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UPSC CDS 1 2022 Math Paper

Question : 17 of 100

Marks: +1, -0

\frac{x-y}{x\sqrt{y} + y\sqrt{x}} = \frac{1}{\sqrt{x}};(x > 0,y > 0)

\frac{x}{y}

Solution:

Concept:

Factorize and simplify the given rational equation using algebraic identities.

Explanation:

Start from

\frac{x-y}{x\sqrt{y} + y\sqrt{x}} = \frac{1}{\sqrt{x}}

Rewrite

x = (\sqrt{x})^2

and

y = (\sqrt{y})^2

in the numerator.

Also,

x\sqrt{y} = \sqrt{x} \cdot \sqrt{x} \cdot \sqrt{y}

and

y\sqrt{x} = \sqrt{y} \cdot \sqrt{y} \cdot \sqrt{x}

So the denominator becomes

\sqrt{x}\sqrt{y}(\sqrt{x} + \sqrt{y})

The numerator becomes

(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})

Cancel the common factor

(\sqrt{x} + \sqrt{y})

from numerator and denominator.

We get

\frac{\sqrt{x} - \sqrt{y}}{\sqrt{x}\sqrt{y}} = \frac{1}{\sqrt{x}}

Multiply both sides by

\sqrt{x}\sqrt{y}

\sqrt{x} - \sqrt{y} = \sqrt{y}

Thus

\sqrt{x} = 2\sqrt{y}

Square both sides:

x = 4y

Therefore

\frac{x}{y} = 4

Answer:

4

(Option C).

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