The domain of the function operatorname{cosec | vedclass

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

Question

(D) The domain of $\operatorname{cosec}^{-1}(y)$ is $y \in (-\infty, -1] \cup [1, \infty)$.For the function $\operatorname{cosec}^{-1}\left(\frac{1+x}

Vedclass · Accepted Answer

$\left[-\frac{1}{2}, \infty\right) - \{0\}$

Vedclass · Answer

(D) The domain of $\operatorname{cosec}^{-1}(y)$ is $y \in (-\infty, -1] \cup [1, \infty)$.For the function $\operatorname{cosec}^{-1}\left(\frac{1+x}{x}\right)$,we must have $\frac{1+x}{x} \in (-\infty, -1] \cup [1, \infty)$.Case $1$: $\frac{1+x}{x} \geq 1$$\frac{1}{x} + 1 \geq 1 \implies \frac{1}{x} \geq 0 \implies x > 0$.Case $2$: $\frac{1+x}{x} \leq -1$$\frac{1}{x} + 1 \leq -1 \implies \frac{1}{x} \leq -2$.Since $\frac{1}{x} \leq -2$,$x$ must be negative. Multiplying by $x$ (which is negative) reverses the inequality:$1 \geq -2x \implies x \geq -\frac{1}{2}$.Thus,$x \in [-\frac{1}{2}, 0)$.Combining both cases,the domain is $x \in [-\frac{1}{2}, 0) \cup (0, \infty)$,which can be written as $[-\frac{1}{2}, \infty) - \{0\}$.

The domain of the function $\operatorname{cosec}^{-1}\left(\frac{1+x}{x}\right)$ is :

Similar Questions

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

The domain of $f(x) = \frac{1}{{\sqrt {{{\log }_{\frac{\pi }{4}}}({{\sin }^{ - 1}}x) - 1} }}$ 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 $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$,then the domain of the function $f(x)$ is

Domain of the function $f(x) = \frac{1}{\sqrt{\ln(\cot^{-1}x)}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module