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Show by LMVT and taking f(x) = log x that ( question attached in image)

Drake , 10 Years ago
Grade 12
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
f(x) = log(x)
Interval = [a, b]
Let ‘c’ be the point in the interval.
Then, by LMVT
f'(c) = \frac{f(b)-f(a)}{b-a}
f'(c) = \frac{log(b)-log(a)}{b-a}
f'(x) = \frac{1}{x}
f''(x) = \frac{-1}{x^{2}}<0
-> f’(x) is always decreasing. So, we have
f'(b) < f'(c) < f'(a)
\frac{1}{b} < \frac{log(\frac{b}{a})}{b-a} < \frac{1}{a}
\frac{b-a}{b} < log(\frac{b}{a}) < \frac{b-a}{a}
1-\frac{a}{b} < log(\frac{b}{a}) < \frac{b}{a}-1
Thanks & Regards
Jitender Singh
IIT Delhi
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