Let f(x) = sin⁻¹ x and g(x) = (x²-x-2)/(2x²-x-6). If g(2) = limₓ→₂ g(x), then the domain of the func

Q: Wait, why are we allowed to just cancel out the $(x-2)$ term? Isn't that division by zero?

That is a brilliant observation! Normally, you would be absolutely right—canceling $(x-2)$ creates a hole at $x=2$. However, the problem explicitly states $g(2) = \lim_{x\to 2} g(x)$. This is the mathematical way of saying, 'We have filled the hole.' The function is defined at $x=2$ by the limit value, making it continuous there. So, for the purpose of this domain calculation, we can treat it as the simplified expression $\frac{x+1}{2x+3}$.

Q: Why do we have to split the inequality $-1 \le \frac{x+1}{2x+3} \le 1$ into two parts? Can't we just solve it all at once?

While it feels tempting to manipulate the whole expression at once, doing so with rational inequalities often leads to sign errors. When you have variables in the denominator, multiplying across can flip the inequality sign depending on whether the denominator is positive or negative. By splitting it into two separate inequalities, you ensure that you are always dealing with a clear 'greater than zero' or 'less than zero' scenario, which is the safest way to use the wavy curve method.

Q: Why is the domain of $\sin^{-1} x$ so restrictive? Why can't it take any value?

Think of $\sin^{-1} x$ as the inverse of the sine function. The sine function, $y = \sin \theta$, oscillates between $-1$ and $1$ for any real input $\theta$. Because the inverse function 'undoes' the sine operation, it can only map back to the values that the sine function actually produces. If you try to input a value like $2$ into $\sin^{-1}$, you are essentially asking, 'What angle has a sine of $2$?' Since the sine of any real angle is bounded between $-1$ and $1$, no such real angle exists. Hence, the domain is strictly $[-1, 1]$.

Question

Let $f(x) = \sin^{-1} x$ and $g(x) = \frac{x^2-x-2}{2x^2-x-6}$. If $g(2) = \lim_{x	o 2} g(x)$, then the domain of the function $f \circ g$ is :

Options:

(A) $(-\infty, -2] \cup [-\frac{4}{3}, \infty)$

(B) $(-\infty, -1] \cup [2, \infty)$

(C) $(-\infty, -2] \cup [-1, \infty)$

(D) $(-\infty, -2] \cup [-\frac{3}{2}, \infty)$

Sigma Percentile · Accepted Answer

### The Architecture of Composite Functions: A Journey into Domain

Welcome, aspiring engineers! Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion. We are looking at the composition of functions, specifically $f(g(x))$, where $f(x) = \sin^{-1} x$ and $g(x)$ is a rational function.

This problem is a classic JEE Advanced challenge because it forces you to synthesize three distinct skills: understanding domain constraints, handling removable discontinuities, and solving rational inequalities. Let us embark on this journey together.

### Phase 1: The Gatekeeper Constraint

Imagine you are building a machine. The first part of the machine is $g(x)$, which takes an input $x$ and churns out a value. The second part of the machine is $f(x) = \sin^{-1} x$, which takes the output of $g(x)$ and processes it further.

For the entire machine to work, the output of $g(x)$ must be acceptable to $f(x)$. We know that the function $f(x) = \sin^{-1} x$ is a very picky gatekeeper. It refuses to accept any input outside the closed interval $[-1, 1]$.

If $g(x)$ produces a value like $1.5$ or $-5$, the machine crashes. Therefore, our fundamental constraint is:

$$-1 \le g(x) \le 1$$

This is the bedrock of our solution. Everything we do from here on is simply ensuring that $g(x)$ stays within these safe boundaries.

### Phase 2: Simplifying the Inner Machine

Now, let us look at $g(x) = \frac{x^2-x-2}{2x^2-x-6}$. Before we start plugging this into our inequality, we must simplify it. A complex rational expression is like a tangled knot; we need to untie it by factoring.

The numerator, $x^2 - x - 2$, factors into $(x-2)(x+1)$. The denominator, $2x^2 - x - 6$, factors into $(2x+3)(x-2)$.

