UPSC NDA I 2019 Mathematics Solved Paper

Concept:L'Hospital's rule applies when a limit is in $\frac{0}{0}$ form: differentiate numerator and denominator separately until the limit is a finite non-zero value.Explanation:Given: $\mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}}$.Substituting $x=0$ gives $\frac{0}{0}$, so L'Hospital's rule is applicable.Differentiate numerator: derivative of $1 - \cos^3(4x)$ is $12 \cos^2(4x) \sin(4x)$.Differentiate denominator: derivative of $x^2$ is $2x$.The limit becomes $\mathop {\lim }\limits_{x \to 0} \frac{12 \cos^2(4x) \sin(4x)}{2x}$.Simplify: $\mathop {\lim }\limits_{x \to 0} \frac{24 \cos^2(4x) \sin(4x)}{4x}$.Rewrite as $24 \cdot \mathop {\lim }\limits_{x \to 0} \cos^2(4x) \cdot \mathop {\lim }\limits_{x \to 0} \frac{\sin(4x)}{4x}$.As $x\to 0$, $\cos(4x) \to 1$ so $\cos^2(4x) \to 1$, and $\frac{\sin(4x)}{4x} \to 1$.Thus the limit equals $24 \times 1 \times 1 = 24$.Answer:C. 24