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Increasing and Decreasing Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIncreasing and Decreasing Functions1 Mark
MCQ
$\text{f}(\text{x})=2\text{x}-\tan^{-1}-\log\Big\{\text{x}+\sqrt{\text{x}^2+1}\Big\}$ is monotonically increasing when:
  • A
    $\text{x}>0$
  • B
    $\text{x}<0$
  • ✓
    $\text{x}\in\text{R}$
  • D
    $\text{x}\in\text{R}-\{0\}$
✓

Answer

Correct option: C.
$\text{x}\in\text{R}$
$\text{f}(\text{x})=2\text{x}-\tan^{-1}-\log\Big\{\text{x}+\sqrt{\text{x}^2+1}\Big\}$
$\Rightarrow\text{f}'(\text{x})=2-\frac{1}{1+\text{x}^2}-\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big(1+\frac{2\text{x}}{2\sqrt{\text{x}^2+1}}\Big)$
$\Rightarrow\text{f}'(\text{x})=2-\frac{1}{1+\text{x}^2}-\frac{1}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\text{f}'(\text{x})=\frac{1+2\text{x}^2}{1+\text{x}^2}-\frac{1}{\sqrt{\text{x}^2+1}}$
$\Rightarrow\text{f}'(\text{x})=\frac{1+2\text{x}^2-\sqrt{\text{x}^2+1}}{1+\text{x}^2}$
Function is increasing monotonically.
$\Rightarrow\frac{1+2\text{x}^2-\sqrt{\text{x}^2+1}}{1+\text{x}^2}>0$
$\Rightarrow1+2\text{x}^2-\sqrt{\text{x}^2+1}>0$
$\Rightarrow1+2\text{x}^2>\sqrt{\text{x}^2+1}$
Squaring on both sides,
$\Rightarrow\big(1+2\text{x}^2\big)>\text{x}^2+1$
$\Rightarrow4\text{x}^4+3\text{x}^2>0$
For all $\text{x}\in\text{R}$

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