(A) 3

(B) 2

(C) 5

(D) 4

Correct Answer: 3

Solution

Given equation:

$$ x|x+3| + |x-1| - 2 = 0 $$

Critical points occur at:

$$ x+3=0 \Rightarrow x=-3, \quad x-1=0 \Rightarrow x=1 $$

So we divide the real line into three intervals:

Case 1: $x \ge 1$

$|x+3| = x+3,\quad |x-1| = x-1$

$$ x(x+3) + (x-1) - 2 = 0 $$

$$ x^2 + 4x - 3 = 0 $$

$$ x = \frac{-4 \pm \sqrt{28}}{2} $$

Both roots are negative, hence no solution in this interval.

Case 2: $-3 \le x < 1$

$|x+3| = x+3,\quad |x-1| = 1-x$

$$ x(x+3) + (1-x) - 2 = 0 $$

$$ x^2 + 2x - 1 = 0 $$

$$ x = -1 \pm \sqrt{2} $$

Both roots lie in $[-3,1)$, so both are valid.

Case 3: $x < -3$

$|x+3| = -(x+3),\quad |x-1| = 1-x$

$$ x(-x-3) + (1-x) - 2 = 0 $$

$$ -x^2 - 4x - 1 = 0 $$

$$ x = -2 \pm \sqrt{3} $$

Only $x = -2 - \sqrt{3}$ satisfies $x < -3$.

Total real solutions:

$$ x = -2-\sqrt3,\; -1-\sqrt2,\; -1+\sqrt2 $$

Hence, the number of real solutions is

$$ \boxed{3} $$