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# Solution for Find the Intervals in Which F(X) = Log (1 + X) − X 1 + X is Increasing Or Decreasing ? - CBSE (Science) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

Find the intervals in which f(x) = log (1 + x) −$\frac{x}{1 + x}$ is increasing or decreasing ?

#### Solution

$f\left( x \right) = \log \left( 1 + x \right) - \frac{x}{1 + x}$

$\text { Domain of f }\left( x \right) \text { is }\left( - 1, \infty \right).$

$f'\left( x \right) = \frac{1}{1 + x} - \left\{ \frac{1 + x - x}{\left( 1 + x \right)^2} \right\}$

$= \frac{1}{1 + x} - \frac{1}{\left( 1 + x \right)^2}$

$= \frac{x}{\left( 1 + x \right)^2}$

$\text { For }f(x) \text { to be increasing, we must have }$

$f'\left( x \right) > 0$

$\Rightarrow \frac{x}{\left( 1 + x \right)^2} > 0$

$\Rightarrow x > 0 \left[ \because \left( 1 + x \right)^2 >0, \text { Domain }:\left( - 1, \infty \right) \right]$

$\Rightarrow x \in \left( 0, \infty \right)$

$\text { So,f(x)is increasing on } \left( 0, \infty \right) .$

$\text { Forf(x) to be decreasing, we must have }$

$f'\left( x \right) < 0$

$\Rightarrow \frac{x}{\left( 1 + x \right)^2} < 0$

$\Rightarrow x < 0 \left[ \because \left( 1 + x \right)^2 >0, \text{Domain }:\left( - 1, \infty \right) \right]$

$\Rightarrow x \in \left( - 1, 0 \right)$

$\text { So,f(x)is decreasing on }\left( - 1, 0 \right).$

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Solution Find the Intervals in Which F(X) = Log (1 + X) − X 1 + X is Increasing Or Decreasing ? Concept: Increasing and Decreasing Functions.
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