Function {sin ^{ - 1}}√ x is defined in interval

English

Gujarati

Hindi

1.Relation and Function

normal

Function ${\sin ^{ - 1}}\sqrt x $ is defined in the interval

A

$(-1, 1)$

B

$[0, 1]$

C

$[-1, 0]$

D

$(-1, 2)$

Solution

(b) Let $y = {\sin ^{ – 1}}\sqrt x \,\, \Rightarrow \,\,\sqrt x = \sin y$

$ \Rightarrow \,\,x = {\sin ^2}y,\,\,\,\,\therefore \,\,0 \le x \le 1$

Standard 12

Mathematics

Share

Similar Questions

Set $A$ has $3$ elements and set $B$ has $4$ elements. The number of injection that can be defined from $A$ to $B$ is

easy

The total number of functions,$f:\{1,2,3,4\} \cdot\{1,2,3,4,5,6\}$ such that $f (1)+ f (2)= f (3)$, is equal to .

hard

(JEE MAIN-2022)

If $f(x) = \frac{x}{{x – 1}}$, then $\frac{{f(a)}}{{f(a + 1)}} = $

easy

If the domain and range of $f(x){ = ^{9 – x}}{C_{x – 1}}$ contains $m$ and $n$ elements respectively, then

normal

If domain of the function $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right)$ is $(\alpha, \beta) \cup(\gamma, \delta]$, then $18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)$ is equal to $….$.

hard

(JEE MAIN-2023)