Q105 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics

For $-1 < x < 1$, if $f(x)=\cos ^2\left(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)$, then $f^{\prime}(x)=$

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

Solution

Given function $f(x)=\cos ^2\left(\tan ^1 \sqrt{\frac{1-x}{1+x}}\right)$, for $-1 < x < 1$. Let $x=\cos 2 \theta$ Then, $\quad \frac{1-x}{1+x}=\frac{1-\cos 2 \theta}{1+\cos 2 \theta}=\tan ^2 \theta$ So, $f(x)=\cos ^2 \theta=\frac{1+\cos 2 \theta}{2}=\frac{1+x}{2}$ So, $\quad f^{\prime}(x)=\frac{1}{2}$.

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

Relevant Questions