Question

Find the value of $m$ such that one root is greater than $2$ and the other root is smaller than $1$ of the quadratic equation $x^2-(m-3)x+m=0$ $(m\in R)$.

• A. $(10,\infty )$
• B. $(9,\infty )$
• C. No solution
• D. $(-\infty ,1)$

Answer 1

The correct option is C No solution

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

When one root is greater than $2$ and other root is smaller than $1$


Condition :

(i) $D>0$

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

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

Now, on solving it,

(i) $D>0$

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

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

$\Rightarrow m^2-10m+9>0$

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

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

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

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

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

$\Rightarrow 4<0$

Hence, $m\in \varnothing$

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

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

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

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

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

Since, in (ii) condition $m$ is null set.

Therefore, entire solution will be null set.

$i.e,m\in \varnothing$