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

Answer 1

The correct option is D $(-\infty ,-1)$

Let $f\left(x\right)=4x^2-20Px+(25P^2+15P-66)$ ...(1)

The roots of (1) are real if

$b^2-4ac=400P^2-16(25P^2+15P-66)=16(66-15P)\ge 0$

$\Rightarrow P\le \frac{22}{5}$ ...(2)

Both roots of (1) are less than $2.$

Therefore, $f\left(2\right)>0$ and sum of roots $<4.$

$\Rightarrow 4.2^2-20P.2+(25P^2+15P-66)>0$ and $-(-20+15)P<4$

$\Rightarrow P^2-P-2>0$ and $P<\frac{4}{5}$

$\Rightarrow (P+1)(P-2)>0$ and $P<\frac{4}{5}$

$\Rightarrow P<-1$ and $P<\frac{4}{5}$

$\Rightarrow P<-1$ ...(3)

From (2) and (3), we get $P<-1\Rightarrow P\in (-\infty ,-1)$