JEE Main 2026 Jan 24 Shift 1, Mathematics Q20: Some Important Properties Of Functions

JEE Main2026Jan 24, Shift 1Mathematics

The number of the real solutions of the equation :

$x ∣ x + 3∣ + ∣ x - 1∣ - 2 = 0$ is

Solution

Split at the breakpoints $x = - 3$ and $x = 1$ .

Case I ( $x \geq 1$ ): $∣ x + 3∣ = x + 3$ , $∣ x - 1∣ = x - 1$ .

$x (x + 3) + (x - 1) - 2 = 0 \Rightarrow x^{2} + 4 x - 3 = 0 \Rightarrow x = - 2 \pm 7$ .

Both roots are less than 1, so both rejected. 0 solutions.

Case II ( $- 3 \leq x < 1$ ): $∣ x + 3∣ = x + 3$ , $∣ x - 1∣ = 1 - x$ .

$x (x + 3) + (1 - x) - 2 = 0 \Rightarrow x^{2} + 2 x - 1 = 0 \Rightarrow x = - 1 \pm 2$ .

Both $\approx 0.41$ and $\approx - 2.41$ lie in $[- 3, 1)$ . 2 solutions.

Case III ( $x < - 3$ ): $∣ x + 3∣ = - (x + 3)$ , $∣ x - 1∣ = 1 - x$ .

$- x (x + 3) + (1 - x) - 2 = 0 \Rightarrow x^{2} + 4 x + 1 = 0 \Rightarrow x = - 2 \pm 3$ .

Only $- 2 - 3 \approx - 3.73$ is less than $- 3$ ; the other is rejected. 1 solution.

Total: 3 real solutions.

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.