$$ \text { Let the function } f(x)=\left\{\begin{array}{cl} \frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & ;\text { if } x \neq 0 \\ 10 & ; \text { if } x=0 \end{array} \text { be continuous at } x=0 .\right. $$ Then $$\alpha$$ is equal to
JEE · Math · previous-year question
- A.10
- B.$$-$$10
- C.5
- D.$$-$$5correct
Answer
D. $$-$$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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