Highly selective Backlog Qs for JEE MainMathematicsQuadratic Equation
Find the maximum and minimum values of x 2 - x + 1 x 2 + x + 1   for real values of x .
Options
- A1 3 , 3
- B1 3 , 1
- C1 3 , 1
- D1 3 , 3
Correct answer
D. 1 3 , 3
Step-by-step solution
Let x 2 - x + 1 x 2 + x + 1 = y ⇒ x 2 + x + 1 y= x 2 - x + 1 ⇒ y - 1 x 2 + y + 1 x + y - 1 = 0. On Solving for x, we get x = - y + 1 ± y + 1 2 - 4 y - 1 2 2 y - 1 = - y + 1 2 y - 1 ± 1 2 y - 1 - 3 y - 3 y - 1 / 3 Now for real values of x we will get real values of y, hence x to be real - 3 y - 3 y - 1 / 3 ≥ 0 ⇒ y ∈ 1 3 , 3 .