JEE Mathematics 2023 Question with Solution

Given: The equation is $x|x|-5|x+2|+6=0$.

Find: The number of real roots.

The sign of the absolute value terms changes at $x=0$ and $x=-2$. So we solve in three intervals:

• $x \ge 0$
• $-2 \le x < 0$
• $x < -2$

For $x \ge 0$, we have $|x|=x$ and $|x+2|=x+2$.

Only $\frac{5+\sqrt{41}}{2}>0$ satisfies the interval condition, so there is one root here.

For $-2 \le x < 0$, we have $|x|=-x$ and $|x+2|=x+2$.

So $x=-1,-4$. Only $x=-1$ lies in $[-2,0)$, so there is one root here.](streamdown:incomplete-link)

For $x<-2$, we have $|x|=-x$ and $|x+2|=-(x+2)$.

Only $\frac{5-\sqrt{89}}{2}<-2$ satisfies the interval condition, so there is one root here.

Therefore,

So, the correct option is B.

Given: The equation is $x|x|-5|x+2|+6=0$.

Find: The number of real roots.

Quick Tip: For equations involving absolute values, split the real line at the sign-change points of the modulus expressions. Here those points are $x=0$ and $x=-2$. In each interval, remove the modulus signs using the correct signs, solve the resulting quadratic, and then keep only those roots that belong to that interval.

This gives exactly one valid root from each interval, so the total number of real roots is $3$. Hence, the correct option is B.