Domain of function f(x) = frac{{{{sin }^{ - 1}}(x - 3)}}{{√ {9 - {x^2}} }} is

English

Gujarati

Hindi

1.Relation and Function

easy

The domain of the function $f(x) = \frac{{{{\sin }^{ - 1}}(x - 3)}}{{\sqrt {9 - {x^2}} }}$ is

A

$[1, 2)$

B

$[2, 3)$

C

$[1, 2]$

D

$[2, 3]$

(AIEEE-2004)

Solution

(b) To define $f(x)$,

$9 – {x^2} > 3 \Rightarrow – 3 < x < 3…..(i)$

$ – 1 \le (x – 3) \le 1 \Rightarrow 2 \le x \le 4…..(ii)$

From $(i)$ and $(ii)$, $2 \le x < 3$ $i.e.,\,\, [2, 3).$

Standard 12

Mathematics

Share

Similar Questions

Let, $f(x)=\left\{\begin{array}{l} x \sin \left(\frac{1}{x}\right) \text { when } x \neq 0 \\ 1 \text { when } x=0 \end{array}\right\}$ and $A=\{x \in R: f(x)=1\} .$ Then, $A$ has

hard

(KVPY-2019)

The function $f\left( x \right) = \left| {\sin \,4x} \right| + \left| {\cos \,2x} \right|$, is a periodic function with period

hard

(JEE MAIN-2014)

If domain of function $f(x) = \sqrt {\ln \left( {m\sin x + 4} \right)} $ is $R$ , then number of possible integral values of $m$ is

normal

Let function $f(x) = {x^2} + x + \sin x – \cos x + \log (1 + |x|)$ be defined over the interval $[0, 1]$. The odd extensions of $f(x)$ to interval $[-1, 1]$ is

hard

Let $f ( x )$ be a quadratic polynomial with leading coefficient $1$ such that $f(0)=p, p \neq 0$ and $f(1)=\frac{1}{3}$. If the equation $f(x)=0$ and $fofofof (x)=0$ have a common real root, then $f(-3)$ is equal to $……..$

normal

(JEE MAIN-2022)