Question

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

• A. 1
• B. 16
• C. 0
• D. 4

Answer 1

The Correct Option is D

The correct answer is (D): $4$

$f(x)=x^2+ax+b$

$g(x)=f(x+1)-f(x-1)$

=${(x+1)^2+a(x+1)+b}-{(x-1)^2+a(x-1)+b}$

$g(x)=4x+2a$

$g(20)=72$

$80+2a=72 ⇒ a=-4$

$∴ f(x)=x^2-4x+b$

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

When $b≥4 f(x)≥0$ for all $x$

$∴$ The minimum value of b is $4$