Question

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

Answer 1

$\bf{\text{let},y=\frac{1-x+x^2}{1+x+x^2}}$

y = (1 - x + x²)/(1 + x + x²)
y + yx + yx² = 1 - x + x²
(y - 1)x² + (y + 1)x + (y - 1) = 0
discriminant , D = b² - 4ac ≥ 0
(y + 1)² - 4(y - 1)² ≥ 0
(y + 1 - 2y + 2)(y + 1 + 2y - 2) ≥ 0
(3 - y)(3y - 1) ≥ 0
1/3 ≤ y ≤ 3
hence, y ∈ [1/3, 3]
so, minimum value is 1/3
option (D) is correct.

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.