MHT-CET Mathematics Full Test-10 QUE-34 Solution

Question-34 $f(x)=(\cos x+i\sin x)(\cos3x+i\sin3x)\ldots(\cos(2n-1)x+i\sin(2n-1)x),\ n\in\mathbb{N}$

Then $f''(x)=$ _ , (Where $i=\sqrt{-1}$ )

(A) $n^2f(x)$

(B) $-n^4f(x)$

(D) $n^4f(x)$

Answer: (B)

Explanation:

f(x)=(\cos x+i\sin x)(\cos3x+i\sin3x)…[\cos(2n-1)x+i\sin(2n-1)x]

=e^{ix}\cdot e^{i3x}…e^{i(2n-1)x}

=e^{i(x+3x+…(2n-1)x)}

$=e^{i(n^2x)}$ …(Sum of n odd terms = $n^2$ )

$f(x)=e^{in^2x}$ …(i)

Differentiating w.r.t. x, we get

f'(x)=in^2e^{in^2x}

Again differentiating w.r.t. x, we get

f''(x)=i^2n^2\times n^2\times e^{in^2x}

=-1\times n^4\times e^{in^2x}

$=-n^4f(x)$ …(From (i))