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

Welcome, fellow traveler on the path to JEE mastery! Today, we are dissecting a problem that looks intimidating at first glance but reveals a beautiful, almost rhythmic structure once we peel back the layers.

We are looking for the domain of $f(x)={\sec }^{-1}(2[x]+1)$. When you see a composite function like this, do not panic. Think of it as an onion; we must work from the outside in, respecting the constraints of each layer as we go.

The Outer Shell - The Secant Inverse Constraint

The outermost function is ${\sec }^{-1}(u)$, where $u=2[x]+1$. Before we even touch the greatest integer part, we must ask: what is the fundamental law governing the secant inverse?

The domain of ${\sec }^{-1}(u)$ is strictly restricted. Because $\sec (y)=u$ implies $\cos (y)=\frac{1}{u}$, and we know that the cosine function is trapped between $-1$ and $1$, the input $u$ must satisfy $|u|\ge 1$.

This is our 'forbidden zone'. If $u$ falls anywhere between $-1$ and $1$, the function simply ceases to exist. Our mission is to ensure that $2[x]+1$ never lands in that forbidden territory.

The Bifurcation - Splitting the Problem

Since our condition is $|2[x]+1|\ge 1$, we are dealing with an absolute value inequality. This naturally splits our problem into two distinct, non-overlapping cases.

We are essentially looking for the 'safe zones' on the number line.

We require $2[x]+1\ge 1$. Let's solve this with the precision of a surgeon.

Subtracting $1$ from both sides gives us $2[x]\ge 0$. Dividing by $2$ leaves us with the elegant condition $[x]\ge 0$.

Now, what does this mean for $x$? The greatest integer function $[x]$ outputs an integer. For $[x]$ to be $0$ or greater, $x$ must be at least $0$. Thus, the first part of our domain is $x\ge 0$.

Now, we look at the other side: $2[x]+1\le -1$. Again, we subtract $1$ to get $2[x]\le -2$.

Dividing by $2$ yields $[x]\le -1$. This is where many students stumble, but you won't.

If the greatest integer of $x$ is $-1$ or less, $x$ can be any negative number. Therefore, this case covers all $x<0$.

The Synthesis - The Beauty of the Union

We have solved both cases. Case 1 gave us the interval $[0,\infty )$, and Case 2 gave us the interval $(-\infty ,0)$.

To find the total domain, we must take the union of these two sets: $x\in [0,\infty )\cup (-\infty ,0)$.

Look at what happens when we combine them. The first set covers everything from zero to positive infinity. The second set covers everything strictly less than zero.

When you unite these, you have covered the entire real number line! There are no gaps. The final domain is $\mathbb{R}$, or $(-\infty ,\infty )$.

Isn't it satisfying? What seemed like a restrictive, broken domain for ${\sec }^{-1}$ was completely bypassed by the clever transformation of the inner function. Keep this in mind for your exam: always visualize the constraints, break them into cases, and trust the algebra. You have got this!