Q88 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics
If $y = \tan^{-1}\left\{\dfrac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}}\right\}$, where $|x| < 1$, then $\dfrac{dy}{dx}$ is equal to
- A$\dfrac{-x}{\sqrt{1 - x^4}}$
- Correct answerB$\dfrac{x}{\sqrt{1 - x^4}}$
- C$\dfrac{-2x}{\sqrt{1 - x^4}}$
- D$\dfrac{2x}{\sqrt{1 - x^4}}$
Solution
Let $x^2 = \cos 2\theta$. Then $\theta = \dfrac{1}{2}\cos^{-1}(x^2)$.
Substituting this into the given expression, we get:
$\sqrt{1 + x^2} = \sqrt{1 + \cos 2\theta} = \sqrt{2\cos^2\theta} = \sqrt{2}\cos\theta$
$\sqrt{1 - x^2} = \sqrt{1 - \cos 2\theta} = \sqrt{2\sin^2\theta} = \sqrt{2}\sin\theta$
Now, substitute these into $y$:
$y = \tan^{-1}\left\{\dfrac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right\}$
Dividing the numerator and the denominator by $\sqrt{2}\cos\theta$:
$y = \tan^{-1}\left\{\dfrac{1 - \tan\theta}{1 + \tan\theta}\right\}$
$y = \tan^{-1}\left\{\tan\left(\dfrac{\pi}{4} - \theta\right)\right\}$
$y = \dfrac{\pi}{4} - \theta$
Substituting back $\theta = \dfrac{1}{2}\cos^{-1}(x^2)$:
$y = \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}(x^2)$
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = 0 - \dfrac{1}{2}\left(\dfrac{-1}{\sqrt{1 - (x^2)^2}}\right) \cdot \dfrac{d}{dx}(x^2)$
$\dfrac{dy}{dx} = \dfrac{1}{2} \cdot \dfrac{1}{\sqrt{1 - x^4}} \cdot 2x$
$\dfrac{dy}{dx} = \dfrac{x}{\sqrt{1 - x^4}}$
Answer: $\dfrac{x}{\sqrt{1 - x^4}}$
Substituting this into the given expression, we get:
$\sqrt{1 + x^2} = \sqrt{1 + \cos 2\theta} = \sqrt{2\cos^2\theta} = \sqrt{2}\cos\theta$
$\sqrt{1 - x^2} = \sqrt{1 - \cos 2\theta} = \sqrt{2\sin^2\theta} = \sqrt{2}\sin\theta$
Now, substitute these into $y$:
$y = \tan^{-1}\left\{\dfrac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right\}$
Dividing the numerator and the denominator by $\sqrt{2}\cos\theta$:
$y = \tan^{-1}\left\{\dfrac{1 - \tan\theta}{1 + \tan\theta}\right\}$
$y = \tan^{-1}\left\{\tan\left(\dfrac{\pi}{4} - \theta\right)\right\}$
$y = \dfrac{\pi}{4} - \theta$
Substituting back $\theta = \dfrac{1}{2}\cos^{-1}(x^2)$:
$y = \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}(x^2)$
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = 0 - \dfrac{1}{2}\left(\dfrac{-1}{\sqrt{1 - (x^2)^2}}\right) \cdot \dfrac{d}{dx}(x^2)$
$\dfrac{dy}{dx} = \dfrac{1}{2} \cdot \dfrac{1}{\sqrt{1 - x^4}} \cdot 2x$
$\dfrac{dy}{dx} = \dfrac{x}{\sqrt{1 - x^4}}$
Answer: $\dfrac{x}{\sqrt{1 - x^4}}$
Hence, the correct answer is option: 'B'.