Question

If both the roots of the quadratic equation $x^2-mx+4=0$ are real and distinct and they lie in the interval $[1,5]$, then $m$ is lying in the interval :

• A. $(4,5)$
• B. $(5,6)$
• C. $(3,4)$
• D. $(4,5]$

Answer 1

The correct option is D $(4,5]$

Given$:x^2-mx+4=0$
Roots are real and distinct
$\therefore D>0$
$\Rightarrow m^2-16>0$
$\Rightarrow (m-4)(m+4)>0$
$\Rightarrow m\in (-\infty ,-4)\cup (4,\infty )\cdots \left(1\right)$
Roots lie in the interval $[1,5]$ $\Rightarrow 1\le \alpha <\beta \le 5$
The possible cases are,


So, the required conditions are,
$f\left(1\right)\ge 0$
$\Rightarrow 1^2-m+4\ge 0$
$\Rightarrow m\le 5$
$\Rightarrow m\in (-\infty ,5]\cdots \left(2\right)$

$f\left(5\right)\ge 0$
$\Rightarrow 25-5m+4\ge 0$
$\Rightarrow m\le \frac{29}{5}$
$\Rightarrow m\in (-\infty ,\frac{29}{5}]\cdots \left(3\right)$

$1<-\frac{b}{2a}<5$
$\Rightarrow 1<\frac{m}{2}<5$
$\Rightarrow 2<m<10$
$\Rightarrow m\in (2,10)\cdots \left(4\right)$

Thus, from equations $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$, we get


$i.e,m\in (4,5]$