MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\log (a + x) - \log a}}{x} + k\mathop {\lim }\limits_{x \to e} \frac{{\log x - 1}}{{x - e}} = 1,$ તો
- ✓$k = e\left( {1 - \frac{1}{a}} \right)$
- B$k = e(1 + a)$
- C$k = e(2 - a)$
- Dસમાન શક્ય નથી.
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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