Question

Let f: R→R be a function such that \(f(x)=\frac{x^2+2x+1}{x^2+1}\). Then

• A. f(x) is many-one in (–∞,–1)
• B. f(x) is one-one in [1, ∞) but not in (–∞, ∞)
• C. f(x) is many-one in (1, ∞)
• D. f(x) is one-one in (–∞, ∞)

Answer 1

The Correct Option is B

Solution: Step 1: Rewriting the function.

The given function is: 

\( f(x) = \frac{x^2 + 2x + 1}{x^2 + 1} \)

We can rewrite it as:

\( f(x) = \frac{(x + 1)^2}{x^2 + 1} = 1 + \frac{2x}{x^2 + 1} \)

Step 2: Analyzing the function.

The function has the form:

\( f(x) = 1 + \frac{2x}{x^2 + 1} \)

By differentiating \( f(x) \), we get:

\( f'(x) = \frac{(x^2 + 1) \cdot 2 - 2x \cdot 2x}{(x^2 + 1)^2} = \frac{2(x^2 + 1) - 4x^2}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2} \)

We analyze the sign of \( f'(x) \):

• For \( x \in [1, \infty) \), \( f'(x) > 0 \), indicating that the function is increasing and thus one-one in this interval.
• For \( x \in (-\infty, 1) \), \( f'(x) < 0 \), indicating that the function is decreasing and thus not one-one in this interval.

Step 3: Conclusion.

Thus, \( f(x) \) is one-one in \( [1, \infty) \) but not in \( (-\infty, \infty) \).

Step 4: Conclusion from the graph.

The graph of the function confirms that \( f(x) \) is one-one in \( [1, \infty) \) and not in \( (-\infty, \infty) \).