For all real values of x, the minimum value of `(1-x+x^2)/(1+x+x^2)`is(A) 0 (B) 1 (C) 3 (D) `1/3`

Options

A. 0
B. 1
C. 3
D. `1/3`

Answer

D

Explanation

Let `y=(1-x+x^(2))/(1+x+x^(2))`
Differentiate w.r.t. x
` (dy)/(dx) = ((1+x+x^(2))(-1+2x)-(1-x+x^(2))(1+2x))/((1+x+x^(2))`
`-1+2x-x+2x^(2)-x^(2)+2x^(3)`
` = (-1-2x+x+2x^(2)-x^(2)-2x^(3))/((1+x+x^(2))^(2))`
` = (-2+2x^(2)-2x^(2)+2x^(2)0)/((1+x+x^(2))^(2))`
` = (2x^(2)-2)/((1+x+x^(2))^(2)) = (2(x^(2)-1))/((1+x+x^(2))^(2))`
Now`(dy)/(dx) = 0 rArr x^(2)=1`
` rArr x = pm 1 `
Now, `(d^(2)y)/(dx^(2))=(2[(1+x+x^(2))^(2)(2x)-(x^(2)-1)(2)(1+x+x^(2))(1+2x)])/((1+x+x^(2))^(4))`
`(4(1+x+x^(2))[(1+x+x^(2))x-(x^(2)-1)(1+2x)])/((1+x+x^(2))^(4))`
`=(4(x+x^(2)+x^(3)-x^(2)-2x^(3)+1+2x))/((1+x+x^(2))^(3))`
` = (4(1+3x-x^(3)))/((1+x+x^(2))^(3))`
at x = 1,
`((d^(2)y)/(dx^(2)))_(x = 1) = (4[1+3(1)-1^(3)])/((1+1+1^(2))^(3))= (4(3))/3^(3)=4/9 gt 0`
at x =- 1,
`((d^(2)y)/(dx^(2)))_(x=-1)=(4[1+3(-1)-(-1)^(3)])/([1+(-1)+(-1)^(2)]^(3))`
` = (4(1-3+1))/((1-1+1)^(3))=4(-1)=-4 lt 0`
`:. ` at x = 1, y will be minimum and minimum value
` y=(1-1+1)/(1+1+1)=1/3`.