Question

If $\alpha ,\beta$ are the roots of the equation $x^2-3x+a=0,aϵR$ and $\alpha <1<\beta$ then-

• A. $aϵ(-\infty ,-2)$
• B. $aϵ(-\infty ,\frac{9}{4})$
• C. $aϵ(2,\frac{9}{4})$
• D. None of the above.

Answer 1

The correct option is A $aϵ(-\infty ,-2)$

$f\left(x\right)=x^2-3x+a=0rootsare\alpha ,\beta \alpha <1<\beta \therefore f\left(1\right)<0f\left(1\right)=1-3+a<0a-2<0\therefore a\in (-\infty ,2)D>0{(-3)}^2-4a>09-4a>04a-9<0a<\frac{9}{4}\therefore a\in (-\infty ,\frac{9}{4})\therefore a\in (-\infty ,-2)$