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Consider a function f:[−1,1]→Yf:[-1,1] \to Y defined as f(x)=xx+2f(x)=\dfrac{x}{x+2} and Y=range(f(x))Y=range(f(x)), then which of following statements are correct?

(A) Function f(x)f(x) is one-one

(B) Function f(x)f(x) is not one-one

(C) Function f(x)f(x) is onto

(D) Function f(x)f(x) is invertible

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

f′(x)=2(x+2)2>0f'(x) = \dfrac{2}{(x+2)^2} > 0 on [−1,1][-1,1], so ff is strictly increasing and hence one-one.

Since the codomain YY is defined to be the range, ff is onto by construction.

Being both one-one and onto, ff is invertible. So (A), (C) and (D) hold.

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