JEE Main202226 Jun 2022Evening ShiftMathematicsFunctionsActual
Let f : ℝ → ℝ be defined as f x = x - 1 and g : R → 1 , - 1 → ℝ be defined as g x = x 2 x 2 - 1 . Then the function f o g is:
Options
- AOne-one but not onto
- Bonto but not one-one
- CBoth one-one and onto
- DNeither one-one nor onto
Correct answer
D. Neither one-one nor onto
Step-by-step solution
f o g x = f g x = f x 2 x 2 + 1 = x 2 x 2 + 1 - 1 = x 2 - x 2 - 1 x 2 + 1 = - 1 x 2 + 1 We know that, 0 ≤ x 2 < ∞ ,   ∀ x ∈ R ⇒ 1 ≤ x 2 + 1 < ∞ ,   ∀ x ∈ R ⇒ 1 ≥ 1 x 2 + 1 > 0 ,   ∀ x ∈ R ⇒ - 1 ≤ - 1 x 2 + 1 < 0 ,   ∀ x ∈ R So, range of f o g x is − 1 ,   0 ⊂ R . Hence, the function f o g x is into function and f o g - x = f g - x = - 1 - x 2 + 1 = - 1 x 2 + 1 =