Animated Solution for Mathematics - Differentiation: The set of all $a\in \mathbb{R}$ for which the equation $x|x-1|+|x+2|+a=0$ has exactly one real root, is

We are tasked with finding the set of all $a\in \mathbb{R}$ such that the equation $x|x-1|+|x+2|+a=0$ has exactly one real root. Let us define the function $f(x)=x|x-1|+|x+2|$.

The given equation can be rewritten as $f(x)=-a$. Geometrically, this implies we are seeking the number of intersection points between the curve $y=f(x)$ and the horizontal line $y=-a$.

If $f(x)$ is strictly monotonic, any horizontal line will intersect the curve exactly once. We proceed by removing the absolute value bars using the critical points $x=1$ and $x=-2$.

In this interval, both $(x-1)$ and $(x+2)$ are non-negative. The function simplifies to:

Calculating the derivative, we find $f^{\prime }(x)=2x$. Since $x\ge 1$, it follows that $f^{\prime }(x)\ge 2$, which is strictly positive. The function is strictly increasing in this interval.

Here, $(x-1)$ is negative, so $|x-1|=-(x-1)$, while $(x+2)$ remains non-negative. The function becomes:

The derivative is $f^{\prime }(x)=-2x+2=2(1-x)$. Since $x<1$, the term $(1-x)$ is positive, ensuring $f^{\prime }(x)>0$. The function continues to increase.

In this region, both expressions are negative. We have:

The derivative is $f^{\prime }(x)=-2x$. Since $x<-2$, we have $-2x>4$, which is strictly positive. The function is increasing throughout this interval as well.

Because $f(x)$ is continuous and strictly increasing across the entire real line, its range is $(-\infty ,\infty )$.

This implies that for any real value of $-a$, there exists exactly one $x$ such that $f(x)=-a$. Consequently, the condition is satisfied for all real values of $a$.

The set of all such $a$ is $(-\infty ,\infty )$.