Question

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

• A. $(-\infty ,4)$
• B. $(-\infty ,2)$
• C. $(-\infty ,3)$
• D. $(-\infty ,1)$

Answer 1

Answer: (C)

$(-\infty ,3)$

The equation $x^2-x+a-3=0$ must have atleast one negative root

For real roots, $D\ge 0\Rightarrow 1-4(a-3)\ge 0$

$\Rightarrow a\le \frac{13}{4}$

$\therefore a\in (-\infty ,\frac{13}{4}]...\left(i\right)$

Both roots will be non-negative, if

$D\ge 0,a-3\ge 0$

$\Rightarrow a\le \frac{13}{4},a\ge 3$

$\therefore a\in [3,\frac{13}{4}]...\left(ii\right)$

Thus, equation will have atleast one negative root if

$a\in (-\infty ,\frac{13}{4}]\cup [3,\frac{13}{4}]$ (Using Eqs. (i) and (ii))

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