Q61 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics

If $y=\tan ^{-1}\left(\frac{\sqrt{x}-x}{1+x^{\frac{3}{2}}}\right)$, then $y^{\prime}(1)$ is equal to

  • A
    0
  • B
    $\frac{1}{2}$
  • C
    -1
  • Correct answer
    D
    $-\frac{1}{4}$

Solution

Given, $\begin{aligned} y & =\tan ^{-1}\left(\frac{\sqrt{x}-x}{1+x^{\frac{3}{2}}}\right) \\ y & =\tan ^{-1}\left(\frac{\sqrt{x}-x}{1+\sqrt{x} \cdot x}\right) \\ y & =\tan ^{-1}(\sqrt{x})-\tan ^{-1} x \\ & {\left[\because \tan ^{-1}\left(\frac{x-y}{1+x y}\right)=\tan ^{-1} x-\tan ^{-1} y\right] }\end{aligned}$ Differentiating w.r.t $x$ $y^{\prime}(x)=\frac{d y}{d x}=\frac{1}{1+x} \times \frac{1}{2 \sqrt{x}}-\frac{1}{1+x^2}$ Now, $\quad y^{\prime}(1)=\frac{1}{2} \times \frac{1}{2}-\frac{1}{2}=-\frac{1}{4}$

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

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