Let f(x) be a function which is differentiable any number of times and `f(2x^(2)-1)=2x^(3)f(x), AA x in R`. Then `f^((2010))(0)=` (Here `f^((n))(x)=n^(th)` order derivative of f at x)

Options

A. `-1`
B. 1
C. 0
D. data is insufficient

Explanation

To solve the problem, we start with the given functional equation:

\[ f(2x^2 - 1) = 2x^3 f(x) \]

for all \( x \in \mathbb{R} \). We need to find the value of the 2010th derivative of \( f \) at \( x = 0 \), denoted as \( f^{(2010)}(0) \).

### Step 1: Substitute specific values for \( x \)

Let's first substitute \( x = 1 \):

\[ f(2(1)^2 - 1) = 2(1)^3 f(1) \]

This simplifies to:

\[ f(1) = 2 f(1) \]

### Step 2: Solve the equation

From the equation \( f(1) = 2 f(1) \), we can rearrange it:

\[ f(1) - 2 f(1) = 0 \implies -f(1) = 0 \implies f(1) = 0 \]

### Step 3: Substitute another value for \( x \)

Next, let's substitute \( x = 0 \):

\[ f(2(0)^2 - 1) = 2(0)^3 f(0) \]

This simplifies to:

\[ f(-1) = 0 \]

### Step 4: Substitute \( x = -1 \)

Now, substitute \( x = -1 \):

\[ f(2(-1)^2 - 1) = 2(-1)^3 f(-1) \]

This simplifies to:

\[ f(1) = -2 f(-1) \]

Since we already found that \( f(1) = 0 \) and \( f(-1) = 0 \), we have:

\[ 0 = -2 \cdot 0 \implies 0 = 0 \]

This does not give us new information.

### Step 5: Generalize the function

Now, let's consider the possibility that \( f(x) = 0 \) for all \( x \). If \( f(x) = 0 \), then:

\[ f(2x^2 - 1) = 0 \quad \text{and} \quad 2x^3 f(x) = 0 \]

Both sides equal zero, confirming that \( f(x) = 0 \) satisfies the functional equation.

### Step 6: Find the derivatives

Since \( f(x) = 0 \) for all \( x \), all derivatives of \( f \) will also be zero:

\[ f^{(n)}(x) = 0 \quad \text{for all } n \geq 1 \]

### Step 7: Evaluate the 2010th derivative at \( x = 0 \)

Thus, we conclude:

\[ f^{(2010)}(0) = 0 \]

### Final Answer

The value of \( f^{(2010)}(0) \) is:

\[ \boxed{0} \]