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. $(0,7)$
• B. $(-\infty ,\frac{7}{3})$
• C. $(\frac{7}{3},\infty )$
• D. $R$

Answer 1

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

Let $f\left(x\right)=x^2-2p(x-4)-15\text{and}\alpha ,\beta$ be roots of $f\left(x\right)=0$.
Now, according to condition, we can plot the graph as:

$f\left(x\right)=x^2-2px+(8p-15)$
Now, the conditions to be satisfied are:
$\left(A\right)D>0$
$\Rightarrow 4p^2-4(8p-15)>0\Rightarrow p^2-8p+15>0$
Now, Discriminant of $p^2-8p+15,$
$D=64-120<0$
$\Rightarrow p^2-8p+15>0\forall p\in R$

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

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

Thus, taking intersection of intervals in $\left(A\right)\&\left(B\right),$ we get the values of $p$ as:
$p\in (-\infty ,\frac{7}{3})$