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AP EAMCET2014MathematicsQuadratic Equation

If x is real, then the minimum value of y= x^2-x+1 x^2+x+1 is

Options

  1. A3
  2. B1 3
  3. C1 3
  4. D2

Correct answer

B. 1 3

Step-by-step solution

Let y= 1-x+x^2 1+x+x^2 On differentiating w.r.t. x , we get aligned d y d x & = (1+x+x^2 )(-1+2 x)- (1-x+x^2 )(1+2 x) (1+x+x^2 )^2 & = -1+2 x-x+2 x^2-x^2+2 x^3-1-2 x+x (1+x+x^2 )^2 & = -2+2 x^2-x^2-2 x^3 (1+x+x^2 )^2 & = 2 x^2-2 (1+x+x^2 )^2 aligned = 2 (x^2-1 ) (1+x+x^2 )^2 Put d y d x =0 x^2=1 x= 1 Now, d^2 y d x^2 = . (1+x+x^2 )(1+2 x) ] (1+x+x^2 )^4 aligned & = 4 (1+x+x^2 ) [ (1+x+x^2 ) x- (x^2-1 )(1+2 x) ] (1+x+x^2 )^4 & = 4 [x+x^2+x^3-x^2-2 x^3+1+2 x ] (1+x+x^2 )^3 & = 4 (1+3 x-x^3 ) (1+x+x^2 )^3 aligned At x

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