Question

Let $x^2-(m-3)x+m,(m\in R)$ be a quadratic expression which is always greater than zero. Then the total number of intergral value(s) of $m$ is .

• A. 7

Answer 1

The correct option is A 7
Given : $x^2-(m-3)x+m>0\forall x\in R,(m\in R)$
By comparing with the standard quadratic equation $y=a{x}^2+bx+c$ we will have the values of $a,b,c,D$ therefore, we have $a=1,b=-(m-3),c=m$

We have given that the value of the given expression is always greater than zero which means $D$ is negative
$\Rightarrow D<0\Rightarrow D=b^2-4ac<0$
$\Rightarrow \left[\right(-(m-3){)}^2-\mathrm{4.1.}m<0]$
$\Rightarrow m^2-6m+9-4m<0$
$\Rightarrow m^2-10m+9<0$
$\Rightarrow (m-1)(m-9)<0$
$\Rightarrow 1<m<9$
$\Rightarrow m=\{2,3,4,5,6,7,8\}$
Or total $7$ values.