Q80 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
If $y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + \cdots \cdots \cdots \infty}}}$, then $\left(\dfrac{dy}{dx}\right)^2$ at $x = \dfrac{\pi}{4}$ is
- A$\dfrac{5}{4}$
- B$\dfrac{-5}{4}$
- Correct answerC$\dfrac{4}{5}$
- D$\dfrac{-4}{5}$
Solution
Given $y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + \cdots \infty}}}$
Squaring both sides, we get:
$y^2 = \tan x + y$
Differentiating both sides with respect to $x$:
$2y \dfrac{dy}{dx} = \sec^2 x + \dfrac{dy}{dx}$
$\dfrac{dy}{dx} (2y - 1) = \sec^2 x$
$\dfrac{dy}{dx} = \dfrac{\sec^2 x}{2y - 1}$
At $x = \dfrac{\pi}{4}$, $\tan x = 1$. Substituting this into the equation for $y$:
$y^2 - y - 1 = 0$
Solving for $y$, we get $y = \dfrac{1 \pm \sqrt{5}}{2}$. Since $y$ represents a principal square root, $y > 0$, so $y = \dfrac{1 + \sqrt{5}}{2}$.
Substituting $y = \dfrac{1 + \sqrt{5}}{2}$ and $x = \dfrac{\pi}{4}$ into the derivative:
$2y - 1 = 2\left(\dfrac{1 + \sqrt{5}}{2}\right) - 1 = \sqrt{5}$
$\sec^2\left(\dfrac{\pi}{4}\right) = (\sqrt{2})^2 = 2$
$\dfrac{dy}{dx} = \dfrac{2}{\sqrt{5}}$
Squaring both sides:
$\left(\dfrac{dy}{dx}\right)^2 = \dfrac{4}{5}$
Answer: $\dfrac{4}{5}$
Squaring both sides, we get:
$y^2 = \tan x + y$
Differentiating both sides with respect to $x$:
$2y \dfrac{dy}{dx} = \sec^2 x + \dfrac{dy}{dx}$
$\dfrac{dy}{dx} (2y - 1) = \sec^2 x$
$\dfrac{dy}{dx} = \dfrac{\sec^2 x}{2y - 1}$
At $x = \dfrac{\pi}{4}$, $\tan x = 1$. Substituting this into the equation for $y$:
$y^2 - y - 1 = 0$
Solving for $y$, we get $y = \dfrac{1 \pm \sqrt{5}}{2}$. Since $y$ represents a principal square root, $y > 0$, so $y = \dfrac{1 + \sqrt{5}}{2}$.
Substituting $y = \dfrac{1 + \sqrt{5}}{2}$ and $x = \dfrac{\pi}{4}$ into the derivative:
$2y - 1 = 2\left(\dfrac{1 + \sqrt{5}}{2}\right) - 1 = \sqrt{5}$
$\sec^2\left(\dfrac{\pi}{4}\right) = (\sqrt{2})^2 = 2$
$\dfrac{dy}{dx} = \dfrac{2}{\sqrt{5}}$
Squaring both sides:
$\left(\dfrac{dy}{dx}\right)^2 = \dfrac{4}{5}$
Answer: $\dfrac{4}{5}$
Hence, the correct answer is option: 'C'.