JEE MAIN - Mathematics (2022 - 29th July Evening Shift - No. 4)
$$ \text { Let the function } f(x)=\left\{\begin{array}{cl} \frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & ;\text { if } x \neq 0 \\ 10 & ; \text { if } x=0 \end{array} \text { be continuous at } x=0 .\right. $$
Then $$\alpha$$ is equal to
10
$$-$$10
5
$$-$$5
Explanation
$$f(x)$$ is continuous at $$x = 0$$
$$\therefore$$ $$f(0) = \mathop {\lim }\limits_{x \to 0} f(x)$$
$$ \Rightarrow 10 = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}(1 + 5x) - {{\log }_e}(1 + \alpha x)} \over x}$$
$$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + 5x)} \over {5x}} \times 5 - {{{{\log }_e}(1 + \alpha x)} \over {\alpha x}} \times \alpha $$
$$ = 1 \times 5 - \alpha $$
$$ \Rightarrow \alpha = 5 - 10 = - 5$$
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