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 $f(x) = \sin^{-1}\left[\log_{2}\left(\frac{x}{2}\right)\right]$ is

Let $f$ be a real valued function defined by $f(x) = \sin^{-1} \left( \frac{1 - |x|}{3} \right) + \cos^{-1} \left( \frac{|x| - 3}{5} \right)$. Then the domain of $f(x)$ is given by:

The range of $\sec ^{-1} x$ is

The domain of $f(x) = \cos^{-1}[x]$ is,where $[x]$ denotes the greatest integer function.

The domain of the derivative of the function $f(x) = \operatorname{Cos}^{-1}(2x - 5) - \operatorname{Sin}^{-1}(x - 2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module