Q83 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics

If $f'(x) = \sin^2 x$ and $y = f\left(\dfrac{2x-1}{x^2+1}\right)$, then $\dfrac{dy}{dx}$ at $x = 1$ is

  • A
    $\dfrac{1}{4}\sin\left(\dfrac{1}{2}\right)$
  • B
    $\dfrac{1}{4}\sin^2\left(\dfrac{1}{2}\right)$
  • C
    $\sin^2\left(\dfrac{1}{4}\right)$
  • Correct answer
    D
    $\dfrac{1}{2}\sin^2\left(\dfrac{1}{2}\right)$

Solution

Given $y = f\left(\dfrac{2x-1}{x^2+1}\right)$

Differentiating with respect to $x$ using the chain rule:

$\dfrac{dy}{dx} = f'\left(\dfrac{2x-1}{x^2+1}\right) \cdot \dfrac{d}{dx}\left(\dfrac{2x-1}{x^2+1}\right)$

$\dfrac{dy}{dx} = f'\left(\dfrac{2x-1}{x^2+1}\right) \cdot \left[ \dfrac{(x^2+1)(2) - (2x-1)(2x)}{(x^2+1)^2} \right]$

Substituting $x = 1$:

$\left. \dfrac{dy}{dx} \right|_{x=1} = f'\left(\dfrac{2(1)-1}{1^2+1}\right) \cdot \left[ \dfrac{(1^2+1)(2) - (2(1)-1)(2(1))}{(1^2+1)^2} \right]$

$\left. \dfrac{dy}{dx} \right|_{x=1} = f'\left(\dfrac{1}{2}\right) \cdot \left[ \dfrac{4 - 2}{4} \right]$

$\left. \dfrac{dy}{dx} \right|_{x=1} = f'\left(\dfrac{1}{2}\right) \cdot \dfrac{1}{2}$

Since $f'(x) = \sin^2 x$, we have $f'\left(\dfrac{1}{2}\right) = \sin^2\left(\dfrac{1}{2}\right)$

$\left. \dfrac{dy}{dx} \right|_{x=1} = \dfrac{1}{2}\sin^2\left(\dfrac{1}{2}\right)$

Answer: $\dfrac{1}{2}\sin^2\left(\dfrac{1}{2}\right)$

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

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