)=\left(\cos x+i\sin x\right)\left(\cos 2x+i\sin 2x\right)...\left(\cos nx+i\sin nx\right) \\ \Rightarrow f\left(x\right)=\left(\cos x+i\sin x\right){\left(\cos x+i\sin x\right)}^2...{\left(\cos x+i\sin x\right)}^n \\ \Rightarrow f\left(x\right)={\left(\cos x+i\sin x\right)}^{1+2+3...........n} \\ \Rightarrow f\left(x\right)={\left(\cos x+i\sin x\right)}^{\frac{n\left(n+1\right)}{2}} \\ \Rightarrow f\left(x\right)={\left(\cos x+i\sin x\right)}^a\left[\mathrm{where}a=\frac{n\left(n+1\right)}{2}\right] \\ \Rightarrow f\left(x\right)=\left(\cos ax+i\sin ax\right)...\left(1\right) \\ \Rightarrow f\left(1\right)=\left(\cos a+i\sin a\right) \\ \Rightarrow 1=\left(\cos a+i\sin a\right)...\left(2\right)\left[\because f\left(1\right)=1\right] \\ \mathrm{Differentiating}\mathrm{eqn}.\left(1\right),\mathrm{we}\mathrm{get}, \\ f\text{'}\left(x\right)=a\left(-\sin ax+i\cos ax\right) \\ \Rightarrow f\text{'}\text{'}\left(x\right)=a^2\left(-\cos ax-i\sin ax\right) \\ \Rightarrow f\text{'}\text{'}\left(x\right)=-a^2\left(\cos ax+i\sin ax\right) \\ \Rightarrow f\text{'}\text{'}\left(x\right)=-{\left\{\frac{n\left(n+1\right)}{2}\right\}}^2\left(\cos ax+i\sin ax\right) \\ \Rightarrow f\text{'}\text{'}\left(1\right)=-{\left\{\frac{n\left(n+1\right)}{2}\right\}}^2\left(\cos a+i\sin a\right) \\ \Rightarrow f\text{'}\text{'}\left(1\right)=-{\left\{\frac{n\left(n+1\right)}{2}\right\}}^2\left[\mathrm{Using}\left(2\right)\right] \end{gathered}$$