The domain of the function f(x) = sin⁻¹( (x² - 3x + 2) / (x² + 2x + 7) ) is :

Q: Wait, why do we check the discriminant of the denominator? Can't I just cross-multiply immediately?

That is a dangerous habit! In inequalities, multiplying by a variable expression is only safe if you know its sign. If the denominator were negative, the inequality signs would flip. By checking the discriminant, we prove the denominator is always positive, allowing us to proceed with confidence.

Q: Why did the second case result in 'all real numbers'? Did I make a mistake?

Not at all! When you solved $2x^2 - x + 9 \ge 0$, you found a quadratic that never touches the x-axis and opens upwards. This means it is positive for every single value of $x$. It is a perfectly valid mathematical result that simplifies your final intersection.

Q: I got $x \ge -1$ from the first part. Is that the final answer?

Yes, because the second part of the inequality was satisfied for all real numbers. When you take the intersection of $x \in [-1, \infty)$ and $x \in \mathbb{R}$, you are left with $x \in [-1, \infty)$. You have successfully navigated the constraints!

Q: Is there a shortcut to check if a quadratic is always positive?

Yes! A quadratic $ax^2 + bx + c$ is always positive if the leading coefficient $a > 0$ and the discriminant $\Delta = b^2 - 4ac < 0$. This tells you the parabola never crosses the x-axis and stays entirely above it.

Q: Why do we use the intersection of the two cases instead of the union?

Great question. The original inequality $-1 \le u \le 1$ is a compound statement. It means $u \ge -1$ AND $u \le 1$ must both be true simultaneously. Therefore, we must find the values of $x$ that satisfy both conditions at the same time, which is the intersection.

Question

The domain of the function $f(x) = \sin^{-1} \left( \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \right)$ is :

Options:

(A) $[1, \infty)$

(B) $(-1, 2]$

(C) $[ -1, \infty)$

(D) $(-\infty, 2]$

Sigma Percentile · Accepted Answer

### The Gateway to Inverse Trigonometry: Understanding Domain Constraints

Welcome, fellow traveler on the JEE journey! Today, we are going to dissect a problem that seems like a simple algebraic exercise but is actually a beautiful lesson in constraints and logical rigor.

We are looking for the domain of the function $f(x) = \sin^{-1} \left( \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \right)$.

### The Conceptual Foundation

Before we touch a single variable, let us pause and respect the function. The inverse sine function, $\sin^{-1}(u)$, is a picky creature. It only accepts inputs $u$ that fall within the closed interval $[-1, 1]$.

If you try to feed it anything outside this range, the function simply ceases to exist in the realm of real numbers. So, our mission is clear: we must ensure that the argument stays trapped within the bounds of $-1$ and $1$.

Mathematically, we write this as:

$$-1 \le \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \le 1$$

### The Trap of the Denominator

Many students rush to cross-multiply here. But stop! In the world of inequalities, multiplying by a variable expression is a high-stakes gamble.

If the denominator $x^2 + 2x + 7$ were negative, the inequality signs would flip, leading you to a completely wrong answer. We must investigate the nature of this quadratic. Let us calculate the discriminant $\Delta = b^2 - 4ac$ for $x^2 + 2x + 7$:

$$\Delta = (2)^2 - 4(1)(7) = 4 - 28 = -24$$

Since $\Delta < 0$ and the leading coefficient is positive, this parabola never touches the $x$-axis and stays strictly above it. It is always positive! This is a massive relief; we can now multiply across the inequality without any fear of flipping the signs.

### Breaking Down the Inequality

With the denominator safely handled, we split our compound inequality into two manageable cases.

**Case 1: The Right Side**

We solve:

$$\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \le 1$$

Multiplying by the denominator, we get $x^2 - 3x + 2 \le x^2 + 2x + 7$. The $x^2$ terms cancel out beautifully, leaving us with $-3x + 2 \le 2x + 7$.

Rearranging, we find $-5x \le 5$, which simplifies to **$x \ge -1$**.

**Case 2: The Left Side**

Now we solve:

$$\frac{x^2 - 3x + 2}{x^2 + 2x + 7} \ge -1$$

This leads to $x^2 - 3x + 2 \ge -(x^2 + 2x + 7)$. Expanding the right side, we get $x^2 - 3x + 2 \ge -x^2 - 2x - 7$.

Moving everything to one side, we arrive at $2x^2 - x + 9 \ge 0$. Again, let us check the discriminant: $\Delta = (-1)^2 - 4(2)(9) = 1 - 72 = -71$.

Just like our denominator, this quadratic is always positive. This means the condition is satisfied for all real numbers $x \in \mathbb{R}$.

### The Final Intersection

We have arrived at the finish line. We need the intersection of our two cases: $x \ge -1$ AND $x \in \mathbb{R}$.

The intersection of these two sets is simply **$x \in [-1, \infty)$**.

Look at the elegance of the result! We started with a complex-looking rational function inside an inverse sine, and through careful logical steps, we reduced it to a simple interval. This is the beauty of JEE mathematics—it is not about brute force; it is about understanding the constraints and navigating them with precision.

Animated Solution for Mathematics - Inverse Trigonometric Functions: The domain of the function $f (x) = sin^{- 1} (\frac{x ^{2} - 3 x + 2}{x ^{2} + 2 x + 7})$ is :

Select Answer:

Visualized Solution

The Given Function

Domain of $sin^{- 1} (u)$

Setting up the Inequality

Checking the Denominator

Always Positive

Splitting into Two Cases

Solving Case 1

Simplifying Case 1

Solving Case 2

Simplifying Case 2

Checking the Second Quadratic

Final Intersection

The Sigma Insight: Domain and Range of Inverse Trigonometric Functions

The Gateway to Inverse Trigonometry

The Conceptual Foundation

The Trap of the Denominator

Breaking Down the Inequality

The Final Intersection

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Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The domain of the function f(x)=sin−1(x2+2x+7x2−3x+2​) is :