JEE Main202331 Jan 2023Evening ShiftMathematicsFunctionsActual
Let f : R - 2 , 6 → R be real valued function defined as f x = x + 2 x + 1 x 2 - 8 x + 12 . Then range of f is
Options
- A- ∞ , - 21 4 ∪ 21 4 , ∞
- B- ∞ , - 21 4 ∪ [ 0 , ∞ )
- C- ∞ , - 21 4 ∪ 0 , ∞
- D- ∞ , - 21 4 ∪ [ 1 , ∞ )
Correct answer
B. - ∞ , - 21 4 ∪ [ 0 , ∞ )
Step-by-step solution
Given, f : R - 2 , 6 → R be real valued function defined as f x = x + 2 x + 1 x 2 - 8 x + 12 Now let, y = x 2 + 2 x + 1 x 2 - 8 x + 12 = x + 1 2 x - 2 x - 6               … 1 Now differentiating the above function we get, ⇒ d y d x = - 2 x + 1 5 x - 16 x - 2 2 x - 6 2 Now by wavy curve method we get, So Graph of y = x + 1 2 x - 2 x - 6 for given domain will be, So, from graph range is y ∈ - ∞ , 21 4 ∪ [ 0 , ∞ )