Question

Find the values of $m$ such that both roots of the quadratic equation $x^2-(m-3)x+m=0$ $(m\in R)$ are smaller than $2$

• A. $(7,10)$
• B. $(-\infty ,1]$
• C. None of the above
• D. $(-\infty ,7)$

Answer 1

The correct option is B $(-\infty ,1]$

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

When both roots of $f\left(x\right)=0$ are smaller than $2.$


Condition :

(i) $D\ge 0$

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

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

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

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

$\Rightarrow m-3<4\Rightarrow m<7$

$m\in (-\infty ,7)$

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

$\Rightarrow (2{)}^2-(m-3\left)\right(2)+m>0$

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

$\Rightarrow -m+10>0\Rightarrow m<10$

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

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

$m\in (-\infty ,1]$