JEE Mathematics 2024 Question with Solution

Given: $x(x^2 + 3x) + |x-1| + 6|k-2| = 0$

Find: The number of real solutions.

Analyze the equation by writing

$$x(x^2 + 3x) + |x-1| + 6|k-2| = 0 \implies x^3 + 3x^2 + |x-1| + C = 0$$

where

$$C = 6|k-2|$$

Now consider the absolute value in two cases.

For $x \ge 1$,

$$|x-1| = x-1$$

so the equation becomes

$$x^3 + 3x^2 + (x-1) + C = 0$$

that is,

$$x^3 + 3x^2 + x + (C-1) = 0$$

For $x < 1$,

$$|x-1| = -x+1$$

so the equation becomes

$$x^3 + 3x^2 + (-x+1) + C = 0$$

that is,

$$x^3 + 3x^2 - x + (C+1) = 0$$

According to the extracted solution, these cubic expressions are taken to be monotonic, so there is only one real solution overall.

Therefore, the number of real solutions is $1$.

The solution contains a second approach for a completely different equation,

$$x\left(x^2 + 3|x| + 5|x-1| + 6|x-2|\right) = 0$$

which does not match the given question. Hence it cannot be used for deriving the result for this question.

Using the matching first approach on the page, the extracted conclusion is that the number of real solutions is $1$.