The domain of f(x) = frac{1}{{sqrt {{{log }_{ | vedclass

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

Question

(C) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive: $\log_{\frac{\pi}{4}}(\sin^{-1} x) - 1 > 0$

Vedclass · Accepted Answer

$\left( 0, \frac{1}{\sqrt{2}} \right)$

Vedclass · Answer

(C) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive: $\log_{\frac{\pi}{4}}(\sin^{-1} x) - 1 > 0$ $\Rightarrow \log_{\frac{\pi}{4}}(\sin^{-1} x) > 1$ Since the base $\frac{\pi}{4} $\sin^{-1} x Also,the argument of the logarithm must be positive: $\sin^{-1} x > 0 \Rightarrow x > 0$ Additionally,for $\sin^{-1} x$ to be defined,$x \in [-1, 1]$. Combining these conditions: $0 Taking the sine of all parts: $\sin(0) $0 Thus,the domain is $x \in \left( 0, \frac{1}{\sqrt{2}} \right)$.

The domain of $f(x) = \frac{1}{{\sqrt {{{\log }_{\frac{\pi }{4}}}({{\sin }^{ - 1}}x) - 1} }}$ is:

Similar Questions

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

Let $[x]$ denote the greatest integer less than or equal to $x$. Then the domain of $f(x) = \sec^{-1}(2[x] + 1)$ is:

If the domain of the function $f(x) = \sec^{-1}\left(\frac{2x}{5x+3}\right)$ is $[\alpha, \beta) \cup (\gamma, \delta]$,then $|3\alpha + 10(\beta + \gamma) + 21\delta|$ is equal to $.......$.

If $y = \tan^{-1} x$,then . . . . . . .

The range of $\sin^{-1}\left(\frac{1+x^2}{2+x^2}\right)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module