Domain of definition of the real valued funct | vedclass

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(A) For the function $f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{6}}$ to be defined,the expression inside the square root must be non-negative: $\sin^{-1

Vedclass · Accepted Answer

$\left[-\frac{1}{4}, \frac{1}{2}\right]$

Vedclass · Answer

(A) For the function $f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{6}}$ to be defined,the expression inside the square root must be non-negative: $\sin^{-1}(2x) + \frac{\pi}{6} \geq 0$ $\sin^{-1}(2x) \geq -\frac{\pi}{6}$ Also,the domain of $\sin^{-1}(	heta)$ is $[-1, 1]$,so $-1 \leq 2x \leq 1$,which implies $-0.5 \leq x \leq 0.5$. The range of $\sin^{-1}(2x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Combining these,we have $-\frac{\pi}{6} \leq \sin^{-1}(2x) \leq \frac{\pi}{2}$. Taking the sine of all parts: $\sin(-\frac{\pi}{6}) \leq 2x \leq \sin(\frac{\pi}{2})$ $-\frac{1}{2} \leq 2x \leq 1$ Dividing by $2$: $-\frac{1}{4} \leq x \leq \frac{1}{2}$ Thus,the domain is $\left[-\frac{1}{4}, \frac{1}{2}ight]$.

Domain of definition of the real valued function $f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{6}}$ is

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