If a twice differentiable function f(x) on `(a,b)` and continuous on [a, b] is such that `f''(x)lt0` for all `x in (a,b)` then for any `c in (a,b),(f(c)-f(a))/(f(b)-f(c))gt`

Options

A. `(b-c)/(c-a)`
B. `(c-a)/(b-c)`
C. `(b-c)(c-a)`
D. `(1)/((b-c)(c-a))`

Explanation

To solve the problem, we need to show that if \( f''(x) < 0 \) for all \( x \) in the interval \( (a, b) \), then for any \( c \) in \( (a, b) \), the following inequality holds:

\[ \frac{f(c) - f(a)}{f(b) - f(c)} > 0 \]

### Step-by-Step Solution:

1. **Understanding the Implication of \( f''(x) < 0 \)**:
Since \( f''(x) < 0 \) for all \( x \in (a, b) \), it implies that \( f'(x) \) is a decreasing function on the interval \( (a, b) \). This means that as \( x \) increases from \( a \) to \( b \), the slope of the function \( f(x) \) is decreasing.

2. **Applying the Mean Value Theorem**:
By the Mean Value Theorem, there exists a point \( u \in (a, c) \) such that:
\[ f'(u) = \frac{f(c) - f(a)}{c - a} \]
Similarly, there exists a point \( v \in (c, b) \) such that:
\[ f'(v) = \frac{f(b) - f(c)}{b - c} \]

3. **Comparing Derivatives**:
Since \( f' \) is decreasing, we have:
\[ f'(u) > f'(c) \quad \text{and} \quad f'(c) > f'(v) \]
This can be expressed as:
\[ f'(u) > f'(c) > f'(v) \]

4. **Setting Up the Inequality**:
From the Mean Value Theorem, we can write:
\[ \frac{f(c) - f(a)}{c - a} > \frac{f(b) - f(c)}{b - c} \]
Rearranging this gives:
\[ (f(c) - f(a))(b - c) > (f(b) - f(c))(c - a) \]

5. **Finalizing the Result**:
Rearranging the terms leads to:
\[ f(c) - f(a) > \frac{(f(b) - f(c))(c - a)}{(b - c)} \]
Since \( b > c \) and \( c > a \), we can conclude that:
\[ \frac{f(c) - f(a)}{f(b) - f(c)} > 0 \]
Thus, we have shown that:
\[ \frac{f(c) - f(a)}{f(b) - f(c)} > 0 \]