Quantitative Aptitude – Quadratic Equation – Let f(x) = x^2+ax+b and g(x)


Slot – 1 – Quantitative Aptitude – Quadratic Equation – Let f(x) = x^2+ax+b and g(x)


Q. 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? ?
  1. 4
  2. 1
  3. 0
  4. 16
Answer: 4

Solutions:
Given, g(x) = f(x+1) – f(x-1)
g(x) = (x+1)^2 + a(x+1) +b – (x-1)^2 – a(x-1) – b
g(x) = 4x +2a
Also ,
g(20) = 72
4*20 + 2a = 72
Or a = -4
Now as f(x) >=0
It means (4ac-b^2)/4a≥0
(4b-a^2)/4≥0
Or (4b-16)/4≥0 ot b≥4 So least value of b =4


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