For continuous function `f: [0, 1]-> R ; A =int_0^1 (x^2* f(x))dx ; B=int_0^1 x * (f(x))^2 dx` then, maximumvalue of `A-B `is `lambda` where `[1/(4lambda)]` is

Explanation

To solve the problem, we need to find the maximum value of \( A - B \) where:

\[ A = \int_0^1 x^2 f(x) \, dx \]
\[ B = \int_0^1 x (f(x))^2 \, dx \]

### Step 1: Express \( A - B \)

We can express \( A - B \) as:

\[ A - B = \int_0^1 x^2 f(x) \, dx - \int_0^1 x (f(x))^2 \, dx \]

This can be rewritten as:

\[ A - B = \int_0^1 \left( x^2 f(x) - x (f(x))^2 \right) \, dx \]

### Step 2: Factor the integrand

We can factor out \( x \) from the integrand:

\[ A - B = \int_0^1 x \left( x f(x) - (f(x))^2 \right) \, dx \]

### Step 3: Analyze the expression \( x f(x) - (f(x))^2 \)

Notice that we can rewrite the expression inside the integral:

\[ x f(x) - (f(x))^2 = f(x) \left( x - f(x) \right) \]

Thus, we have:

\[ A - B = \int_0^1 x f(x) \left( x - f(x) \right) \, dx \]

### Step 4: Determine the maximum value of \( A - B \)

To maximize \( A - B \), we need to maximize the product \( f(x)(x - f(x)) \). The function \( g(x) = x - f(x) \) will be maximized when \( f(x) \) is at its midpoint between 0 and 1.

Setting \( f(x) = kx \) for some \( k \) in the interval [0, 1], we can find the maximum value of \( kx(1 - kx) \).

### Step 5: Find the maximum of \( kx(1 - kx) \)

The maximum of the function \( kx(1 - kx) \) occurs at \( x = \frac{1}{2k} \) when \( k \) is chosen such that \( 0 < k < 1 \).

The maximum value occurs when:

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

Thus, we have:

\[ A - B \leq \int_0^1 \frac{1}{4} x \, dx = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \]

### Step 6: Conclusion

The maximum value of \( A - B \) is \( \frac{1}{12} \). Therefore, we have:

\[ \lambda = \frac{1}{12} \]

Finally, we need to find \( \frac{1}{4\lambda} \):

\[ \frac{1}{4\lambda} = \frac{1}{4 \cdot \frac{1}{12}} = \frac{12}{4} = 3 \]

### Final Answer

Thus, the value of \( \frac{1}{4\lambda} \) is:

\[ \boxed{3} \]