Domain of function f(x)= ^ -1 5 x is - Vidyadip

MCQ

Domain of function $f(x)=\sin ^{-1} 5 x$ is

A
$\left(-\frac{1}{5}, \frac{1}{5}\right)$
✓
$\left[-\frac{1}{5}, \frac{1}{5}\right]$
C
R
D
$\left(0, \frac{1}{5}\right)$

✓

Answer

Correct option: B.

$\left[-\frac{1}{5}, \frac{1}{5}\right]$

(B)
$-1 \leq 5 x \leq 1 \Rightarrow \frac{-1}{5} \leq x \leq \frac{1}{5}$
Hence, domain is $\left[\frac{-1}{5}, \frac{1}{5}\right]$.

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 "MCQ" from Functions→Functions — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+1}-x\right)$ is equal to

The first three of four given numbers are in $G.P.$ and their last three are $A.P.$ with common difference $6.$ If first and fourth numbers are equal, then the first number is:

If $\frac{|\text{x}-2|}{\text{x}-2}\geq0,$ then :

if x, y, z are any three real numbers, then tan (x - y) + tan (y - z) + tan (z - x) is equal to

The equation of the circle drawn with the two foci of $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ end $-$ point of a diameter is

The equation of the tangent to the ellipse $9 x^2+16 y^2=144$ at the point $(4 \cos \theta, 3 \sin \theta)$ is

If $x+\frac{1}{x}=2 \cos \theta$, then $x^3+\frac{1}{x^3}=$

If $\sqrt{ a + ib }=x+ i y$, then possible value of $\sqrt{ a - ib }$ is $( a , b , x, y \in R )$

Axis of the parabola $x^2-4 x-3 y+10=0$ is

If ${^\text{m}}\text{C}_{\text{1}}={^\text{n}}\text{C}_{\text{2}},$ is then :