The function \( f:R\to[0,1) \) defined by \( f(x)=\displaystyle\frac{x^2}{x^2+1} \) is :(Where \( R \) is set of real numbers)
(A) One-one
(B) one-one and onto
(C) onto
(D) Not one-one
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) and (D) only
One-one: f depends only on \( x^2 \), so
\( \displaystyle f(1)=\frac{1}{2}=f(-1) \)
Two different inputs give the same output, so f is not one-one.
Onto [0, 1): for any y in [0, 1), solve \( y=\frac{x^2}{x^2+1} \):
\( \displaystyle y(x^2+1)=x^2 \)
\( \displaystyle x^2(1-y)=y \)
\( \displaystyle x^2=\frac{y}{1-y} \)
For \( 0\le y\lt 1 \), \( \frac{y}{1-y}\ge 0 \), so \( x=\sqrt{\frac{y}{1-y}} \) is a real number with f(x) = y. Every element of the codomain is an image, so f is onto.
So f is onto [0, 1) but not one-one.
Checking the statements
(A) One-one — Incorrect. f(1) = f(−1) = 1/2.
(B) one-one and onto — Incorrect. f is onto but not one-one, since f(1) = f(−1).
(C) onto — Correct. Every y in [0, 1) is the image of \( x=\sqrt{\frac{y}{1-y}} \).
(D) Not one-one — Correct. f(1) = f(−1) = 1/2.