If the functions f(x) and g(x) are continuous on [a,b] and differentiable on (a,b) then in the interval (a,b) the equation
`|{:(f'(x),f(a)),(g'(x),g(a)):}|=(1)/(a-b)=|{:(f(a),f(b)),(g(a),g(b)):}|`

Options

A. has at least one root
B. has exactly one root
C. has at most one root
D. no root

Explanation

To solve the given problem, we will use the Mean Value Theorem (MVT), specifically Lagrange's Mean Value Theorem, which states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c\) in \((a, b)\) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

### Step-by-Step Solution:

1. **Apply the Mean Value Theorem to \(f(x)\)**:
- By applying MVT to \(f(x)\) on \([a, b]\), we get:
\[ f'(c_1) = \frac{f(b) - f(a)}{b - a} \]
for some \(c_1 \in (a, b)\).

2. **Apply the Mean Value Theorem to \(g(x)\)**:
- Similarly, applying MVT to \(g(x)\) on \([a, b]\), we have:
\[ g'(c_2) = \frac{g(b) - g(a)}{b - a} \]
for some \(c_2 \in (a, b)\).

3. **Set up the combined equation**:
- We want to analyze the expression:
\[ |(f'(x), f(a)), (g'(x), g(a))| = \frac{1}{b - a} |(f(a), f(b)), (g(a), g(b))| \]
This can be interpreted as a determinant involving the values of the functions and their derivatives.

4. **Formulate the determinant**:
- The left-hand side can be expressed in determinant form as:
\[ |f(a), f'(x); g(a), g'(x)| \]
And the right-hand side can be expressed as:
\[ \frac{1}{b - a} |f(a), f(b); g(a), g(b)| \]

5. **Cross-multiply**:
- Cross-multiplying gives us:
\[ (b - a) |f(a), f'(x); g(a), g'(x)| = |f(a), f(b); g(a), g(b)| \]

6. **Existence of roots**:
- Since both \(f(x)\) and \(g(x)\) are continuous and differentiable, the determinants will yield at least one real solution in the interval \((a, b)\).

7. **Conclusion**:
- Therefore, the combined equation will also have at least one real root in the interval \((a, b)\).

### Final Answer:
The equation has at least one real root in the interval \((a, b)\).

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