Sometimes functions are defined like `f(x)=max{sinx,cosx}`, then `f(x)` is splitted like `f(x)={{:(cosx, x in (0,(pi)/(4)]),(sinx, x in ((pi)/(4),(pi)/(2)]):}` etc.
If `f(x)=max{x^(2),2^(x)}`,then if `x in (0,1)`, `f(x)=`

Options

A. `x^(2)`
B. `2^(x)`
C. both are equal
D. none of these

Answer

To solve the problem where \( f(x) = \max\{x^2, 2^x\} \) and we need to find \( f(x) \) for \( x \in (0, 1) \), we will follow these steps:

### Step 1: Understand the Functions
We are given two functions:
1. \( g(x) = x^2 \)
2. \( h(x) = 2^x \)

### Step 2: Find the Intersection Points
To determine where \( f(x) \) switches from one function to the other, we need to find the points where \( x^2 = 2^x \).

### Step 3: Solve the Equation
We can solve the equation \( x^2 = 2^x \) graphically or numerically.

1. At \( x = 0 \):
\[ 0^2 = 0 \quad \text{and} \quad 2^0 = 1 \quad \Rightarrow \quad 0 < 1 \]

2. At \( x = 1 \):
\[ 1^2 = 1 \quad \text{and} \quad 2^1 = 2 \quad \Rightarrow \quad 1 < 2 \]

3. At \( x = 2 \):
\[ 2^2 = 4 \quad \text{and} \quad 2^2 = 4 \quad \Rightarrow \quad 4 = 4 \]

From the above evaluations, we can see that \( x^2 < 2^x \) for \( x \in (0, 1) \) and they intersect at \( x = 2 \).

### Step 4: Determine the Maximum Function in the Interval
Since \( x^2 < 2^x \) for \( x \in (0, 1) \), we conclude that:
\[ f(x) = 2^x \quad \text{for} \quad x \in (0, 1) \]

### Step 5: Conclusion
Thus, for \( x \in (0, 1) \):
\[ f(x) = 2^x \]

### Final Answer
The final answer is:
\[ f(x) = 2^x \quad \text{for} \quad x \in (0, 1) \]