Let f be a continuous function satisfying `f '(l n x)=[1` for `0< x<= 1, x` for `x > 1` and `f (0) = 0` then `f(x)` can be defined as

Options

A. `f(x)={{:(1,if,xle1),(1-e^(x),if,xgt1):}`
B. `f(x)={{:(1,if,xle1),(e^(x)-1,if,xgt1):}`
C. `f(x)={{:(1,if,xlt1),(e^(x),if,xgt1):}`
D. `f(x)={{:(x,if,xle1),(e^(x)-1,if,xgt1):}`

Explanation

To solve the problem, we need to find the function \( f(x) \) that satisfies the given conditions. We are given that:

1. \( f'( \ln x ) = 1 \) for \( 0 < x \leq 1 \)
2. \( f'(x) = x \) for \( x > 1 \)
3. \( f(0) = 0 \)

Let's break down the solution step by step.

### Step 1: Find \( f(x) \) for \( 0 < x \leq 1 \)

Since \( f'(\ln x) = 1 \) for \( 0 < x \leq 1 \), we can integrate this to find \( f(\ln x) \).

\[ f(\ln x) = \int 1 \, dx = x + C \]

Where \( C \) is a constant of integration. However, we need to express \( f(x) \) in terms of \( x \). We can substitute \( y = \ln x \), which gives \( x = e^y \). Thus, we have:

\[ f(y) = e^y + C \]

### Step 2: Find the constant \( C \)

We know that \( f(0) = 0 \). To find \( C \), we need to evaluate \( f(0) \). Since \( \ln(1) = 0 \), we can use \( x = 1 \):

\[ f(0) = f(\ln(1)) = f(0) = e^0 + C = 1 + C \]

Setting this equal to zero gives:

\[ 1 + C = 0 \implies C = -1 \]

Thus, for \( 0 < x \leq 1 \):

\[ f(\ln x) = e^{\ln x} - 1 = x - 1 \]

So we can write:

\[ f(x) = x - 1 \quad \text{for } 0 < x \leq 1 \]

### Step 3: Find \( f(x) \) for \( x > 1 \)

For \( x > 1 \), we have \( f'(x) = x \). Integrating this gives:

\[ f(x) = \int x \, dx = \frac{x^2}{2} + D \]

Where \( D \) is another constant of integration.

### Step 4: Determine the constant \( D \)

To find \( D \), we need to ensure continuity at \( x = 1 \). We already found \( f(1) \) from the previous part:

\[ f(1) = 1 - 1 = 0 \]

Now we need to evaluate \( f(1) \) from the second part:

\[ f(1) = \frac{1^2}{2} + D = \frac{1}{2} + D \]

Setting these equal gives:

\[ 0 = \frac{1}{2} + D \implies D = -\frac{1}{2} \]

Thus, for \( x > 1 \):

\[ f(x) = \frac{x^2}{2} - \frac{1}{2} \]

### Final Definition of \( f(x) \)

Combining both parts, we have:

\[ f(x) = \begin{cases} x - 1 & \text{for } 0 < x \leq 1 \\ \frac{x^2}{2} - \frac{1}{2} & \text{for } x > 1 \end{cases} \]

### Summary

The function \( f(x) \) can be defined as:

\[ f(x) = \begin{cases} x - 1 & \text{for } 0 < x \leq 1 \\ \frac{x^2}{2} - \frac{1}{2} & \text{for } x > 1 \end{cases} \]