UPSC NDA I 2021 Mathematics Solved Paper

Concept:The limit yields $\frac{0}{0}$ form, so L'Hôpital's rule is applied to find the value of $a$.Explanation:Substitute $x = a$ in the given limit:$$\lim_{x \to a} \frac{a^x - x^a}{x^x - a^a} = \frac{a^a - a^a}{a^a - a^a} = \frac{0}{0}$$This is an indeterminate form. Differentiate numerator and denominator separately:Derivative of numerator $a^x - x^a$ is $a^x \ln a - a x^{a-1}$.Derivative of denominator $x^x - a^a$ is $x^x (\ln x + 1)$ (since $a^a$ is constant).Now apply L'Hôpital's rule:$$\lim_{x \to a} \frac{a^x \ln a - a x^{a-1}}{x^x (\ln x + 1)}$$Plug $x = a$:$$\frac{a^a \ln a - a \cdot a^{a-1}}{a^a (\ln a + 1)} = \frac{a^a \ln a - a^a}{a^a (\ln a + 1)} = \frac{\ln a - 1}{\ln a + 1}$$Set this equal to $-1$ (given):$$\frac{\ln a - 1}{\ln a + 1} = -1$$Multiply both sides: $\ln a - 1 = -\ln a - 1$Simplify: $2\ln a = 0$, so $\ln a = 0$Thus $a = e^0 = 1$.Answer:$a = 1$