Question

Let $A$ denote the number of integers satisfying the inequality $3\sqrt{6+x-x^2}>4x-2$.$B$ denote the value of $a$ for which the minimum value of the function $P\left(x\right)=x^2-4ax-a^4$ assume the greatest value.

Answer 1

$3\sqrt{6+x-x^2}>4x-2$

$36+9x-9x^2>16x^2-16x+4$

$25x^2-25x-32<0⟹x\in (-0.74,1.74)$

$A=2$

$P\left(x\right)=x^2-4ax-a^4$

Minimum value of $P\left(x\right)$ is $\frac{-\left(\right(4a{)}^2-4\left(1\right)(-a^4))}{4\left(1\right)}=-a^4-4a^2$

For $a=0,P\left(x\right)$ assume the greatest value

$B=0$