We see a common factor of $(x-2)$ in both the numerator and the denominator. Usually, this would imply a hole at $x=2$. However, the problem statement gives us a gift: $g(2) = \lim_{x	o 2} g(x)$.

This tells us that the function is continuous at $x=2$. We can safely cancel the $(x-2)$ terms, leaving us with the elegant, simplified form:

$$g(x) = \frac{x+1}{2x+3}$$

This simplification makes our life significantly easier. We are no longer dealing with a quadratic mess; we are dealing with a simple linear-over-linear rational function.

### Phase 3: The Inequality War

Now, we return to our constraint: $-1 \le \frac{x+1}{2x+3} \le 1$. To solve this, we must split it into two separate, manageable battles.

**Battle 1: The Right Side**
We solve $\frac{x+1}{2x+3} \le 1$. Subtracting $1$ from both sides, we get:

$$\frac{x+1}{2x+3} - 1 \le 0 \implies \frac{x+1 - (2x+3)}{2x+3} \le 0 \implies \frac{-x-2}{2x+3} \le 0$$

Multiplying by $-1$ flips the inequality sign, giving us:

$$\frac{x+2}{2x+3} \ge 0$$

Using the wavy curve method, we identify critical points at $x = -2$ and $x = -\frac{3}{2}$. Testing the intervals, we find the solution set for this part is $(-\infty, -2] \cup (-\frac{3}{2}, \infty)$.

**Battle 2: The Left Side**
We solve $\frac{x+1}{2x+3} \ge -1$. Adding $1$ to both sides, we get:

$$\frac{x+1}{2x+3} + 1 \ge 0 \implies \frac{x+1 + 2x+3}{2x+3} \ge 0 \implies \frac{3x+4}{2x+3} \ge 0$$

Here, our critical points are $x = -\frac{4}{3}$ and $x = -\frac{3}{2}$. Testing the intervals, we find the solution set for this part is $(-\infty, -\frac{3}{2}) \cup [-\frac{4}{3}, \infty)$.

### Phase 4: The Intersection

We are almost there! We have two solution sets, but for the composite function to be defined, **both** conditions must be true simultaneously. We need the intersection of our two battle results.

Looking at the number line:
1. The first set covers $(-\infty, -2]$ and $(-\frac{3}{2}, \infty)$.
2. The second set covers $(-\infty, -\frac{3}{2})$ and $[-\frac{4}{3}, \infty)$.

When we overlap these, we see that for $x < -\frac{3}{2}$, the intersection is $(-\infty, -2]$. For $x > -\frac{3}{2}$, the intersection is $[-\frac{4}{3}, \infty)$.

Combining these, we arrive at our final domain:

$$(-\infty, -2] \cup [-\frac{4}{3}, \infty)$$

### Conclusion

Look at what we have achieved! We started with a daunting composite function and, through systematic simplification and logical partitioning, we found the exact set of values that keep the function alive.

This is the essence of JEE Advanced mathematics: it is not about memorizing formulas, but about understanding the flow of constraints. You have mastered the art of the domain today. Keep this clarity, keep this rigor, and you will conquer any problem that comes your way.

Animated Solution for Mathematics - Functions: Let $f (x) = sin^{- 1} x$ and $g (x) = \frac{x ^{2} - x - 2}{2 x ^{2} - x - 6}$ . If $g (2) = lim_{x \to 2} g (x)$ , then the domain of the function $f \circ g$ is :

Select Answer:

Visualized Solution

Introduction to the Problem

The Domain Constraint of $sin^{- 1} x$

Factoring $g (x)$

Simplifying $g (x)$ and Continuity

Setting up the Rational Inequality

Solving Inequality Part 1

Critical Points for Part 1

Solving Inequality Part 2

Critical Points for Part 2

Visualizing the Intersection

The Final Domain Calculation

Conclusion and Key Takeaway

The Sigma Insight: Composite Functions

The Architecture of Composite Functions

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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