Q98 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
If $y = f\left(\dfrac{3 + 2x}{3 - 2x}\right)$, where $f(x) = \tan(\log x)$ and $\dfrac{dy}{dx} = \left(\dfrac{A}{B + C x^2}\right) \cdot \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$, then the respective values of A, B and C are ......
- A$3, 2, -4$
- B$12, -9, 4$
- C$6, 6, -2$
- Correct answerD$12, 9, -4$
Solution
Given $y = f\left(\dfrac{3 + 2x}{3 - 2x}\right)$ and $f(x) = \tan(\log x)$
$y = \tan\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
Differentiating with respect to $x$ using the chain rule:
$\dfrac{dy}{dx} = \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right) \cdot \dfrac{d}{dx}\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
$\dfrac{dy}{dx} = \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right) \cdot \dfrac{3 - 2x}{3 + 2x} \cdot \dfrac{d}{dx}\left(\dfrac{3 + 2x}{3 - 2x}\right)$
$\dfrac{dy}{dx} = \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right) \cdot \dfrac{3 - 2x}{3 + 2x} \cdot \left(\dfrac{2(3 - 2x) - (-2)(3 + 2x)}{(3 - 2x)^2}\right)$
$\dfrac{dy}{dx} = \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right) \cdot \dfrac{3 - 2x}{3 + 2x} \cdot \dfrac{12}{(3 - 2x)^2}$
$\dfrac{dy}{dx} = \dfrac{12}{(3 + 2x)(3 - 2x)} \cdot \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
$\dfrac{dy}{dx} = \dfrac{12}{9 - 4x^2} \cdot \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
Comparing with the given expression $\dfrac{dy}{dx} = \left(\dfrac{A}{B + C x^2}\right) \cdot \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
We get $A = 12$, $B = 9$, $C = -4$.
Answer: $12, 9, -4$
$y = \tan\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
Differentiating with respect to $x$ using the chain rule:
$\dfrac{dy}{dx} = \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right) \cdot \dfrac{d}{dx}\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
$\dfrac{dy}{dx} = \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right) \cdot \dfrac{3 - 2x}{3 + 2x} \cdot \dfrac{d}{dx}\left(\dfrac{3 + 2x}{3 - 2x}\right)$
$\dfrac{dy}{dx} = \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right) \cdot \dfrac{3 - 2x}{3 + 2x} \cdot \left(\dfrac{2(3 - 2x) - (-2)(3 + 2x)}{(3 - 2x)^2}\right)$
$\dfrac{dy}{dx} = \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right) \cdot \dfrac{3 - 2x}{3 + 2x} \cdot \dfrac{12}{(3 - 2x)^2}$
$\dfrac{dy}{dx} = \dfrac{12}{(3 + 2x)(3 - 2x)} \cdot \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
$\dfrac{dy}{dx} = \dfrac{12}{9 - 4x^2} \cdot \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
Comparing with the given expression $\dfrac{dy}{dx} = \left(\dfrac{A}{B + C x^2}\right) \cdot \sec^2\left(\log\left(\dfrac{3 + 2x}{3 - 2x}\right)\right)$
We get $A = 12$, $B = 9$, $C = -4$.
Answer: $12, 9, -4$
Hence, the correct answer is option: 'D'.