Question

For all real values of x, the minimum value of `(1 - x + x^2)/(1+x+x^2)` is ______.

विकल्प

• 0

• 1

• 3

• `1/3`

Answer 1

For all real values of x, the minimum value of `(1 - x + x^2)/(1+x+x^2)` is `underline(1/3)`.

Explanation:

Let, y = `(1 - x + x^2)/(1 + x = x^2)`

Differentiating both sides with respect to x,

`dy/dx = ((-1 + 2x)(1 + x + x^2) - (1 - x + x^2)(1 + 2x))/((1 + x + x^2)^2)`

`= ((- 1 - x + x^2) + (2x + 2x^2 + 2x^3) - [(1 - x + x^2 + 2x - 2x^2 = 2x^3)])/((1 + x + x^2)^2)`

`= (-1 - x - x^2 + 2x + 2x^2 + 2x^3 - 1 + x - x^2 - 2x + 2x^2 - 2x^3)/((1 + x + x^2)^2)`

`= (-2 + 2x^2)/((1 + x + x^2)^2)`

`= (2 (x^2 - 1))/((1 + x + x^2)^2)`

`= (2(x - 1)(x + 1))/((1 + x + x^2)^2)`

For highest and lowest value, `dy/dx = 0`

`therefore (2(x - 1)(x + 1))/((1 + x + x^2)^2) = 0/1` 

`=>` (x - 1)(x + 1) = 0

`therefore` x = 1, -1

The sign of `dy/dx` at x = 1 changes from negative to positive when the point moves through x = 1.

`therefore` y is minimum at the point x = 1.

Minimum value, f(1) `= (1 - 1 + 1)/(1 + 1 + 1) = 1/3`