Functions – Modulus functions – JEE Main 24 January 2025 Shift 2

Solution:

LHS $= x^2 + 3x + 2 = (x+1)(x+2)$. This is a parabola opening upwards with roots at $-1$ and $-2$.
RHS $= \min\{|x-3|, |x+2|\}$.
To simplify the RHS, we find where $|x-3| = |x+2|$.
The midpoint of $3$ and $-2$ is $x = 0.5$.
For $x < 0.5$, $|x+2| < |x-3|$, so RHS $= |x+2|$.
For $x > 0.5$, $|x-3| < |x+2|$, so RHS $= |x-3|$.

Number of solutions = 2.

Ans. (1)