JEE Mathematics 2025 Question with Solution

Given: Solve $x^2 + 3x + 2 = \min \left( |x - 3|, |x + 2| \right)$.

Find: The number of real solutions.

The solution depends on which absolute value is smaller.

Since $|x-3| < |x+2|$ when $x < \frac{1}{2}$, take

$$x^2 + 3x + 2 = |x-3|$$

For $x < 3$, this becomes

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

So,

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

which gives

$$x = -2 \pm \sqrt{5}$$

Checking these values against $|x-3| < |x+2|$, neither satisfies the required minimum condition.

Now for $|x+2| < |x-3|$, we have $x > \frac{1}{2}$, so take

$$x^2 + 3x + 2 = |x+2|$$

For $x \ge -2$,

$$x^2 + 3x + 2 = x + 2$$

which gives

$$x^2 + 2x = 0$$

so the roots are $x=0$ and $x=-2$, but neither lies in $x > \frac{1}{2}$.

For $x < -2$,

$$x^2 + 3x + 2 = -(x+2)$$

so

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

thus

$$(x+2)^2 = 0$$

and hence $x=-2$, which again does not satisfy $x > \frac{1}{2}$.

Finally, check the boundary value $x=-2$ explicitly:

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

and

$$\min(|x-3|, |x+2|) = \min(5,0) = 0$$

So the equation is satisfied at $x=-2$.

Therefore, there is exactly one real solution. The correct option is A.

Given: $x^2 + 3x + 2 = \min \left( |x - 3|, |x + 2| \right)$

Find: How many real values of $x$ satisfy the equation.

Use the comparison of distances from $3$ and $-2$.

1. If $|x-3| < |x+2|$, then the minimum is $|x-3|$. This happens for $x < \frac{1}{2}$.
2. If $|x+2| < |x-3|$, then the minimum is $|x+2|$. This happens for $x > \frac{1}{2}$.

For the first case,

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

which gives

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

Hence,

$$x = -2 \pm \sqrt{5}$$

But on checking, neither root makes $|x-3|$ smaller than $|x+2|$, so both are rejected.

For the second case, if $x \ge -2$, then $|x+2| = x+2$ and

$$x^2 + 3x + 2 = x+2$$

so

$$x^2 + 2x = 0$$

This gives $x=0$ or $x=-2$, neither valid for $x > \frac{1}{2}$.

If $x < -2$, then $|x+2| = -(x+2)$ and

$$x^2 + 3x + 2 = -(x+2)$$

so

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

Thus $x=-2$, again not valid for $x > \frac{1}{2}$.

Now explicitly test $x=-2$:

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

and

$$\min(|x-3|,|x+2|)=\min(5,0)=0$$

So $x=-2$ is a valid solution.

Therefore, the equation has exactly $1$ real solution.