Q82 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics

If $\sqrt{y+x} + \sqrt{y-x} = c$ such that $\dfrac{dy}{dx} = f(x) - \sqrt{[f(x)]^2 - 1}$, then $f(x) =$ ____

  • Correct answer
    A
    $\dfrac{y}{x}$
  • B
    $-\dfrac{x}{y}$
  • C
    $-\dfrac{y}{x}$
  • D
    $\dfrac{x}{y}$

Solution

Given equation: $\sqrt{y+x} + \sqrt{y-x} = c$

Differentiating both sides with respect to $x$:

$\dfrac{1}{2\sqrt{y+x}} \left(\dfrac{dy}{dx} + 1\right) + \dfrac{1}{2\sqrt{y-x}} \left(\dfrac{dy}{dx} - 1\right) = 0$

$\dfrac{dy}{dx} \left( \dfrac{1}{\sqrt{y+x}} + \dfrac{1}{\sqrt{y-x}} \right) = \dfrac{1}{\sqrt{y-x}} - \dfrac{1}{\sqrt{y+x}}$

$\dfrac{dy}{dx} \left( \dfrac{\sqrt{y-x} + \sqrt{y+x}}{\sqrt{y^2-x^2}} \right) = \dfrac{\sqrt{y+x} - \sqrt{y-x}}{\sqrt{y^2-x^2}}$

$\dfrac{dy}{dx} = \dfrac{\sqrt{y+x} - \sqrt{y-x}}{\sqrt{y+x} + \sqrt{y-x}}$

Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{y+x} - \sqrt{y-x}$:

$\dfrac{dy}{dx} = \dfrac{(\sqrt{y+x} - \sqrt{y-x})^2}{(\sqrt{y+x})^2 - (\sqrt{y-x})^2}$

$\dfrac{dy}{dx} = \dfrac{(y+x) + (y-x) - 2\sqrt{y^2-x^2}}{(y+x) - (y-x)}$

$\dfrac{dy}{dx} = \dfrac{2y - 2\sqrt{y^2-x^2}}{2x}$

$\dfrac{dy}{dx} = \dfrac{y}{x} - \dfrac{\sqrt{y^2-x^2}}{x}$

$\dfrac{dy}{dx} = \dfrac{y}{x} - \sqrt{\left(\dfrac{y}{x}\right)^2 - 1}$

Comparing this with the given equation $\dfrac{dy}{dx} = f(x) - \sqrt{[f(x)]^2 - 1}$, we get:

$f(x) = \dfrac{y}{x}$

Answer: $\dfrac{y}{x}$

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

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