Q103 / 180 · KCET 2011 · Mathematics

$\frac{d}{d x}\left[\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]$ is

  • Correct answer
    A
    $\frac{1}{4}$
  • B
    $\frac{1}{2}$
  • C
    $-\frac{1}{2}$
  • D
    $-\frac{3}{4}$

Solution

$\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right\}$ put $x=2 \cos \theta$ $\Rightarrow \frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\left.\frac{2+\cos \theta \cdot 2}{2-\cos \theta \cdot 2}\right)}\right\}\right.$ $=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)\right\}$ $=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}}\right)\right\}$ $=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1}(\cot \theta / 2)\right)\right\}$ $=\frac{d}{d x}\left\{\cos ^{2} \frac{\theta}{2}\right\}=\frac{1}{2} \cdot \frac{d}{d x}(1+\cos \theta)$ $=\frac{d}{d x}\left(\frac{1}{2}+\frac{1}{2} \cdot \frac{x}{2}\right)$ $=\frac{d}{d x}\left(\frac{1}{2}+\frac{x}{4}\right)$ $=(0+1 / 4)=1 / 4$

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

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