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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Inverse Trigonometric Functions

If the domain of the function f(x)= ^ -1 ( 1 x^ 2 -2 x-2 ) , is...

If the domain of the function f(x)=sin⁡−1(1x2−2x−2)f(x)=\sin ^{-1}\left(\frac{1}{x^{2}-2 x-2}\right), is (−∞,α]∪[β,γ]∪[δ,∞)(-\infty, \alpha] \cup[\beta, \gamma] \cup[\delta, \infty), then α+β+γ+δ\alpha+\beta+\gamma+\delta is equal to

Options

  1. A2
  2. B3
  3. C4
  4. D5

Correct answer

C. 4

Step-by-step solution

For sin⁡−1(1x2−2x−2)\sin^{-1}\left(\frac{1}{x^2-2x-2}\right) to be defined, we need \left|\frac{1}{x^2-2x-2}\right| \leq 1,i.e.,, i.e.,|x^2-2x-2| \geq 1.<br>Case1:<br>.<br>Case 1:<br>x^2-2x-2 \geq 1 \Rightarrow x^2-2x-3 \geq 0<br><br>\Rightarrow (x-3)(x+1) \geq 0<br><br>\Rightarrow x \leq -1 or x≥3x \geq 3.
Case 2:
x2−2x−2≤−1⇒x2−2x−1≤0x^2-2x-2 \leq -1 \Rightarrow x^2-2x-1 \leq 0
⇒(x−1)2≤2\Rightarrow (x-1)^2 \leq 2
⇒1−2≤x≤1+2\Rightarrow 1-\sqrt{2} \leq x \leq 1+\sqrt{2}.
Domain =(−∞,−1]∪[1−2,1+2]∪[3,∞)= (-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty).
So α=−1, β=1−2, γ=1+2, δ=3\alpha = -1,\ \beta = 1-\sqrt{2},\ \gamma = 1+\sqrt{2},\ \delta = 3.
α+β+γ+δ=−1+1−2+1+2+3=4\alpha + \beta + \gamma + \delta = -1 + 1 - \sqrt{2} + 1 + \sqrt{2} + 3 = 4.
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