Question

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

• A. \(\left[-\frac{1}{2}, \infty\right) - \{0\}\)
• B. \(\left(-\frac{1}{2}, \infty\right) - \{0\}\)
• C. \(\left[-\frac{1}{2}, 0\right) \cup [1, \infty)\)
• D. \(\left(-1, -\frac{1}{2}\right] \cup (0, \infty)\)

Hint:

Remember the domains of inverse trigonometric functions. For \(\sin^{-1}(x)\) and \(\cos^{-1}(x)\), the domain is \(|x| \le 1\). For \(\sec^{-1}(x)\) and \(\text{cosec}^{-1}(x)\), the domain is \(|x| \ge 1\). For \(\tan^{-1}(x)\) and \(\cot^{-1}(x)\), the domain is all real numbers.

Answer 1

The Correct Option is A

Step 1: Understanding the Question
We need to find the domain of the function \(f(x) = \text{cosec}^{-1}\left(\frac{1+x}{x}\right)\). The domain of the inverse cosecant function, \(\text{cosec}^{-1}(t)\), is defined for all values of \(t\) such that \(|t| \geq 1\).
Step 2: Key Formula or Approach
For the given function, the argument is \(t = \frac{1+x}{x}\). Therefore, we need to solve the inequality:
\[ \left|\frac{1+x}{x}\right| \geq 1 \] Also, the denominator cannot be zero, so \(x \neq 0\).
Step 3: Detailed Explanation
The inequality \(|t| \geq 1\) is equivalent to \(t \geq 1\) or \(t \leq -1\). We solve these two cases for \(t = \frac{1+x}{x}\).
Case 1: \(\frac{1+x}{x} \geq 1\)
\[ \frac{1+x}{x} - 1 \geq 0 \] \[ \frac{1+x-x}{x} \geq 0 \] \[ \frac{1}{x} \geq 0 \] This is true when \(x>0\). So, the solution for this case is \(x \in (0, \infty)\).
Case 2: \(\frac{1+x}{x} \leq -1\)
\[ \frac{1+x}{x} + 1 \leq 0 \] \[ \frac{1+x+x}{x} \leq 0 \] \[ \frac{1+2x}{x} \leq 0 \] To solve this inequality, we find the critical points by setting the numerator and denominator to zero. Critical points are \(x = -1/2\) and \(x = 0\). Using the wavy curve method or testing intervals, we find the solution is \(x \in [-1/2, 0)\).
Step 4: Final Answer
The domain of the function is the union of the solutions from both cases.
Domain = \([-1/2, 0) \cup (0, \infty)\).
This can also be written as \(\left[-\frac{1}{2}, \infty\right) - \{0\}\).