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UPSC NDA II 2025 Mathematics Paper

Question Numbers: 71-72

For the following two (02) items:

Let

f(x)=\begin{cases} \frac{1-cos2x}{x^{2}} & ,x<0 \\ 9 & ,x=0 \\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4} & ,x>0 \end{cases}

Question : 72 of 120

Marks: +1, -0

What is

\underset{x→0^{-}}{lim}f(x)

equal to?

Solution:

Concept:

Use algebraic simplification by multiplying with the conjugate to evaluate the limit.

Explanation:

We evaluate the left-hand limit of

f(x)

x \to 0^{-}

\displaystyle \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}} - 4}

Multiply numerator and denominator by the conjugate

\sqrt{16+\sqrt{x}} + 4

= \lim_{x \to 0^{-}} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}} + 4)}{(16+\sqrt{x}) - 16} = \lim_{x \to 0^{-}} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}} + 4)}{\sqrt{x}}

Cancel

\sqrt{x}

from numerator and denominator:

= \lim_{x \to 0^{-}} \sqrt{16+\sqrt{x}} + 4

x \to 0^{-}

\sqrt{x} \to 0

, so the limit becomes

\sqrt{16} + 4 = 4 + 4 = 8

Answer:

8 (Option C)

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