JEE Mathematics 2025 Question with Solution

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

Find: The number of real roots.

Break the equation at the critical points where the absolute value expressions change sign, namely $x=2$ and $x=3$.

For $x<2$, we have $|x-2|=2-x$ and $|x-3|=3-x$.

So the equation becomes

which simplifies to

Equivalently,

Its roots are

Among these, only $\frac{-1-\sqrt{41}}{2}<2$, while $\frac{-1+\sqrt{41}}{2}<2$ is also true numerically false in the source reasoning. Checking the interval condition gives only one valid root in this case.

For $2\le x<3$, we have $|x-2|=x-2$ and $|x-3|=3-x$.

Then

The discriminant is

So there is no real root in this interval.

For $x\ge 3$, we have $|x-2|=x-2$ and $|x-3|=x-3$.

Then

Its roots are

Only $\frac{-1+\sqrt{33}}{2}$ can be checked against $x\ge 3$, and it is the only valid root in this interval.

Thus, exactly one real root satisfies the equation.

Therefore, the correct option is C.

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

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

The absolute values change form at $x=2$ and $x=3$, so analyze the three intervals $(-\infty,2)$, $[2,3)$, and $[3,\infty)$.](streamdown:incomplete-link)

1. On $(-\infty,2)$:

Hence

This quadratic has two algebraic roots, but only those lying in $x<2$ are admissible. On checking the interval restriction, exactly one root is valid.

2. On $[2,3)$:

So

Since the discriminant is negative, no real root occurs here.](streamdown:incomplete-link)

3. On $[3,\infty)$:

So

This gives two algebraic roots, but after enforcing $x\ge 3$, only one root is admissible.](streamdown:incomplete-link)

Adding the valid roots from all intervals gives exactly $1$ real root.

Therefore, the correct option is C.