CAT 2020 Slot 2 QA Question 2: Let $f(x) = x^2 + ax + b$ and $g(x) = f(x + 1) - f(x - 1)$. If $f(x) \geq 0$ for all real $x$, and $g(20) = 72$, then the smallest possible value of $b$ is

Question

Let $f(x) = x^2 + ax + b$ and $g(x) = f(x + 1) - f(x - 1)$. If $f(x) \geq 0$ for all real $x$, and $g(20) = 72$, then the smallest possible value of $b$ is

Accepted Answer

First simplify $g(x)$.

$f(x+1) = (x+1)^2 + a(x+1) + b$
$f(x-1) = (x-1)^2 + a(x-1) + b$
$g(x) = f(x+1) - f(x-1) = 4x + 2a$
$g(20) = 80 + 2a = 72$
$a = -4$

Thus,
$f(x) = x^2 - 4x + b$
$f(x) = (x-2)^2 + b - 4$

For $f(x)$ to be non-negative for every real $x$, its minimum value must be non-negative.

Let f(x)=x2+ax+bf(x) = x^2 + ax + bf(x)=x2+ax+b and g(x)=f(x+1)−f(x−1)g(x) = f(x + 1) - f(x - 1)g(x)=f(x+1)−f(x−1). If f(x)≥0f(x) \geq 0f(x)≥0 for all real xxx, and g(20)=72g(20) = 72g(20)=72, then the smallest possible value of bbb is

Let $f(x) = x^2 + ax + b$ and $g(x) = f(x + 1) - f(x - 1)$ . If $f(x) \geq 0$ for all real $x$ , and $g(20) = 72$ , then the smallest possible value of $b$ is