ExamHabit
Practise free

JEE Main 2017

JEE Main 2017 Mathematics question on Sets Relations and Functions

The function f:R→[−12,12]f: R \to \left[-\dfrac{1}{2}, \dfrac{1}{2}\right] defined as f(x)=x1+x2f(x) = \dfrac{x}{1+x^2} is
  1. A.
    invertible
  2. B.
    injective but not surjective
  3. C.
    surjective but not injective
  4. D.
    neither injective nor surjective
Show answer and solution

Correct option: C

Solution

f(x) = x/(1+x²) is not injective because f(x) = f(1/x) for x ≠ 0 (e.g., f(2) = f(1/2) = 2/5), so different inputs give the same output. For surjectivity, the range equals [-1/2, 1/2]: as x→±∞, f→0; the maximum value 1/2 is achieved at x=1 and minimum -1/2 at x=-1, and by continuity every value in [-1/2, 1/2] is attained. Since the codomain is exactly [-1/2, 1/2], f is surjective but not injective.

[JEE Main 2017]

Practise Sets Relations and Functions the way it is asked

A timed set of real JEE Main Sets Relations and Functions questions, graded as the exam grades. No sign-up.

Start a Sets Relations and Functions test

More: all JEE Main Sets Relations and Functions questions · JEE Main 2017 · Mathematics chapter list