Question

If $\alpha, \beta$ are the roots of the quadratic equation $x^2-2 p(x-4)-15=0$, then the set of values of $p$ for which one root is less than $1 \&$ the other root is greater than 2 is:

• A. $\left(\frac{7}{3}, \infty\right)$
• B. $\left(-\infty, \frac{7}{3}\right)$
• C. $x \in R$
• D. $\left(-\infty, \frac{11}{4}\right)$

Answer 1

Answer: (B)

$x^2-2 p x+(8 p-15)=0$
$f(1)<0$ and $f(2)<0$
$f(1)=1-2 p+8 p-15<0$
$ p< 7 / 3$
$f(2)=4-4 p+8 p-15<0$
$4 p-11<0 \Rightarrow p<\frac{11}{4}$
$p \in(-\infty, 7 / 3)$ Ans.