Question

Find the possible values of ${}^{\prime }{a}^{\prime }$ such that the inequality $3-x^2=|x-a|$ has atleast one negative solution.

Answer 1

$\begin{array}{c}\left(3-x^2\right)=\left|x-a\right|\\ Ifa=-\sqrt{3}\\ ifa=0\\ Lety=3-x^2\\ y^{\prime }=-2x\\ 1=-2x\\ x=\frac{-1}{2}\\ y=3-\frac{1}{4}\\ =\frac{11}{4}\\ So,a>-\frac{1}{2}-\frac{11}{4}\\ a>-\frac{13}{4}anda<3\\ So,-\frac{13}{4}<a<3\end{array}$