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MHTCET 10 May 2024 Shift 2 Paper

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Question : 116 of 150

Marks: +1, -0

If

x^{2} y^{2}=\sin ^{-1} x+\cos ^{-1} x

\frac{d y}{d x}

x=1

y=2

is

$\frac{1}{2}$
2
$-\frac{1}{2}$
-2

Solution:

Concept:

Use the standard identity

\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}

, then differentiate implicitly.

Explanation:

Given:

x^{2} y^{2} = \sin^{-1} x + \cos^{-1} x

.

Since

\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}

, the equation becomes

x^{2} y^{2} = \frac{\pi}{2}

.

Differentiate both sides with respect to

x

:

\frac{d}{dx}\left(x^{2} y^{2}\right) = 0

Using the product rule:

2x y^{2} + x^{2} \cdot 2y \frac{dy}{dx} = 0

Simplify:

2x y^{2} + 2x^{2} y \frac{dy}{dx} = 0

Isolate

\frac{dy}{dx}

:

2x^{2} y \frac{dy}{dx} = -2x y^{2}

\frac{dy}{dx} = -\frac{2x y^{2}}{2x^{2} y} = -\frac{y}{x}

At

x=1

and

y=2

:

\left.\frac{dy}{dx}\right|_{x=1, y=2} = -\frac{2}{1} = -2

Answer:

\frac{dy}{dx} = -2

, which matches option D.

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