Q53 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

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

For any quadratic expression $\mathrm{P}(\mathrm{x})$ $\begin{aligned} & \mathrm{P}(\mathrm{x})>0 \forall \mathrm{x} \Rightarrow \mathrm{D} < 0, \mathrm{a}>0 \\ & \mathrm{D}=(2 \mathrm{p})^2-4(8-2 \mathrm{p}) < 0 \\ & 4 \mathrm{p}^2-32+8 \mathrm{p} < 0 \\ & (\mathrm{p}+4)(\mathrm{p}-2) < 0 \\ & \therefore \mathrm{p} \in(-4,2)\end{aligned}$

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

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