Consider a function \( f:[-1,1]\to Y \) defined as \( f(x)=\displaystyle\frac{x}{x+2} \) and \( Y=range(f(x)) \) , then which of following statements are correct?
(A) Function \( f(x) \) is one-one
(B) Function \( f(x) \) is not one-one
(C) Function \( f(x) \) is onto
(D) Function \( f(x) \) is invertible
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
On [−1, 1] the denominator x + 2 is never zero. For one-one, suppose \( f(x_1)=f(x_2) \):
\( \displaystyle \frac{x_1}{x_1+2}=\frac{x_2}{x_2+2} \)
\( x_1x_2+2x_1=x_1x_2+2x_2 \)
\( x_1=x_2 \)
The co-domain Y is the range of f, so every element of Y is an image.
So f is one-one, and it is onto because Y is its range.
Checking the statements
(A) Function \( f(x) \) is one-one — Correct. \( f(x_1)=f(x_2) \) gives \( x_1=x_2 \).
(B) Function \( f(x) \) is not one-one — Incorrect. f is one-one, since \( f(x_1)=f(x_2) \) gives \( x_1=x_2 \).
(C) Function \( f(x) \) is onto — Correct. The co-domain Y is the range of f.
(D) Function \( f(x) \) is invertible — Correct. f is one-one and onto, so it is a bijection and has an inverse.