The function of `f: R to R ` defined by
`f(x)=2^(x)+x^(|x|)`, is

Options

A. one-one and onto
B. many-one and onto
C. one-one and into
D. many-one and into

Explanation

To analyze the function \( f: \mathbb{R} \to \mathbb{R} \) defined by

\[ f(x) = 2^x + x^{|x|} \]

we will determine its properties, specifically whether it is one-one (injective) and onto (surjective).

### Step 1: Split the function based on the value of \( x \)

The function involves the absolute value of \( x \), so we need to consider two cases: when \( x < 0 \) and when \( x \geq 0 \).

- For \( x < 0 \):
\[ f(x) = 2^x + x^{-x} = 2^x + \frac{1}{|x|^x} \]

- For \( x \geq 0 \):
\[ f(x) = 2^x + x^x \]

### Step 2: Determine the range of the function

#### Case 1: \( x < 0 \)

For \( x < 0 \):
- \( 2^x \) approaches \( 0 \) as \( x \) decreases (i.e., becomes more negative).
- \( x^{-x} \) (or \( \frac{1}{|x|^x} \)) is positive and increases as \( x \) approaches \( 0 \) from the left.

Thus, as \( x \) approaches \( 0 \) from the left, \( f(x) \) approaches \( 1 \) (since \( 2^0 = 1 \) and \( x^{-x} \) approaches \( 1 \)). As \( x \) decreases further, \( f(x) \) will be positive but will not reach \( 0 \).

#### Case 2: \( x \geq 0 \)

For \( x \geq 0 \):
- Both \( 2^x \) and \( x^x \) are non-negative and increase as \( x \) increases.
- At \( x = 0 \), \( f(0) = 1 + 1 = 2 \).
- As \( x \) approaches infinity, \( f(x) \) also approaches infinity.

### Step 3: Determine if the function is onto

The function \( f(x) \) for \( x < 0 \) gives positive values but does not include \( 0 \). For \( x \geq 0 \), the function starts from \( 2 \) and goes to infinity. Therefore, the range of \( f(x) \) is \( (0, \infty) \), which does not cover all real numbers \( \mathbb{R} \).

Since the range of \( f(x) \) does not equal the codomain \( \mathbb{R} \), the function is not onto.

### Step 4: Determine if the function is one-one

To check if the function is one-one, we can find the derivative \( f'(x) \) and check its sign.

For \( x < 0 \):
\[ f'(x) = \frac{d}{dx}(2^x) + \frac{d}{dx}(x^{-x}) = 2^x \ln(2) - x^{-x} \ln(-x) \]
For \( x < 0 \), both terms are positive, indicating that \( f'(x) > 0 \). Thus, \( f(x) \) is strictly increasing for \( x < 0 \).

For \( x \geq 0 \):
\[ f'(x) = \frac{d}{dx}(2^x) + \frac{d}{dx}(x^x) = 2^x \ln(2) + x^x (\ln(x) + 1) \]
Both terms are positive for \( x \geq 0\) (since \( 2^x > 0 \) and \( x^x > 0 \) for \( x \geq 0 \)). Thus, \( f'(x) > 0 \) for \( x \geq 0 \) as well.

Since \( f'(x) > 0 \) for all \( x \in \mathbb{R} \), the function is strictly increasing, which implies it is one-one.

### Conclusion

The function \( f(x) = 2^x + x^{|x|} \) is one-one but not onto.