Download App

Increasing and Decreasing Functions — Maths STD 12 Science — Question

Maharashtra 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 $\text{x}\in\text{R}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Increasing and Decreasing Functions→Increasing and Decreasing Functions — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

The value of $\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}+\cos\text{x}}\text{ dx}$ is:
→
Function $f(x)=x^2-3 x+4$ has minimum value at $x=\ldots \ldots$
→
The function $f : R \rightarrow R, f(x) = x^2$ is:
→
If a plane cuts intercepts $-6,3,4$ on the co-ordinate axes, then the length of the perpendicular from the origin to the plane is
→
$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot^3\text{x}}\text{ dx}$ is equal to:
→
The minimum value of $\text{x}\log_{\text{e}}\text{x}$ is equal to :
→
If $\text{y}=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\tan\frac{\text{x}}{2}\Big),\text{a}>\text{b}>0,$then:
→
$\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=$
→
If the angles $A , B , C$ of a triangle are in A.P. and the sides $a, b, c$ opposite to these angles are in G. P., then $a^2, b^2, c^2$ are in
→
$\int_0^\pi x \sin x d x=$
→