Question

If $x^2-x+a-3<0$ for atleast one negative value of $x$, then complete set of values of $a$ is

Answer 1

Given $x^2-x+a-3<0$

For at least one negative value of x.

The coefficient of $x^2$ is positive. The expression on the left-hand side must be positive at infinitely many points. It can be negative at least one point if the smaller root were negative.

$\frac{1-\sqrt{13-4a}}{2}<0$

$\Rightarrow \sqrt{13-4a}>1$

$\Rightarrow 13-4a>1$

$\Rightarrow 4a>12$

$\Rightarrow a<3$

$\Rightarrow a\in (-\infty ,3)$