Question

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

• A. None of the above
• B. $(9,\infty )$
• C. $(10,\infty )$
• D. $(-\infty ,1)$

Answer 1

The correct option is C $(10,\infty )$

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

When exactly one root of $f\left(x\right)=0$ lies in the interval $(1,2)$


Condition :

(i) $D>0$

(ii) $f\left(1\right).f\left(2\right)<0$

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) $f\left(1\right).f\left(2\right)<0$

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

$\Rightarrow (1-m+3+m)(4-2m+6+m)<0$

$\Rightarrow 4(10-m)<0$

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

Now, taking both the above condition, we get

$m\in (10,\infty )$