Basic Maths and Logarithm
Modulus Equations
JEE Main
Grade 11
Question:
The number of real solutions of $|x - 1| \cdot |x + 3| = 8$ is:
(1) $0$
(2) $1$
(3) $2$
(4) $4$
Step-by-Step Solution
Key Concept: $(x - 1)(x + 3) = \pm 8$. Case 1 ($= +8$): $x^2 + 2x - 11 = 0 \Rightarrow x = -1 \pm 2\sqrt{3}$ (two real roots). Case 2 ($= -8$): $x^2 + 2x + 5 = 0 \Rightarrow \Delta = -16 < 0$ (no real roots). Total: 2 solutions.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (3)