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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

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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

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

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

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

\[ = \frac{x}{1 + x}\]

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

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

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

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

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

\[ = \frac{x}{1 - x}\]

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

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

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

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

  Is there an error in this question or solution?
Solution 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.
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