If $x^2+2 p x-2 p+8>0$ for all real values of $x$, then the set of all possible values of $p$ is

• A

  $(2,4)$
• B

  $(-\infty,-4)$
• C

  $(2, \infty)$
• Correct answer
  D

  $(-4,2)$

Solution

Given, $x^2+2 p x-2 p+8>0$, for all real values of $x$. $$ \begin{aligned} \therefore \quad & D=b^2-4 a c < 0 \\ & \quad[\because \text { if a }>0 \text { and } D < 0, \text { then } f(x)>0 \text { for all } x \in R] \\ \Rightarrow & (2 p)^2+4(2 p-8) < 0 \Rightarrow 4 p^2+8 p-32 < 0 \end{aligned} $$ $$ \begin{aligned} & \Rightarrow \quad p^2+2 p-8 < 0 \Rightarrow p^2+4 p-2 p-8 < 0 \\ & \Rightarrow \quad(p+4)(p-4 < 0 \end{aligned} $$
$$ \Rightarrow \quad-4 < p < 2 \Rightarrow p \in(-4,2) $$

Hence, the correct answer is option: 'D'.