If `f:R->R` is a twice differentiable function such that `f''(x) > 0` for all `x in R,` and `f(1/2)=1/2` and `f(1)=1,` then

Options

A. (a) `0ltf'(1)le1/2`
B. (b) `f'(1)le0`
C. (c)`f'(1) > 1`
D. (d) `1/2ltf'(1)le1`

Explanation

To solve the problem, we need to analyze the given conditions about the function \( f \) and its derivatives. Let's break down the solution step by step.

### Step 1: Understand the Given Conditions
We know that:
- \( f: \mathbb{R} \to \mathbb{R} \) is a twice differentiable function.
- \( f''(x) > 0 \) for all \( x \in \mathbb{R} \).
- \( f\left(\frac{1}{2}\right) = \frac{1}{2} \) and \( f(1) = 1 \).

### Step 2: Apply the Mean Value Theorem
According to the Mean Value Theorem (MVT), since \( f \) is continuous on the closed interval \(\left[\frac{1}{2}, 1\right]\) and differentiable on the open interval \(\left(\frac{1}{2}, 1\right)\), there exists some \( c \in \left(\frac{1}{2}, 1\right) \) such that:

\[ f(1) - f\left(\frac{1}{2}\right) = (1 - \frac{1}{2}) f'(c) \]

Substituting the known values:

\[ 1 - \frac{1}{2} = \frac{1}{2} f'(c) \]

This simplifies to:

\[ \frac{1}{2} = \frac{1}{2} f'(c) \]

### Step 3: Solve for \( f'(c) \)
From the equation above, we can deduce:

\[ f'(c) = 1 \]

### Step 4: Analyze the Implications of \( f''(x) > 0 \)
Since \( f''(x) > 0 \) for all \( x \in \mathbb{R} \), this indicates that \( f'(x) \) is strictly increasing.

### Step 5: Relate \( f'(c) \) and \( f'(1) \)
Since \( c < 1 \) and \( f'(x) \) is strictly increasing, we have:

\[ f'(c) < f'(1) \]

Given that \( f'(c) = 1 \), we can conclude:

\[ 1 < f'(1) \]

### Step 6: Conclusion
This means that \( f'(1) \) must be strictly greater than 1. Therefore, the correct option is:

- \( f'(1) \) is strictly greater than 1.

### Final Answer
Thus, the answer is that \( f'(1) > 1 \).

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