Question

If $p\left(x\right)=-x^2+\mathrm{px}-p-8\mathrm{and}p\left(x\right)<0$ for all real values of x, then the value of p cannot be

• A. 7
• B. -3
• C. 0
• D. 10

Answer 1

The correct option is D 10

$p\left(x\right)=-x^2+\mathrm{px}-p-8\mathrm{and}p\left(x\right)<0,$ for all real values of x.

⇒p(x) lies below x-axis

⇒ No real roots for p(x) exist

⇒ Discriminant, D < 0

$\Rightarrow p^2-4(-1)(-p-8)<0$

$\Rightarrow p^2-4p-32<0$

$\Rightarrow (p-8)(p+4)<0$

Using wavy curve method:

$\Rightarrow -4<p<8$

$\Rightarrow p\ne 10$

Hence, the correct answer is option (d).