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MHT CET202121 Sep 2021Evening ShiftMathematicsApplication of DerivativesActual

For all real x , the minimum value of the function f(x)= 1-x+x^2 1+x+x^2 is

Options

  1. A1 3
  2. B0
  3. C3
  4. D1

Correct answer

A. 1 3

Step-by-step solution

We have f(x)= 1-x+x^2 1+x+x^2 f^ (x)= (1+x+x^2 )(2 x-1)- (1-x+x^2 )(2 x+1) (1+x+x^2 )^2 = (2 x+2 x^2+2 x^3-x-1-x^2 )- (2 x-2 x^2+2 x^3+1-x+x^2 ) (1+x+x^2 )^2 = (x+x^2+2 x^3-1 )- (x-x^2+2 x^3+1 ) (1+x+x^2 )^2 = 2 x^2-2 (1+x+x^2 )^2 and when f^ (x)=0 , we get 2 ( x ^2-1 )=0 x = 1 When x =1, f ( x )= 1 3 and when x =-1, f ( x )=3 Hence minimum value of f(x) is 1 3 .

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