_ x 0 (a + x) - a x + k _ x e x - 1 x - e = 1, then - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 0} \frac{{\log (a + x) - \log a}}{x} + k\mathop {\lim }\limits_{x \to e} \frac{{\log x - 1}}{{x - e}} = 1,$ then

✓
$k = e\left( {1 - \frac{1}{a}} \right)$
B
$k = e(1 + a)$
C
$k = e(2 - a)$
D
The equality is not possible

✓

Answer

Correct option: A.

$k = e\left( {1 - \frac{1}{a}} \right)$

a
(a) Let $f(x) = \log x\,\, \Rightarrow \,\,f'\,(x) = \frac{1}{x}$

Therefore, given function $ = f'(a) + kf'(e) = 1$

$ \Rightarrow \,\,\frac{1}{a} + \frac{k}{e} = 1\,\, $

$\Rightarrow \,\,k = e\,\left( {\frac{{a - 1}}{a}} \right)$

Aliter : Apply $ L-$ Hospital’s rule to find both the limits.

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