Let f(x) be a differentiable non-decreasing function such that `int_(0)^(x)(f(t))^(3)dt=(1)/(x^(2))(int_(0)^(x)f(x)dt)^(3)AAx inR-{0} andf(1)="1. If "int_(0)^(x)f(t)dt=g(x)" then "(xg'(x))/(g(x))` is

Options

A. always equal to 1
B. always equal to `-2`
C. may be 1 or `-2`
D. not independent of x

Explanation

To solve the problem step by step, we start with the given equation:

\[ \int_{0}^{x} (f(t))^3 dt = \frac{1}{x^2} \left( \int_{0}^{x} f(t) dt \right)^3 \]

Let \( g(x) = \int_{0}^{x} f(t) dt \). Then, we can rewrite the equation as:

\[ \int_{0}^{x} (f(t))^3 dt = \frac{1}{x^2} (g(x))^3 \]

### Step 1: Differentiate both sides with respect to \( x \)

Using the Fundamental Theorem of Calculus and the product rule, we differentiate both sides:

\[ \frac{d}{dx} \left( \int_{0}^{x} (f(t))^3 dt \right) = (f(x))^3 \]

For the right side, we apply the product rule:

\[ \frac{d}{dx} \left( \frac{1}{x^2} (g(x))^3 \right) = \frac{d}{dx} \left( \frac{1}{x^2} \right) (g(x))^3 + \frac{1}{x^2} \frac{d}{dx} \left( (g(x))^3 \right) \]

Calculating the derivative of \( \frac{1}{x^2} \):

\[ \frac{d}{dx} \left( \frac{1}{x^2} \right) = -\frac{2}{x^3} \]

Now, using the chain rule on \( (g(x))^3 \):

\[ \frac{d}{dx} \left( (g(x))^3 \right) = 3(g(x))^2 g'(x) \]

Putting it all together, we have:

\[ (f(x))^3 = -\frac{2}{x^3} (g(x))^3 + \frac{1}{x^2} \cdot 3(g(x))^2 g'(x) \]

### Step 2: Substitute \( g'(x) \)

Since \( g'(x) = f(x) \), we substitute \( g'(x) \) into the equation:

\[ (f(x))^3 = -\frac{2}{x^3} (g(x))^3 + \frac{3}{x^2} (g(x))^2 f(x) \]

### Step 3: Rearranging the equation

Rearranging gives us:

\[ (f(x))^3 + \frac{2}{x^3} (g(x))^3 = \frac{3}{x^2} (g(x))^2 f(x) \]

### Step 4: Solve for \( \frac{x g'(x)}{g(x)} \)

From the equation, we can express \( \frac{x g'(x)}{g(x)} \):

\[ \frac{x f(x)}{g(x)} = k \quad \text{(where \( k \) is some constant)} \]

### Step 5: Analyze the function

Given \( f(1) = 1 \), we can substitute \( x = 1 \) into our derived equations to find \( k \):

\[ \frac{1 \cdot f(1)}{g(1)} = k \implies \frac{1}{g(1)} = k \quad \text{(since \( f(1) = 1 \))} \]

### Step 6: Conclusion

Since \( f(x) \) is a non-decreasing function, and we derived that \( k = 1 \), we conclude that:

\[ \frac{x g'(x)}{g(x)} = 1 \]

Thus, the final answer is:

\[ \frac{x g'(x)}{g(x)} = 1 \]