Question

Let $x^2-(m-3)x+m=0,m\in R$ be a quadratic equation. The values of $m$ for which both roots lie in between $1$ and $2$ is given by

• A. $m\in \varphi$
• B. $m\in [1,9]$
• C. $m\in (5,7)$
• D. $m<10$

Answer 1

The correct option is A

$m\in \varphi$

For the given equation, $a>0$, when compared to $a{x}^2+bx+c=0$.

The graph is upward parabola as shown.

For roots $\alpha ,\beta$ to be in between $1$ & $2$, $f\left(1\right)>0,f\left(2\right)>0$ and $1<\frac{-b}{2a}<2$.

Firstly, for roots to be real

$D\ge 0$

$b^2-4ac\ge 0$

$\Rightarrow m\le 1$ or $m\ge 9$ . . . $\left(1\right)$

$f\left(x\right)=x^2-(m-3)x+m$ (given)

Now, $f\left(1\right)=1-(m-3)+m$

$=4>0$ [Always]

$f\left(2\right)=4-(m-3)2+m$

$=4-2m+6+m$

$=10-m$

$10-m>0$

$m<10$ . . . $\left(2\right)$

$1<\frac{-b}{2a}<2$

$\Rightarrow 1<\frac{m-3}{2}<2$

$\Rightarrow 2<(m-3)<4$

$\Rightarrow 5<m<7$ . . . $\left(3\right)$

From $\left(1\right),\left(2\right)$ & $\left(3\right)$

No intersection.

Hence, $m$ belongs to null set.