Statement-1 : `f : R rarr R, f(x) = x^(2) log(|x| + 1)` is an into function.
and
Statement-2 : `f(x) = x^(2) log(|x|+1)` is a continuous even function.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.
C. Statement -1 is False, Statement -2 is False
D. Statement -1 is False, Statement -2 is True

Answer

To determine the validity of the statements regarding the function \( f(x) = x^2 \log(|x| + 1) \), we will analyze both statements step by step.

### Step 1: Analyze Statement 2
**Statement 2:** \( f(x) = x^2 \log(|x| + 1) \) is a continuous even function.

1. **Check Continuity:**
- The function \( x^2 \) is a polynomial function, which is continuous everywhere.
- The function \( \log(|x| + 1) \) is also continuous for all \( x \) because the argument \( |x| + 1 \) is always positive.
- Since the product of two continuous functions is continuous, \( f(x) \) is continuous.

2. **Check if the function is even:**
- A function is even if \( f(-x) = f(x) \) for all \( x \).
- Calculate \( f(-x) \):
\[ f(-x) = (-x)^2 \log(|-x| + 1) = x^2 \log(|x| + 1) = f(x) \]
- Since \( f(-x) = f(x) \), the function is even.

**Conclusion for Statement 2:** The function \( f(x) = x^2 \log(|x| + 1) \) is indeed a continuous even function.

### Step 2: Analyze Statement 1
**Statement 1:** \( f(x) = x^2 \log(|x| + 1) \) is an into function.

1. **Definition of an into function:**
- A function is called an "into" function if its range is not equal to its co-domain. In this case, the co-domain is \( \mathbb{R} \).

2. **Determine the range of \( f(x) \):**
- The term \( x^2 \) is always non-negative, i.e., \( x^2 \geq 0 \).
- The term \( \log(|x| + 1) \) is also non-negative since \( |x| + 1 \geq 1 \) implies \( \log(|x| + 1) \geq 0 \).
- Therefore, \( f(x) \geq 0 \) for all \( x \).
- The minimum value of \( f(x) \) occurs at \( x = 0 \):
\[ f(0) = 0^2 \log(0 + 1) = 0 \]
- As \( |x| \) increases, \( f(x) \) also increases without bound because both \( x^2 \) and \( \log(|x| + 1) \) increase.
- Thus, the range of \( f(x) \) is \( [0, \infty) \).

3. **Compare range and co-domain:**
- The co-domain is \( \mathbb{R} \) (which includes negative values).
- The range \( [0, \infty) \) does not cover negative values.
- Therefore, the range is not equal to the co-domain.

**Conclusion for Statement 1:** The function \( f(x) = x^2 \log(|x| + 1) \) is indeed an into function.

### Final Conclusion
Both statements are correct:
- **Statement 1** is true: \( f(x) \) is an into function.
- **Statement 2** is true: \( f(x) \) is a continuous even function.