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CAT 2020 Quant Question [Slot 2] with Solution 22

Question

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 $g(20) = 72$, then the smallest possible value of b is

1. 1
2. 16
3. 0
4. 4
Option: 4
Solution:

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

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

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

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

$g(20) = 72$

$80 + 2a = 72 \Rightarrow a = - 4$

$\therefore f(x) = {x^2} - 4x + b$

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

when $b \ge 4f(x) \ge 0$ for all x

$\therefore$ The minimum value of b is 4

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