UPSC NDA I and II 2020 Math Solved Paper

Concept:Use direct factorization and L'Hôpital's Rule to evaluate the two limits and equate them.Explanation:First, evaluate the left-hand limit (LHS):$\underset{x→1}{lim}\frac{x^{4}-1}{x-1}$ Factor $x^{4}-1$ as $(x^{2}-1)(x^{2}+1)$, then factor further $x^{2}-1 = (x-1)(x+1)$.Cancel the $(x-1)$ factor with the denominator:$\underset{x→1}{lim}\frac{(x-1)(x+1)(x^{2}+1)}{x-1} = \underset{x→1}{lim}(x+1)(x^{2}+1)$.Substitute $x=1$: $(1+1)(1^{2}+1) = (2)(2) = 4$. Thus, LHS = $4$. Now evaluate the right-hand limit (RHS):$\underset{x→k}{lim}\frac{x^{3}-k^{3}}{x^{2}-k^{2}}$. This is of the form $\frac{0}{0}$, so apply L'Hôpital's Rule: differentiate numerator and denominator separately with respect to $x$.Derivative of numerator: $3x^{2}$. Derivative of denominator: $2x$.The limit becomes $\underset{x→k}{lim}\frac{3x^{2}}{2x} = \underset{x→k}{lim}\frac{3x}{2} = \frac{3k}{2}$.Set LHS equal to RHS: $4 = \frac{3k}{2}$. Multiply both sides by $2$: $8 = 3k$. Divide by $3$: $k = \frac{8}{3}$.Answer:$k = \frac{8}{3}$ (Option C).