UPSC NDA 2 2022 Mathematics Solved Paper

Concept:The given limit represents the derivative of $\sin^2 x$ with respect to $x$, evaluated using L'Hospital's rule because the expression takes the $\frac{0}{0}$ form as $h \to 0$.Explanation:We need $\displaystyle\lim_{h \to 0} \frac{\sin^2(x+h)-\sin^2 x}{h}$.As $h \to 0$, both numerator and denominator approach $0$, so the limit is of type $\frac{0}{0}$.Apply L'Hospital's rule: differentiate numerator and denominator separately with respect to $h$.The derivative of the numerator $\sin^2(x+h)$ is $2\sin(x+h)\cos(x+h)$.The derivative of the denominator $h$ is $1$.Thus the limit becomes $\displaystyle\lim_{h \to 0} \frac{2\sin(x+h)\cos(x+h)}{1}$.Now evaluate as $h \to 0$: $\sin(x+h) \to \sin x$ and $\cos(x+h) \to \cos x$.So the limit equals $2\sin x \cos x = \sin 2x$.Answer:Option C: $\sin 2x$.