Examine whether the following functions are one-one, many-one or into, onto `f(x)=x^(2)+lnx`

Explanation

To determine whether the function \( f(x) = x^2 + \ln x \) is one-one, many-one, into, or onto, we will follow these steps:

### Step 1: Determine the Domain
The function \( f(x) = x^2 + \ln x \) consists of two parts: \( x^2 \) and \( \ln x \).

- The term \( x^2 \) is defined for all real numbers.
- The term \( \ln x \) is defined only for \( x > 0 \).

Thus, the domain of \( f(x) \) is:
\[ \text{Domain} = (0, \infty) \]

### Step 2: Determine the Co-domain
The co-domain of the function is the set of all possible output values. Since \( x^2 \) can take any non-negative value and \( \ln x \) can take any real value as \( x \) approaches \( 0 \) from the right, the co-domain is:
\[ \text{Co-domain} = \mathbb{R} \]

### Step 3: Check if the Function is One-One or Many-One
To determine if the function is one-one or many-one, we will find the derivative of \( f(x) \) and analyze it.

1. **Differentiate the function**:
\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(\ln x) = 2x + \frac{1}{x} \]

2. **Set the derivative to zero** to find critical points:
\[ 2x + \frac{1}{x} = 0 \]
Rearranging gives:
\[ 2x^2 + 1 = 0 \]
This equation has no real solutions since \( 2x^2 \) is always non-negative and cannot equal \(-1\).

3. **Analyze the sign of the derivative**:
- For \( x > 0 \), both \( 2x \) and \( \frac{1}{x} \) are positive, hence \( f'(x) > 0 \).
- Since \( f'(x) > 0 \) for all \( x > 0 \), the function is strictly increasing.

Since the function is strictly increasing and has no critical points, it is **one-one**.

### Step 4: Check if the Function is Into or Onto
- **Into**: A function is into if its range is not equal to its co-domain.
- **Onto**: A function is onto if its range is equal to its co-domain.

1. **Find the range of \( f(x) \)**:
- As \( x \) approaches \( 0^+ \), \( f(x) \) approaches \( -\infty \) (since \( \ln x \to -\infty \)).
- As \( x \) approaches \( \infty \), \( f(x) \) approaches \( \infty \) (since both \( x^2 \) and \( \ln x \) approach \( \infty \)).

Thus, the range of \( f(x) \) is:
\[ \text{Range} = (-\infty, \infty) \]

Since the range equals the co-domain, the function is **onto**.

### Final Conclusion
The function \( f(x) = x^2 + \ln x \) is:
- **One-One**
- **Onto**