Let $f(x) = x^2 + ax + b$ and $g(x) = f(x+1) - f(x-1)$. If $f(x) \ge 0$ for all real $x$ and $g(10) = 44$, find the minimum value of $b$.
Answer and solution
Correct answer: A
Solution
$g(x) = f(x+1) - f(x-1) = 4x + 2a$.
$g(10) = 40 + 2a = 44 \Rightarrow a = 2$.
For $f(x) \ge 0$ for all real $x$, the discriminant must be $\le 0$:
$a^2 - 4b \le 0 \Rightarrow 4 - 4b \le 0 \Rightarrow b \ge 1$.
Minimum value of $b = 1$.
$g(10) = 40 + 2a = 44 \Rightarrow a = 2$.
For $f(x) \ge 0$ for all real $x$, the discriminant must be $\le 0$:
$a^2 - 4b \le 0 \Rightarrow 4 - 4b \le 0 \Rightarrow b \ge 1$.
Minimum value of $b = 1$.