Question

If one root of $x^2-2p(x-4)-15=0$ is less than $1$ and the other root is greater than $2,$ then the range of $p$ is

• A. $(\frac{7}{3},\infty )$
• B. $(-\infty ,\frac{7}{3})$
• C. $R$
• D. $(-\frac{7}{3},\frac{7}{3})$

Answer 1

The correct option is B $(-\infty ,\frac{7}{3})$

Let $f\left(x\right)=x^2-2p(x-4)-15$
According to condition,


Now, $f\left(1\right)<0;f\left(2\right)<0$
$f\left(1\right)<0\Rightarrow 1-2p(-3)-15<0\Rightarrow 6p<14\Rightarrow p<\frac{7}{3}\cdots \left(1\right)$

$f\left(2\right)<0\Rightarrow 4-2p(2-4)-15<0\Rightarrow 4p-11<0$
$\Rightarrow p<\frac{11}{4}\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$p\in (-\infty ,\frac{7}{3})$