Question

The largest positive integral of $\alpha$ for which the inequality $3-|x-\alpha |>x^2$ is satisfied by at least one negative $x$ is

Answer 1

$3-|x-\alpha |>x^2$
$\Rightarrow 3-x^2>|x-\alpha |$
As we increase the value of $\alpha ,$ the graph of $y=|x-\alpha |$ moves from left to right.

Extreme Case $\text{I}:$


Extreme Case $\text{II}:$


Let $f\left(x\right)=3-x^2$ and $g\left(x\right)=|x-\alpha |$
$|x-\alpha |=\{\begin{array}{cc}\alpha -x,& x<\alpha \\ x-\alpha ,& x\ge \alpha \end{array}$

$3-x^2=x-\alpha$
$\Rightarrow x^2+x-\alpha -3=0$
$\Delta =0\Rightarrow \alpha =-\frac{13}{4}$

$3-x^2=\alpha -x$
$\Rightarrow x^2-x+\alpha -3=0$
$\Delta =0\Rightarrow \alpha =\frac{13}{4}$

So, if the above inequality has solution (either positive or negative),
then $\alpha \in (-\frac{13}{4},\frac{13}{4})\cdots \left(1\right)$

For the negative solution, the right $y-$intercept of $g\left(x\right)$ should be less than $3.$
$\Rightarrow \alpha <3\cdots \left(2\right)$


From $\left(1\right)$ and $\left(2\right),$ $\alpha \in (-\frac{13}{4},3)$
Least positive integral value of $\alpha$ is $2.$