KVPY2020MathematicsFunctions
Let R be the set of all real numbers and f : R → R be a continuous function. Suppose | f ( x ) - f ( y ) | ≥ | x - y | for all real numbers x and y . Then
Options
- Af is one-one, but need not be onto
- Bf is onto, but need not be one-one
- Cf need not be either one-one or onto
- Df is one-one and onto
Correct answer
D. f is one-one and onto
Step-by-step solution
Let f ( x ) = f ( y ) So, | f ( x ) - f ( y ) | ≥ | x - y | ⇒ 0 ≥ | x - y | ⇒ x - y = 0 ⇒ x = y ⇒ f is one-one Since, f is continuous So f ( 0 ) is finite Now, | f ( x ) - f ( 0 ) | ≥ | x - 0 | ⇒ lim x → ∞ | f ( x ) - f ( 0 ) | ≥ lim x → ∞ | x | ⇒ lim x → ∞ f ( x ) = ∞ ⇒ f is unbounded ⇒ f is surjective