If `f(x),g(x),h(x) and phi(x)` are polynomial in x, `(int_1^x f(x) h(x) dx) (int_1^x g(x) phi(x) dx) - (int_1^x g(x) h(x) dx)` is divisible by `(x-1)^ λ.` Find maximum value of `lambda.`

Answer

To solve the problem, we need to analyze the expression given and determine the maximum value of \( \lambda \) such that the expression is divisible by \( (x - 1)^\lambda \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We have the expression:
\[ I(x) = \left( \int_1^x f(t) h(t) \, dt \right) \left( \int_1^x g(t) \phi(t) \, dt \right) - \left( \int_1^x g(t) h(t) \, dt \right) \]
We need to find the maximum \( \lambda \) such that \( I(x) \) is divisible by \( (x - 1)^\lambda \).

2. **Setting Up the Integrals**:
Let:
\[ F_1(x) = \int_1^x f(t) h(t) \, dt, \quad F_2(x) = \int_1^x g(t) \phi(t) \, dt, \quad F_3(x) = \int_1^x g(t) h(t) \, dt \]
Then we can rewrite our expression as:
\[ I(x) = F_1(x) F_2(x) - F_3(x) \]

3. **Evaluating at \( x = 1 \)**:
Since \( I(1) = 0 \) (because all integrals from 1 to 1 are zero), we have:
\[ F_1(1) = 0, \quad F_2(1) = 0, \quad F_3(1) = 0 \]
This implies:
\[ F_1(1) F_2(1) - F_3(1) = 0 \implies 0 \cdot 0 - 0 = 0 \]

4. **Differentiating the Expression**:
We differentiate \( I(x) \) using the product and chain rules:
\[ I'(x) = F_1'(x) F_2(x) + F_1(x) F_2'(x) - F_3'(x) \]
Evaluating at \( x = 1 \):
\[ I'(1) = f(1)h(1)g(1)\phi(1) - g(1)h(1) = 0 \]
This gives us another condition.

5. **Further Differentiation**:
We differentiate again:
\[ I''(x) = F_1''(x) F_2(x) + 2F_1'(x) F_2'(x) + F_1(x) F_2''(x) - F_3''(x) \]
Evaluating at \( x = 1 \):
\[ I''(1) = 0 \]

6. **Continuing the Process**:
We continue differentiating until we find a non-zero value. After differentiating four times, we find:
\[ I^{(4)}(1) \neq 0 \]
This indicates that \( I(x) \) has a root of multiplicity 4 at \( x = 1 \).

7. **Conclusion**:
Since \( I(x) \) is divisible by \( (x - 1)^4 \) but not by \( (x - 1)^5 \), we conclude that the maximum value of \( \lambda \) is:
\[ \lambda = 4 \]