Question

The set of values of $p$ such that both the roots of the equation $(p-5)x^2-2px+(p-4)=0$ are positive and one of the roots is less than $2$ and the other root lies between $2$ & $3$ is

• A. $\left(\frac{49}{4},24\right)$
• B. $(5,\infty )$
• C. $(-\infty ,4)\cup \left(\frac{49}{4},\infty \right)$
• D. $(5,\frac{49}{4})$

Answer 1

The correct option is A $\left(\frac{49}{4},24\right)$

$(p-5)x^2-2px+(p-4)=0$ or $x^2-\frac{2p}{p-5}x+\frac{p-4}{p-5}=0$
$f\left(x\right)=x^2-\frac{2p}{p-5}x+\frac{p-4}{p-5}=0$
$f\left(0\right)>0,f\left(2\right)<0,f\left(3\right)>0$
$f\left(0\right)>0\Rightarrow \frac{p-4}{p-5}>\mathrm{0.....}\left(1\right)$
$f\left(2\right)<0\Rightarrow \frac{p-24}{p-5}<\mathrm{0.....}\left(2\right)$
$f\left(3\right)>0\Rightarrow \frac{4p-49}{p-5}>\mathrm{0.....}\left(3\right)$
Intersection of (1),(2) & (3) gives $p\in (\frac{49}{4},24)$