Functions Question 141
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{\log (a+x)-\log a}{x}+k\underset{x\to e}{\mathop{\lim }},\frac{\log x-1}{x-e}=1, $ then
[IIT Screening]
Options:
A) $ k=e( 1-\frac{1}{a} ) $
B) $ k=e(1+a) $
C) $ k=e(2-a) $
D) The equality is not possible
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Answer:
Correct Answer: A
Solution:
Let $ f(x)=\log x\Rightarrow {f}’,(x)=\frac{1}{x} $ Therefore, given function $ ={f}’(a)+k{f}’(e)=1 $
$ \Rightarrow \frac{1}{a}+\frac{k}{e}=1\Rightarrow k=e,( \frac{a-1}{a} ) $ Aliter : Apply L-Hospital?s rule to find both the limits.
BETA
