Which of the following statements is/are true ?

Options

A. `f : R rarr R, f(x) = log [x+ sqrt(1+x^(2))]` is an odd function
B. `f : R rarr R`, f(0) = 3, then f(x) must not be an odd function
C. `f : R rarr R, f(x) = x((e^(x) - e^(-x))/(e^(x) + e^(-x)))` is an onto function
D. Graph of f(x) = x sin x is bounded between lines y = x = -x

Answer

To determine which of the given statements are true, we will analyze each statement step by step.

### Step 1: Analyze the first statement
**Statement 1:** \( f(x) = \log(x) + \sqrt{1 + x^2} \) is an odd function.

1. **Definition of an odd function:** A function \( f(x) \) is odd if \( f(-x) = -f(x) \) for all \( x \).
2. **Calculate \( f(-x) \):**
\[ f(-x) = \log(-x) + \sqrt{1 + (-x)^2} = \log(-x) + \sqrt{1 + x^2} \]
Since \( \log(-x) \) is not defined for \( x > 0 \), we cannot directly evaluate \( f(-x) \) for positive values of \( x \). However, we can still analyze the function for negative values.
3. **Check if \( f(-x) = -f(x) \):**
We can see that:
\[ f(-x) = \log(-x) + \sqrt{1 + x^2} \]
and
\[ -f(x) = -(\log(x) + \sqrt{1 + x^2}) = -\log(x) - \sqrt{1 + x^2} \]
Since \( f(-x) \) does not equal \(-f(x)\), we conclude that \( f(x) \) is not an odd function.

**Conclusion for Statement 1:** **False**

### Step 2: Analyze the second statement
**Statement 2:** If \( f(0) = 3 \), then \( f(x) \) cannot be an odd function.

1. **Using the definition of an odd function:** If \( f(x) \) is odd, then \( f(0) \) must equal \( -f(0) \).
2. **Substituting \( f(0) = 3 \):**
\[ f(0) = -f(0) \implies 3 = -3 \]
This is a contradiction. Therefore, if \( f(0) \) is not zero, \( f(x) \) cannot be an odd function.

**Conclusion for Statement 2:** **True**

### Step 3: Analyze the third statement
**Statement 3:** \( f(x) = \frac{x(e^x - e^{-x})}{e^x + e^{-x}} \) is an onto function.

1. **Definition of onto function:** A function is onto if every element in the codomain has a pre-image in the domain.
2. **Check the limits of \( f(x) \):**
- As \( x \to \infty \):
\[ f(x) \to \frac{x(e^x)}{e^x} = x \to \infty \]
- As \( x \to -\infty \):
\[ f(x) \to \frac{x(-e^{-x})}{e^{-x}} = -x \to -\infty \]
3. **Behavior of the function:** Since the function approaches both \( \infty \) and \( -\infty \), it does cover all real values.

**Conclusion for Statement 3:** **True**

### Step 4: Analyze the fourth statement
**Statement 4:** The graph of \( f(x) = x \sin(x) \) is bounded between the lines \( y = x \) and \( y = -x \).

1. **Check the behavior of \( f(x) = x \sin(x) \):**
- The sine function oscillates between -1 and 1.
- Therefore, \( f(x) \) oscillates between \( -x \) and \( x \).
2. **Graphical representation:** The graph of \( f(x) \) will always lie between the lines \( y = x \) and \( y = -x \).

**Conclusion for Statement 4:** **True**

### Final Conclusions:
- **Statement 1:** False
- **Statement 2:** True
- **Statement 3:** True
- **Statement 4:** True

### Summary of True Statements:
- Statements 2, 3, and 4 are true.