Q75 / 160 · TS EAMCET 2010 · Mathematics

If $f(x)=(\cos x)(\cos 2 x) \ldots(\cos n x)$ then $f^{\prime}(x)+\sum_{r=1}^n(r \tan r x) f(x)$ is equal to

  • A
    $f(x)$
  • Correct answer
    B
    0
  • C
    $-f(x)$
  • D
    $2 f(x)$

Solution

$f(x)=(\cos x)(\cos 2 x) \ldots(\cos n x)$ $f^{\prime}(x)=-\sin x \cdot \cos 2 x \ldots \cos n x$ $+\cos x \frac{d}{d x}\{\cos 2 x \cdot \cos 3 x \ldots \cos n x\}$ $f^{\prime}(x)=-\sin x \cdot \cos 2 x \ldots \cos n x$ $+\cos x[-2 \sin 2 x \cdot \cos 3 x \ldots \cos n x$ $\left.+\cos 2 x \frac{d}{d x} \cos 3 x \cdot \cos 4 x \ldots \cos n x\right]$ $f^{\prime}(x) \Rightarrow-(\sin x \cdot \cos 2 x \ldots \cos n x)$ $-(2 \cos x \cdot \sin 2 x \cdot \cos 3 x \ldots \cos n x)$ $+\cos x \cdot \cos 2 x \frac{d}{d x}(\cos 3 x \cdot \cos 4 x \cdot \cos n x)$ $f^{\prime}(x) \Rightarrow-(\sin x \cdot \cos 2 x \ldots \cos n x)$ $-(2 \cos x \cdot \sin 2 x \ldots \cos n x)$ $-(3 \cos x \cdot \cos 2 x \cdot \sin 3 x \ldots \cos n x)$ $-(n \cos x \cdot \cos 2 x \ldots \sin n x)$ So, $\Rightarrow f^{\prime}(x)+\sum_{r=1}^n(r \tan r x) f(x)$ $=f^{\prime}(x)+\{\tan x+2 \tan 2 x+3 \tan 3 x$ $+\ldots+n \tan n x\} f(x)$ $=f^{\prime}(x)+f(x) \tan x+2 f(x) \tan 2 x$ $+\ldots+n f(x) \tan n x$ $=f^{\prime}(x)+[(\sin x \cdot \cos 2 x \ldots \cos n x)$ $+(2 \cos x \cdot \sin 2 x \ldots \cos n x)$ $+\ldots+(n \cos x \cdot \cos 2 x \ldots \sin n x)]$ $=f^{\prime}(x)-f^{\prime}(x)$ $\Rightarrow 0$ Hence, $f^{\prime}(x)+\sum_{r=1}^n(r \tan r x) f(x)=0$

Hence, the correct answer is option: 'B'.

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