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प्रश्न
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`
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उत्तर
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`
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