Question 14.

f(x) = $x^2$ + ax + b

$f(x+1)=x^2+2x+1+ax+a+b$

$f(x−1)=x^2−2x+1+ax−a+b$

g(x)=f(x+1)−f(x−1)=4x+2a

Now g(20)=72 from this we get a=−4; f(x) = $x^2$ − 4x + b

For this expression to be greater than zero it has to be a perfect square which is possible for b≥ 4

Hence the smallest value of 'b' is 4.