5. Continuity and Differentiation
easy

फलन $f(x) = {e^x},a = 0,b = 1$ के लिए मध्यमान प्रमेय में  $c$ का मान होगा

A

$log\, x$

B

$\log (e - 1)$

C

$0$

D

$1$

Solution

(b) यहाँ $\frac{{f(b) – f(a)}}{{b – a}} = f'(c)$

==> $\frac{{{e^b} – {e^a}}}{{b – a}} = f'(c)$

==>$\frac{{e – 1}}{{1 – 0}} = {e^c} \Rightarrow c = \log (e – 1)$.

Standard 12
Mathematics

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