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

 

Options

• \[\frac{n\left( n + 1 \right)}{2}\]

• \[\left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2\]

• \[- \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2\]

• none of these

Answer 1

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

Here, 

\[f\left( x \right) = \left( \cos x + i \sin x \right)\left( \cos2x + i \sin2x \right) . . . \left( \cos nx + i \sin nx \right)\]

\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right) \left( \cos x + i \sin x \right)^2 . . . \left( \cos x + i \sin x \right)^n \]

\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right)^{1 + 2 + 3 . . . . . . . . . . . n} \]

\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right)^\frac{n\left( n + 1 \right)}{2} \]

\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right)^a \left[ \text { where a } = \frac{n\left( n + 1 \right)}{2} \right]\]

\[ \Rightarrow f\left( x \right) = \left( \cos ax + i \sin ax \right) . . . \left( 1 \right)\]

\[ \Rightarrow f\left( 1 \right) = \left( \cos a + i \sin a \right)\]

\[ \Rightarrow 1 = \left( \cos a + i \sin a \right) . . . \left( 2 \right) \left[ \because f\left( 1 \right) = 1 \right]\]

\[\text { Differentiating eqn } . \left( 1 \right),\text {  we get }, \]

\[f'\left( x \right) = a\left( - \sin ax + i \cos ax \right)\]

\[ \Rightarrow f''\left( x \right) = a^2 \left( - \cos ax - i \sin ax \right)\]

\[ \Rightarrow f''\left( x \right) = - a^2 \left( \cos ax + i \sin ax \right)\]

\[ \Rightarrow f''\left( x \right) = - \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2 \left( \cos ax + i \sin ax \right)\]

\[ \Rightarrow f''\left( 1 \right) = - \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2 \left( \cos a + i \sin a \right)\]

\[ \Rightarrow f''\left( 1 \right) = - \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2 \left[ \text{ Using } \left( 2 \right) \right]\]