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Solution for The Function F ( X ) = X 1 + | X | is (A) Strictly Increasing (B) Strictly Decreasing (C) Neither Increasing Nor Decreasing (D) None of Thes - CBSE (Science) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

Question

The function $f\left( x \right) = \frac{x}{1 + \left| x \right|}$ is

(a) strictly increasing
(b) strictly decreasing
(c) neither increasing nor decreasing
(d) none of these

Solution

(a)  strictly increasing

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

$\text { Case 1: When }x > 0, \left| x \right| = x$

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

$= \frac{x}{1 + x}$

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

$= \frac{1}{\left( 1 + x \right)^2} > 0, \forall x \in R$

$\text { So,f }\left( x \right) \text { is strictly increasing when }x> 0.$

$\text { Case 2: When }x < 0, \left| x \right| = - x$

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

$= \frac{x}{1 - x}$

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

$= \frac{1}{\left( 1 - x \right)^2} > 0, \forall x \in R$

$\text { So,f }\left( x \right) \text { is strictly increasing when }x <0.$

$\text { Thus,f }\left( x \right) \text { is strictly increasing on R } .$

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Solution for question: The Function F ( X ) = X 1 + | X | is (A) Strictly Increasing (B) Strictly Decreasing (C) Neither Increasing Nor Decreasing (D) None of Thes concept: Increasing and Decreasing Functions. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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