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\\Let \: f:\left ( -1,\infty \right )\rightarrow R\; \:\: be\; \: \:de\! f ined\:\; by \: \; f(0)=1\:and\; f(x)=\frac{1}{x}\log_{e}\left ( 1+x \right ),x\neq 0
Then the function f:
Option: 1 Decreases in (-1,0) and increases in (0,\infty)
Option: 2 increases in (-1,\infty).
Option: 3 increases in (-1,0) and decreases in (0,\infty)
Option: 4 decreases in (-1,\infty).

Answers (1)

best_answer

f^{\prime}(x)=\frac{\frac{x}{1+x}-\ln (1+x)}{x^{2}}=\frac{x-(1+x) \ln (1+x)}{x^{2}(1+x)}

Here x^2 is always positive

Hence 

f'(x) depends on 

\frac{x-(1+x) \ln (1+x)}{(1+x)}

\frac{1+x-(1+x) \ln (1+x)-1}{(1+x)}

-\left (\frac{1}{1+x}+\ln(1+x)-1 \right )

-\left (\frac{1}{1+x}+\ln(1+x)-\ln e \right )

-\left [\frac{1}{1+x}+\ln\left (\frac{1+x}e \right ) \right ]

Which is negative foe all values of x\in(-1,\infty)

Posted by

Suraj Bhandari

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