CAT 2020 Slot 2 Question 53

Question 53

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

Solution

$$f\left(x\right)=\ x^2+ax+b$$ 

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

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

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

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.


View Video Solution


Create a FREE account and get:

  • All Quant CAT Formulas and shortcuts PDF
  • 20+ CAT previous papers with solutions PDF
  • Top 500 CAT Solved Questions for Free

Comments

Register with

OR

Boost your Prep!

Download App