Q81 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
If $y = \cos^{-1}\left(\dfrac{1 - 4^x}{1 + 4^x}\right)$, then $\dfrac{dy}{dx}$ at $x = 1$ is
- Correct answerA$\dfrac{4\log 2}{5}$
- B$\dfrac{\log 2}{5}$
- C$\dfrac{2\log 8}{5}$
- D$\dfrac{5\log 8}{2}$
Solution
Given $y = \cos^{-1}\left(\dfrac{1 - 4^x}{1 + 4^x}\right)$
Let $2^x = \tan \theta$. Then $4^x = (2^x)^2 = \tan^2 \theta$.
Substituting this in the given expression:
$y = \cos^{-1}\left(\dfrac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\right)$
$y = \cos^{-1}(\cos 2\theta)$
Since $2^x > 0$, $\tan \theta > 0 \Rightarrow \theta \in \left(0, \dfrac{\pi}{2}\right) \Rightarrow 2\theta \in (0, \pi)$.
Thus, $y = 2\theta = 2\tan^{-1}(2^x)$.
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = 2 \cdot \dfrac{1}{1 + (2^x)^2} \cdot \dfrac{d}{dx}(2^x)$
$\dfrac{dy}{dx} = \dfrac{2}{1 + 4^x} \cdot 2^x \log 2$
At $x = 1$:
$\dfrac{dy}{dx} = \dfrac{2}{1 + 4^1} \cdot 2^1 \log 2 = \dfrac{4\log 2}{5}$
Answer: $\dfrac{4\log 2}{5}$
Let $2^x = \tan \theta$. Then $4^x = (2^x)^2 = \tan^2 \theta$.
Substituting this in the given expression:
$y = \cos^{-1}\left(\dfrac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\right)$
$y = \cos^{-1}(\cos 2\theta)$
Since $2^x > 0$, $\tan \theta > 0 \Rightarrow \theta \in \left(0, \dfrac{\pi}{2}\right) \Rightarrow 2\theta \in (0, \pi)$.
Thus, $y = 2\theta = 2\tan^{-1}(2^x)$.
Differentiating with respect to $x$:
$\dfrac{dy}{dx} = 2 \cdot \dfrac{1}{1 + (2^x)^2} \cdot \dfrac{d}{dx}(2^x)$
$\dfrac{dy}{dx} = \dfrac{2}{1 + 4^x} \cdot 2^x \log 2$
At $x = 1$:
$\dfrac{dy}{dx} = \dfrac{2}{1 + 4^1} \cdot 2^1 \log 2 = \dfrac{4\log 2}{5}$
Answer: $\dfrac{4\log 2}{5}$
Hence, the correct answer is option: 'A'.