Animated Solution for Mathematics - Inverse Trigonometric Functions: The domain of the function ${\text{cosec}}^{-1}\left(\frac{1+x}{x}\right)$ is :

Welcome, future engineers! Today, we are going to peel back the layers of a function that might look intimidating at first glance: $f(x)={\text{cosec}}^{-1}\left(\frac{1+x}{x}\right)$.

Many students see inverse trigonometric functions and immediately panic, thinking about complex graphs or obscure identities. But here is the secret: domain problems are not about memorizing graphs; they are about understanding the 'rules of existence' for a function.

Think of the domain as the guest list for a party. We need to find every single $x$ that is allowed to enter the function without causing a mathematical disaster, such as division by zero.

Let's start with the gatekeeper: the inverse cosecant function. We know that for any input $u$, the function $y={\text{cosec}}^{-1}(u)$ is only defined if the magnitude of $u$ is at least $1$.

Mathematically, this is expressed as $|u|\ge 1$. This is our first constraint and the bedrock upon which the rest of our solution is built.

We set our argument $u=\frac{1+x}{x}$ into this inequality:

Now, look at that expression. We can rewrite $\frac{1+x}{x}$ as $\frac{1}{x}+\frac{x}{x}$, which simplifies beautifully to $\frac{1}{x}+1$.

Suddenly, the inequality becomes:

This is a vital lesson in JEE preparation: always look for the simplest form of an expression before you start heavy calculations. Complexity is often just a mask for simplicity.

When we deal with absolute value inequalities of the form $|A|\ge 1$, we are essentially saying that $A\ge 1$ or $A\le -1$. This splits our problem into two distinct paths.

Path 1: $\frac{1}{x}+1\ge 1$.

Subtracting $1$ from both sides, we get $\frac{1}{x}\ge 0$. For a fraction to be non-negative with a positive numerator, the denominator $x$ must be positive. Therefore, $x>0$.

Path 2: $\frac{1}{x}+1\le -1$.

Subtracting $1$ from both sides gives $\frac{1}{x}\le -2$. Do not cross-multiply; instead, bring the $-2$ to the left side:

We use the Wavy Curve Method for the rational inequality $\frac{1+2x}{x}\le 0$. The critical points are $x=-1/2$ (where the numerator is zero) and $x=0$ (where the denominator is zero).

Testing the intervals, we find that the expression is negative between $-1/2$ and $0$. Since the inequality is "less than or equal to," we include $-1/2$ but must exclude $0$ because division by zero is undefined.

Combining our two paths, $x>0$ and $x\in [-1/2,0)$, we arrive at the final domain:

x \in \left