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

CAT 2021 Online Course @ INR 8999 only

CAT 2020 Quant questions with Solutions