Range of f(x) = sin^{-1} (sqrt{x^2 + x + 1}) | vedclass

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

Question

(D) Let $g(x) = x^2 + x + 1$. Since $x^2 + x + 1 = (x + 1/2)^2 + 3/4$,the minimum value of $g(x)$ is $3/4$ at $x = -1/2$. As $x 	o \pm \infty$,$g(x)

Vedclass · Accepted Answer

$\left[ \frac{\pi}{3}, \frac{\pi}{2} \right]$

Vedclass · Answer

(D) Let $g(x) = x^2 + x + 1$. Since $x^2 + x + 1 = (x + 1/2)^2 + 3/4$,the minimum value of $g(x)$ is $3/4$ at $x = -1/2$. As $x 	o \pm \infty$,$g(x) 	o \infty$. However,for $\sin^{-1}(\sqrt{g(x)})$ to be defined,we must have $0 \le \sqrt{g(x)} \le 1$,which implies $0 \le g(x) \le 1$. Thus,$3/4 \le x^2 + x + 1 \le 1$. Taking the square root,we get $\sqrt{3}/2 \le \sqrt{x^2 + x + 1} \le 1$. Applying the $\sin^{-1}$ function,we get $\sin^{-1}(\sqrt{3}/2) \le \sin^{-1}(\sqrt{x^2 + x + 1}) \le \sin^{-1}(1)$. This results in $\pi/3 \le f(x) \le \pi/2$. Therefore,the range is $\left[ \frac{\pi}{3}, \frac{\pi}{2} ight]$.

Range of $f(x) = \sin^{-1} (\sqrt{x^2 + x + 1})$ is -

Similar Questions

If $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$,then the domain of the function $f(x)$ is

The domain of ${\sin ^{ - 1}}({\log _3}x)$ is

The domain of $\sin^{-1} \left[ \log_3 \left( \frac{x}{3} \right) \right]$ 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:

Considering only the principal values of the inverse trigonometric functions,the domain of the function $f(x) = \cos^{-1}\left(\frac{x^{2}-4x+2}{x^{2}+3}\right)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module