The domain of the function f(x) = frac{sin^{- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To define the function $f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$,we must satisfy two conditions:$1$. The argument of the square root in the

Vedclass · Accepted Answer

$[2, 3)$

Vedclass · Answer

(B) To define the function $f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$,we must satisfy two conditions:$1$. The argument of the square root in the denominator must be strictly positive:$9 - x^2 > 0$$x^2 $-3 $2$. The argument of the inverse sine function must lie in the interval $[-1, 1]$:$-1 \le x - 3 \le 1$Adding $3$ to all parts:$2 \le x \le 4$  --- $(ii)$Taking the intersection of conditions $(i)$ and $(ii)$:$x \in (-3, 3) \cap [2, 4]$$x \in [2, 3)$Thus,the domain of the function is $[2, 3)$.

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

Similar Questions

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \sin^{-1} \left( \frac{x+[x]}{3} \right)$ is $[\alpha, \beta)$, then $\alpha^2 + \beta^2$ is equal to:

The domain of the function $f(x) = \sin^{-1}\left(\log_2\left(\frac{x^2}{2}\right)\right)$ is

The domain of the function $f(x) = \sin^{-1}\left(\frac{x^{2}-3x+2}{x^{2}+2x+7}\right)$ is:

If the domain of the function $f(x) = \frac{\cos^{-1} \sqrt{x^2-x+1}}{\sqrt{\sin^{-1}(\frac{2x-1}{2})}}$ is the interval $(\alpha, \beta]$,then $\alpha + \beta$ is equal to:

The function $f(x) = \sin^{-1}(\sqrt{x})$ is defined in the interval:

Vedclass Test Series

Exam Paper Generator

Online Exam Module