JEE MAIN - Mathematics (2023 - 13th April Morning Shift - No. 7)
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 :
$$(-\infty, \infty)$$
$$(-6, \infty)$$
$$(-\infty,-3)$$
$$(-6,-3)$$
Explanation
$$
x|x-1|+|x+2|+a=0
$$
Case I : If $x<-2$ then
$$ -x^2+x-x-2+a=0 $$
$$ a=x^2+2 $$
$y=x^2+2$ is decreasing $\forall x \in(-\infty,-2)$
Case II : If $-2 \leq x<1$ then
$$ \begin{aligned} & -x^2+x+x+2+a=0 \\\\ & a=x^2-2 x-2 \end{aligned} $$
$y=x^2-2 x-2$ is decreasing $\forall x \in[-2,1)$.
Case III: If $x \geq 1$ then
$$ \begin{aligned} & x^2-x+x+2+a=0 \\\\ & a=-\left(x^2+2\right) \end{aligned} $$
$y=-\left(x^2+2\right)$ is decreasing $\forall x \in[1, \infty)$
$\therefore$ Exactly one real root $\forall x \in R$
Case I : If $x<-2$ then
$$ -x^2+x-x-2+a=0 $$
$$ a=x^2+2 $$
$y=x^2+2$ is decreasing $\forall x \in(-\infty,-2)$
Case II : If $-2 \leq x<1$ then
$$ \begin{aligned} & -x^2+x+x+2+a=0 \\\\ & a=x^2-2 x-2 \end{aligned} $$
$y=x^2-2 x-2$ is decreasing $\forall x \in[-2,1)$.
Case III: If $x \geq 1$ then
$$ \begin{aligned} & x^2-x+x+2+a=0 \\\\ & a=-\left(x^2+2\right) \end{aligned} $$
$y=-\left(x^2+2\right)$ is decreasing $\forall x \in[1, \infty)$
$\therefore$ Exactly one real root $\forall x \in R$
Comments (0)
