Write the correct alternative in the following:
If $f(x)=(\cos x+i \sin x)(\cos 2 x+i \sin 2 x)(\cos 3 x+i \sin 3 x) \ldots(\cos n x+i \sin n x)$ and $f(1)=1$, then $f^{\prime \prime}(1)$ is equal to
A. $\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
B. $\left\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right\}^{2}$
C. $-\left\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right\}^{2}$
D. none of these
Given:
$f(x)=(\cos x+i \sin x)(\cos 2 x+i \sin 2 x)(\cos 3 x+i \sin 3 x) \ldots(\cos n x+i \sin n x)$
Since $e^{i x}=\cos x+i \sin x$
So, $f(x)=e^{i x} \times e^{i 2 x} \times e^{i 3 x} \times e^{i 4 x} \times \ldots \times e^{i n x}$
$f(x)=e^{i x(1+2+3+4+\cdots+n)}$
$=e^{i x \frac{n(n+1)}{2}}$
$f(1)=e^{\frac{i n(n+1)}{2}}$
$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{ix} \frac{\mathrm{n}(\mathrm{n}+1)}{2} \mathrm{e}^{\mathrm{ix} \frac{\mathrm{n}(\mathrm{n}+1)}{2}}$
$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{i}^{2} \mathrm{x}^{2}\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2} \mathrm{e}^{\mathrm{ix} \frac{\mathrm{n}(\mathrm{n}+1)}{2}}$
$\mathrm{f}^{\prime}(\mathrm{x})=-\mathrm{x}^{2}\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2} \mathrm{e}^{\mathrm{ix} \frac{\mathrm{n}(\mathrm{n}+1)}{2}}$
$\mathrm{f}^{\prime}(1)=-1^{2}\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2} \times 1$
$=-\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2}$