UPSC NDA I 2021 Mathematics Solved Paper

Concept: Indeterminate Forms: Any expression whose value cannot be defined, like $\frac{0}{0}, \pm \frac{\infty}{\infty}, 0^{0}, \infty^{0}$ etc. - L'Hospital's Rule: For the differentiable functions $f(x)$ and $g(x)$, the $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$, if $f(x)$ and $g(x)$are both 0 or $\pm \infty$ (i.e. an Indeterminate Form) is equal to the $\lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$ if it exists. - For the indeterminate form $\infty - \infty$, first rationalize by multiplying with the conjugate and then divide the terms by the highest powerof the variable to get terms so that $\frac{1}{x} \to 0$ as $x \to \infty$ - For the indeterminate form $\frac{0}{0}$, first try to rationalize by multiplying with the conjugate, or simplify by cancelling some terms in the numerator and denominator. Else, usethe L'Hospital's rule. Calculation: $\lim\limits_{x \to a} \frac{a^{x}-x^{a}}{x^{x}-a^{a}}$ $= \frac{a^{a}-a^{2}}{a^{a}-a^{a}}$ $= \frac{0}{0}$, an indeterminate form. Applying L'Hospital's rule, we get: $\lim\limits_{x \to a} \frac{a^{x}-x^{a}}{x^{x}-a^{a}}$ $= \lim\limits_{x \to a} \frac{a^{x} \log a - a x^{2-1}}{x^{x}(\log x+1)}$ $= \frac{a^{a} \log a - a \cdot a^{a-1}}{a^{n}(\log a+1)}$ $= \frac{\log a - 1}{\log a + 1}$ According to the question, $\frac{\log a - 1}{\log a + 1} = -1$. $\Rightarrow \log a - 1 = -\log a - 1$ $\Rightarrow 2 \log a = 0$ $\Rightarrow a = 1$