(A) 4

(B) 2

(C) 5

(D) 3

Correct Answer: 4

Explanation

For \( \sin^{-1}(u) \) to be defined, the argument must satisfy:

$$ -1 \le u \le 1 $$

Here,

$$ u=\frac{1}{x^2-2x-2} $$

So,

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

This gives two inequalities:

$$ \frac{1}{x^2-2x-2} \le 1 \quad \text{and} \quad \frac{1}{x^2-2x-2} \ge -1 $$

Solving,

$$ x^2-2x-3 \ge 0 $$ $$ x^2-2x-1 \le 0 $$

Factorising:

$$ (x-3)(x+1)\ge 0 $$ $$ (x-1)^2 \le 2 $$

This gives:

$$ x \le -1 \;\; \text{or} \;\; x \ge 3 $$ $$ 1-\sqrt{2} \le x \le 1+\sqrt{2} $$

Combining both:

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

So,

$$ \alpha=-1,\quad \beta=1-\sqrt{2},\quad \gamma=1+\sqrt{2},\quad \delta=3 $$

Now,

$$ \alpha+\beta+\gamma+\delta = (-1)+(1-\sqrt{2})+(1+\sqrt{2})+3 $$ $$ =4 $$

Hence, the correct answer is 4.

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