Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 367 Question 367 of 401 Hard MCQ
\[
\text { Let the function } f(x)=\left\{\[ log e ( 1 + 5 x ) − log e ( 1 + α x ) x ; if x ≠ 0 10 ; if x = 0 \begin{array}{|ccc|}\frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & ;\text { if } x \neq 0 \\ 10 & ; \text { if } x=0\end{array} x l o g e ( 1 + 5 x ) − l o g e ( 1 + α x ) 10 ; if x = 0 ; if x = 0 \] \text { be continuous at } x=0 .\right.
\]
Then \[\alpha\] is equal to
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f ( x ) f(x) f ( x ) is continuous at x = 0 x = 0 x = 0
\[\therefore\] \[f(0) = \mathop {\lim }\limits_{x \to 0} f(x)\]
\[ \Rightarrow 10 = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}(1 + 5x) - {{\log }_e}(1 + \alpha x)} \over x}\]
\[ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + 5x)} \over {5x}} \times 5 - {{{{\log }_e}(1 + \alpha x)} \over {\alpha x}} \times \alpha \]
\[ = 1 \times 5 - \alpha \]
\[ \Rightarrow \alpha = 5 - 10 = - 5\] Answer the question to unlock