If the domain of the function f(x) = sin⁻¹(1/(x²-2x-2)), is (-∞, α] ∪ [β, γ] ∪ [δ, ∞), then α + β +

Q: Wait, why does $-1 \le \frac{1}{u} \le 1$ imply $|u| \ge 1$? I thought it might be something else.

That is a brilliant question! Think about the magnitude. If $u$ is a fraction like $0.5$, its reciprocal is $2$, which is outside the allowed range of $[-1, 1]$. To keep the reciprocal inside $[-1, 1]$, the denominator $u$ must be 'large'—specifically, its magnitude must be at least $1$. If $|u| 1$, which violates our domain constraint. So, $|u| \ge 1$ is the only way to keep the reciprocal within the safe zone.

Q: Why do we split the inequality into two cases? Can't we just square both sides?

You *could* square both sides, but that often leads to higher-degree polynomials and potential sign errors. Splitting into $|u| \ge 1 \implies u \ge 1$ or $u \le -1$ is the standard, safest way to handle absolute value inequalities. It keeps the math linear and manageable, allowing you to use the wavy curve method effectively.

Q: Why do we use the wavy curve method? Is it really necessary?

The wavy curve method (or sign scheme) is your best friend in JEE Advanced. It allows you to visualize the sign of a polynomial across the entire real number line instantly. Without it, you would have to test arbitrary points in every interval, which is slow and prone to errors. Embrace the curve; it is the most elegant way to solve inequalities.

Question

If the domain of the function $f(x) = \sin^{-1}(\frac{1}{x^2-2x-2})$, is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to

Options:

(A) 2

(B) 3

(C) 4

(D) 5

Sigma Percentile · Accepted Answer

### The Gatekeeper Function

Every function has its own personality, and the inverse sine function, $\sin^{-1}(u)$, is a strict gatekeeper. It refuses to accept any input $u$ that falls outside the closed interval $[-1, 1]$.

If you try to feed it a value like $2$ or $-5$, it simply shuts down. So, our first mission is to ensure that the argument stays within these boundaries:

$$-1 \le \frac{1}{x^2-2x-2} \le 1$$

This is our starting line.

### The Algebraic Pivot

Staring at a reciprocal inequality can be intimidating. However, mathematics is often about finding a simpler perspective.

If the reciprocal of a number is trapped between $-1$ and $1$, it implies that the magnitude of the number itself must be at least $1$. Specifically:

$$|x^2-2x-2| \ge 1$$

Because if the denominator were a small fraction, say $0.5$, the reciprocal would be $2$, which is outside our gatekeeper's limit. By shifting to the absolute value form, we have transformed a complex double inequality into a much cleaner, more manageable condition.

### The Parabolic Dance

An absolute value inequality $|u| \ge 1$ splits into two distinct realities: $u \ge 1$ or $u \le -1$. Let us explore these two cases.

**Case 1:** $x^2-2x-2 \ge 1$. Bringing the $1$ to the left, we get $x^2-2x-3 \ge 0$.

Factoring this quadratic is a joy—it becomes $(x-3)(x+1) \ge 0$. Using the wavy curve method, we see the function is positive in the regions $(-\infty, -1] \cup [3, \infty)$.

**Case 2:** $x^2-2x-2 \le -1$. Bringing the $-1$ over, we get $x^2-2x-1 \le 0$.

This quadratic does not factorize with simple integers, so we summon the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

Plugging in our coefficients, we find the roots to be $1 \pm \sqrt{2}$. Since we need the region where the parabola is less than or equal to zero, our solution is the interval $[1-\sqrt{2}, 1+\sqrt{2}]$.

### The Synthesis

We have our pieces. The domain is the union of these intervals:

$$(-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty)$$

Comparing this to the format $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, we identify our constants: $\alpha = -1$, $\beta = 1-\sqrt{2}$, $\gamma = 1+\sqrt{2}$, and $\delta = 3$.

The final step is the sum:

$$\alpha + \beta + \gamma + \delta = -1 + (1-\sqrt{2}) + (1+\sqrt{2}) + 3$$

Notice the elegance here? The $-\sqrt{2}$ and $+\sqrt{2}$ cancel out perfectly, leaving us with $-1 + 1 + 1 + 3 = 4$.

The final answer is **4**.

Animated Solution for Mathematics - Inverse Trigonometric Functions: If the domain of the function $f (x) = sin^{- 1} (\frac{1}{x ^{2} - 2 x - 2})$ , is $(- \infty, α] \cup [β, γ] \cup [δ, \infty)$ , then $α + β + γ + δ$ is equal to

Select Answer:

Visualized Solution

The Core Constraint of Inverse Sine

Converting to Absolute Value

Splitting into Two Cases

Solving Case 1 (Part A)

Solving Case 1 (Part B)

Domain from Case 1

Solving Case 2 (Part A)

Solving Case 2 (Part B)

Domain from Case 2

Combining the Intervals

Mapping to the Given Format

Calculating the Final Sum

The Sigma Insight: Domain and Range of Inverse Trigonometric Functions

The Gatekeeper Function

The Algebraic Pivot

The Parabolic Dance

The Synthesis

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