QJEEM22Ma09f7: JEE Main 2022 Mathematics - Class 12 JEEM22Ma09f7 Solution: Then is equal to

$f(x)$ is continuous at $x = 0$

$\therefore$ $f(0) = \mathop {\lim }\limits_{x \to 0} f(x)$

$\Rightarrow 10 = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}(1 + 5x) - {{\log }_e}(1 + \alpha x)} \over x}$

$= \mathop {\lim }\limits_{x \to 0} {{\log (1 + 5x)} \over {5x}} \times 5 - {{{{\log }_e}(1 + \alpha x)} \over {\alpha x}} \times \alpha$

$= 1 \times 5 - \alpha$

$\Rightarrow \alpha = 5 - 10 = - 5$