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UPSC NDA II 2025 Mathematics Paper

Question Numbers: 76-77

For the following two (02) items:

Consider the function

f(x)=\frac{x}{1-x}(x>0,x≠1)

Question : 77 of 120

Marks: +1, -0

What is

(1-x)f(\sqrt{x})+xf(\sqrt{x}+1)

equal to?

Solution:

Concept:

This problem uses substitution and algebraic simplification. Given

f(x) = \frac{x}{1-x}

, we find

f(\sqrt{x})

and

f(\sqrt{x}+1)

, then compute the given expression.

Explanation:

First, find

f(\sqrt{x})

f(\sqrt{x}) = \frac{\sqrt{x}}{1 - \sqrt{x}}

Next, find

f(\sqrt{x}+1)

f(\sqrt{x}+1) = \frac{\sqrt{x}+1}{1 - (\sqrt{x}+1)} = \frac{\sqrt{x}+1}{-\sqrt{x}} = -\frac{\sqrt{x}+1}{\sqrt{x}}

Now substitute into

(1-x)f(\sqrt{x}) + x f(\sqrt{x}+1)

(1-x) \cdot \frac{\sqrt{x}}{1-\sqrt{x}} + x \cdot \left(-\frac{\sqrt{x}+1}{\sqrt{x}}\right) = \frac{(1-x)\sqrt{x}}{1-\sqrt{x}} - x \cdot \frac{\sqrt{x}+1}{\sqrt{x}}

Simplify the second term:

x \cdot \frac{\sqrt{x}+1}{\sqrt{x}} = \sqrt{x}(\sqrt{x}+1) = x + \sqrt{x}

For the first term, note

1-x = (1-\sqrt{x})(1+\sqrt{x})

. So:

\frac{(1-\sqrt{x})(1+\sqrt{x})\sqrt{x}}{1-\sqrt{x}} = \sqrt{x}(1+\sqrt{x}) = \sqrt{x} + x

Thus the entire expression becomes:

(\sqrt{x} + x) - (x + \sqrt{x}) = 0

Answer:

The value is

0

, which corresponds to option D.

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