The range of the function f(x) = 1 - x^2 1 + | x | is - Vidyadip

MCQ

The range of the function $f(x) = \frac{{\sqrt {1 - {x^2}} }}{{1 + \left| x \right|}}$ is

✓
$\left[ {0,1} \right]$
B
$\left[ {0,\frac{1}{{\sqrt 2 }}} \right]$
C
$\left[ {0,\frac{1}{2}} \right]$
D
$\left[ {0,\frac{{\sqrt 3 }}{2}} \right]$

✓

Answer

Correct option: A.

$\left[ {0,1} \right]$

a
Put $x = sin \theta $

$y = \frac{{\left| {\sin \theta } \right|}}{{1 + \left| {\cos \theta } \right|}}$

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 "M.C.Q (1 Marks)" from 1. relation and function→1. relation and function — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

What is $ \tan ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } +\tan ^{ -1 }{ \left( \frac { 1 }{ 3 } \right) }$equal to?

$ \frac { \pi }{ 3 }$
$ \frac { \pi }{ 4 }$
$ \frac { \pi }{ 6 }$
$ \frac { \pi }{ 9 }$

If A is an invertible matrix, then which of the following is not true:

$(\text{A}^2)^\text{-1}=(\text{A}^{-1})^2$
$|\text{A}^{-1}|=|\text{A}|^{-1}$
$(\text{A}^\text{T})^\text{-1}=(\text{A}^{-1})^\text{T}$
$|\text{A}|\neq0$

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

For a real number $x$ let $[x]$ denote the largest integer less than or equal to $x$ and $\{x\}=x-[x]$. Let $n$ be a positive integer. Then, $\int \limits_0^n \cos (2 \pi[x]\{x\}) d x$ is equal to

Find adjoint of each of the matrices $\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$

R is a relation on the set Z of integers and it is given by (x, y) ∈ R ⇔ | x - y | ≤ 1. Then, R is:

Reflexive and transitive.
Reflexive and symmetric.
Symmetric and transitive.
An equivalence relation.

If the matrix is a square matrix and it contains 36 elements, then the order of the matrix is:

4 × 4
8 × 8
6 × 6
3 × 3

$\int\frac{\text{xdx}}{(\text{x-1)}(\text{x-2})}$ equals:

$\log\begin{vmatrix}\frac{(\text{x}-1)^2}{\text{x}-2}\end{vmatrix}+\text{c}$
$\log\begin{vmatrix}\frac{(\text{x}-2)^2}{\text{x}-2}\end{vmatrix}+\text{c}$
$\log\begin{vmatrix}\Big(\frac{\text{x}-1}{\text{x}-2}\Big)^2\end{vmatrix}+\text{c}$
$\log|(\text{x}-1)(\text{x}-2)+\text{c}$

The value of $ \cos^{-1}\left (\cot \left (\dfrac {\pi}{2}\right )\right ) + \cos^{-1} \left (\sin \left (\dfrac {2\pi}{3}\right )\right )$ is:

$ \dfrac {2\pi}{3}$
2
3
π

If $\int\limits_0^1 {\left( {4{x^3} - f(x)} \right)f(x)dx = \frac{4}{7}} $ ,then the area of region bounded by $y = f(x)$ , $x-$ axis and ordinate $x = 1$ and $x = 2$ , is