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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
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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