UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Use the trigonometric identity $1 - \cos \theta = 2\sin^2(\theta/2)$.A critical step is evaluating $\sqrt{\sin^2 \theta} = |\sin \theta|$, not $\sin \theta$.Since the sign of $\sin 2x$ depends on whether $x$ approaches $0$ from the left or right, we must check both one-sided limits.The standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ is also used.Explanation:Start with $\lim\limits_{x \to 0} \frac{x}{\sqrt{1 - \cos 4x}}$.Apply the identity: $1 - \cos 4x = 2\sin^2 2x$.Thus, $\sqrt{1 - \cos 4x} = \sqrt{2\sin^2 2x} = \sqrt{2} \cdot |\sin 2x|$.The expression becomes $\lim\limits_{x \to 0} \frac{x}{\sqrt{2} |\sin 2x|}$.For the right-hand limit ($x \to 0^+$), $\sin 2x > 0$, so $|\sin 2x| = \sin 2x$.Limit $= \lim\limits_{x \to 0^+} \frac{x}{\sqrt{2} \sin 2x} = \frac{1}{\sqrt{2}} \lim\limits_{x \to 0^+} \frac{x}{\sin 2x}$.$= \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{1}{2\sqrt{2}}$.For the left-hand limit ($x \to 0^-$), $\sin 2x < 0$, so $|\sin 2x| = -\sin 2x$.Limit $= \lim\limits_{x \to 0^-} \frac{x}{\sqrt{2} (-\sin 2x)} = -\frac{1}{\sqrt{2}} \lim\limits_{x \to 0^-} \frac{x}{\sin 2x}$.$= -\frac{1}{\sqrt{2}} \cdot \frac{1}{2} = -\frac{1}{2\sqrt{2}}$.Since the left-hand limit ($-\frac{1}{2\sqrt{2}}$) and the right-hand limit ($\frac{1}{2\sqrt{2}}$) are different, the overall limit does not exist.Answer:Option D. Limit does not exist.