CAT 2020 Slot 2 QA Question & Solution

AlgebraMedium

Let $f(x)=x^{2}+ax+b$ and $g(x)=f(x+1)-f(x-1)$. If $f(x)\geq0$ for all real x, and $g(20)=72$. then the smallest possible value of b is

$f\left(x\right)=\ x^2+ax+b$

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

$f\left(x-1\right)=x^2-2x+1+ax-a+b$

$g(x)=f(x+1)-f(x-1)= 4x+2a$

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

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

Hence the smallest value of 'b' is 4.