Question

Find the values of $m$ such that both the roots of the quadratic equation $x^2-(m-3)x+m=0$ $(m\in R)$ lies in the interval $(1,2).$

• A. $\varphi$
• B. None of the above.
• C. $(5,7)$
• D. $(9,\infty )$

Answer 1

The correct option is A $\varphi$

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

When both the roots of $f\left(x\right)$ lies in the interval $(1,2)$


Condition :

(i) $D\ge 0$

(ii) $f\left(1\right)>0$ & $f\left(2\right)>0$

(iii) $1<-\frac{b}{2a}<2$

Now, on solving it,

(i) $D\ge 0$

$\Rightarrow (m-3{)}^2-4m\ge 0$

$\Rightarrow m^2-6m+9-4m\ge 0$

$\Rightarrow m^2-10m+9>\ge$

$\Rightarrow (m-1)(m-9)\ge 0$

$m\in (-\infty ,1]\cup [9,\infty )$

(ii) $f\left(1\right)>0$

$\Rightarrow 1^2-(m-3).1+m>0$

$\Rightarrow 1-m+3+m>0$

$\Rightarrow 4>0\Rightarrow m\in R$

& $f\left(2\right)>0$

$\Rightarrow 2^2-(m-3).2+m>0$

$\Rightarrow 4-2m+6+m>0$

$\Rightarrow m<10\Rightarrow m\in (-\infty ,10)$

$\therefore f\left(1\right)>0,f\left(2\right)>0$

$m\in (-\infty ,10)$

(iii) $1<-\frac{b}{2a}<2$

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

$\Rightarrow 2<m-3<4$

$\Rightarrow 5<m<7$

$m\in (5,7)$

Now, taking intersection of all the three above conditions, we get

There is no common solution, $\therefore m\in \varphi$