The function f ( x ) = lo g ( 1 + x ) − 2 + x 2 x is increasing on

Given f ( x ) = lo g ( 1 + x ) − 2 + x 2 x f ′ ( x ) = 1 + x 1 − ( 2 + x ) 2 ( 2 + x ) ( 2 ) − 2 x = 1 + x 1 − ( 2 + x ) 2 4 = ( 1 + x ) ( 2 + x ) 2 ( 2 + x ) 2 − 4 − 4 x = ( 1 + x ) ( 2 + x ) 2 x 2 > 0 for all x ∈ ( 0 , ∞ ) Thus, given function f ( x ) is increasing on ( 0 , ∞ ) .

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMonotonicity-Increasing-DecreasingEasy2 minPYQ_2023

MathematicsEasysingle choice

The function $f (x) = lo g (1 + x) - \frac{2 x}{2 + x}$ is increasing on

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Solution:

Given $f (x) = lo g (1 + x) - \frac{2 x}{2 + x}$
$f^{'} (x) = \frac{1}{1 + x} - \frac{( 2 + x ) ( 2 ) - 2 x}{( 2 + x ) ^{2}} = \frac{1}{1 + x} - \frac{4}{( 2 + x ) ^{2}} = \frac{( 2 + x ) ^{2} - 4 - 4 x}{( 1 + x ) ( 2 + x ) ^{2}} = \frac{x ^{2}}{( 1 + x ) ( 2 + x ) ^{2}} > 0 for all x \in (0, \infty)$
Thus, given function $f (x)$ is increasing on $(0$ , $\infty)$ .

Stream:BITSATSubject:MathematicsTopic:Application of DerivativeSubtopic:Monotonicity-Increasing-Decreasing

⏱ 2mℹ️ Source: PYQ_2023

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