Question

Let f: [0,∞) → \(\R\) be a function defined by \(f(x)=\frac{x+1}{x+2}\) for all \(x\isin\R\). Then f is

• A. one-one and onto.
• B. one-one but not onto.
• C. onto but not one-one.
• D. neither one-one nor onto.

Answer 1

The Correct Option is B

To determine the nature of the function \( f(x) = \frac{x+1}{x+2} \), we need to investigate if it is one-one (injective) and/or onto (surjective) over its domain, which is \( [0, \infty) \).

Step 1: Test for One-One (Injectivity)

A function is one-one if \( f(a) = f(b) \) implies \( a = b \) for all \( a, b \in [0, \infty) \).

Assume \( f(a) = f(b) \):

\[\frac{a+1}{a+2} = \frac{b+1}{b+2}\]

This implies:

\[(a+1)(b+2) = (b+1)(a+2)\]

Expanding both sides, we get:

\[ab + 2a + b + 2 = ab + 2b + a + 2\]

Simplifying, we find:

\[2a + b = 2b + a \implies a = b\]

Thus, the function \( f(x) \) is one-one.

Step 2: Test for Onto (Surjectivity)

A function is onto if for every \( y \in \R \), there exists an \( x \in [0, \infty) \) such that \( f(x) = y \).

Rewriting \( y = f(x) \):

\[y = \frac{x+1}{x+2} \implies y(x+2) = x+1\]

This implies:

\[yx + 2y = x + 1\]

Rearranging gives:

\[x(y - 1) = 1 - 2y\]

For \( x \) to be defined, \( y - 1 \neq 0 \), which means \( y \neq 1 \). Therefore, \( x \) exists only if \( y < 1 \). This shows that not every \( y \in \R \) has a pre-image in \([0, \infty)\).

Thus, the function is not onto because it cannot map to values \( y \geq 1 \).

Conclusion

Since the function is one-one but not onto, the correct answer is:

one-one but not onto.