Question

The value of P for which both the roots of the equation $4x^2-20px+(25P^2+15p-66)=0$ are less than 2 lies in

• A. $\left(\frac{4}{5},2\right)$
• B. $\left(2,\infty \right)$
• C. $\left(-1,-\frac{4}{5}\right)$
• D. $\left(-\infty ,-1\right)$

Answer 1

Answer: (D)

Let $f(x)=4x^2-20Px+(25P^2+15P-66)=0\dots (i)$
As roots of (i) are real $\therefore D\ge 0$
$\therefore 400P^2-16(25P^2+15P-66)\ge 0$
$\therefore P\ge \frac{22}{5}$
Now, roots of (i) are less than 2
$\therefore f(2)>0$ and $\alpha +\beta <4$
$\therefore 16-40P+25P^2+15P-66>0$ and $\frac{20}{4}P<4$
$\Rightarrow P^2-P-2>0$ and $P<\frac{4}{5}$
$\therefore P>2$ or $P<-1$ and $P<\frac{4}{5}$
i.e. $P>2$, or $P<-1$ and $P<\frac{4}{5}$
$\therefore P<-1$
$\therefore P\in \left(-\infty ,-1\right)$