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प्रश्न
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
विकल्प
\[\frac{n\left( n + 1 \right)}{2}\]
\[\left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2\]
\[- \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2\]
none of these
उत्तर
(c) \[- \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2\]
Here,
\[f\left( x \right) = \left( \cos x + i \sin x \right)\left( \cos2x + i \sin2x \right) . . . \left( \cos nx + i \sin nx \right)\]
\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right) \left( \cos x + i \sin x \right)^2 . . . \left( \cos x + i \sin x \right)^n \]
\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right)^{1 + 2 + 3 . . . . . . . . . . . n} \]
\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right)^\frac{n\left( n + 1 \right)}{2} \]
\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right)^a \left[ \text { where a } = \frac{n\left( n + 1 \right)}{2} \right]\]
\[ \Rightarrow f\left( x \right) = \left( \cos ax + i \sin ax \right) . . . \left( 1 \right)\]
\[ \Rightarrow f\left( 1 \right) = \left( \cos a + i \sin a \right)\]
\[ \Rightarrow 1 = \left( \cos a + i \sin a \right) . . . \left( 2 \right) \left[ \because f\left( 1 \right) = 1 \right]\]
\[\text { Differentiating eqn } . \left( 1 \right),\text { we get }, \]
\[f'\left( x \right) = a\left( - \sin ax + i \cos ax \right)\]
\[ \Rightarrow f''\left( x \right) = a^2 \left( - \cos ax - i \sin ax \right)\]
\[ \Rightarrow f''\left( x \right) = - a^2 \left( \cos ax + i \sin ax \right)\]
\[ \Rightarrow f''\left( x \right) = - \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2 \left( \cos ax + i \sin ax \right)\]
\[ \Rightarrow f''\left( 1 \right) = - \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2 \left( \cos a + i \sin a \right)\]
\[ \Rightarrow f''\left( 1 \right) = - \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2 \left[ \text{ Using } \left( 2 \right) \right]\]
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