The function f(x)= x 1+|x| is: - Vidyadip

MCQ

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

✓
Stritcly increasing.
B
Stritcly decreasing.
C
Neither increasing nor decreasing.
D
None of these.

✓

Answer

Correct option: A.

Stritcly increasing.

$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$
Case $I:$
When $x > 0, |x| = x$
$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$
$=\frac{\text{x}}{1+\text{x}}$
$\Rightarrow\text{f}'(\text{x})=\frac{(1+\text{x})1-\text{x}(1)}{(1+\text{x})^2}$
$=\frac{1}{(1+\text{x})^2}>0,\forall\ \text{x}\in\text{R}$
So, $f(x)$ is strictly increasing when $x > 0.$
Case $II:$
When $x < 0, |x| = -x$
$\text{f}(\text{x})=\frac{\text{x}}{1+|\text{x}|}$
$=\frac{\text{x}}{1+\text{x}}$
$\Rightarrow\text{f}'(\text{x})=\frac{(1-\text{x})1-\text{x}(-1)}{(1-\text{x})^2}$
$=\frac{1}{(1-\text{x})^2}>0,\forall\ \text{x}\in\text{R}$
So, $f(x)$ is strictly increasing when $x < 0.$
Thus, $f(x)$ is strictly increasing on $R.$

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