Question

If $f\left(x\right)$ is given by $f\left(x\right)=(\cos x+i\sin x)(\cos 3x+i\sin 3x)....\left(\cos \right(2n-1)x+i\sin (2n-1\left)x\right)$, then $f^{\prime \prime }\left(x\right)$ is equal to

• A. $n^{3}f\left(x\right)$
• B. $-n^{4}f\left(x\right)$
• C. $-n^{2}f\left(x\right)$
• D. $n^{4}f\left(x\right)$

Answer 1

The correct option is B $-n^{4}f\left(x\right)$

$f\left(x\right)=(\cos x+i\sin x)(\cos 3x+i\sin 3x)....\left(\cos \right(2n-1)x+i\sin (2n-1\left)x\right)$
$f\left(x\right)=\cos \{x+3x+5x+......+(2n-1)x\}+i\sin \{x+3x+5x+........+(2n-1)x\}$
$f\left(x\right)=\cos n^{2}x+i\sin n^{2}x$
$f^{\prime }\left(x\right)={-n}^2\left(\sin n^{2}x\right)+n^2\left(i\cos n^{2}x\right)$
$f^{\prime \prime }\left(x\right)=-n^4\cos n^{2}x-n^{4}i\sin n^{2}x$
$f^{\prime \prime }\left(x\right)=-n^4(\cos n^{2}x+i\sin n^{2}x)$
$f^{\prime \prime }\left(x\right)=-n^{4}f\left(x\right)$