Let `f(x)` be a fourth differentiable function such `f(2x^2-1)=2xf(x)AA x in R,` then `f^(iv)(0)` is equal

Options

A. 0
B. 1
C. -1
D. Data insufficient]

Answer

To solve the problem, we need to find the value of \( f^{(iv)}(0) \) given the functional equation:

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

where \( f(x) \) is a fourth differentiable function.

### Step 1: Assume a form for \( f(x) \)

Since the equation involves polynomial expressions, we can assume that \( f(x) \) is a polynomial. Let's assume:

\[ f(x) = ax + b \]

where \( a \) and \( b \) are constants.

### Step 2: Substitute into the functional equation

Now we substitute \( f(x) \) into the functional equation:

\[ f(2x^2 - 1) = a(2x^2 - 1) + b = 2ax^2 - a + b \]

On the right-hand side, we have:

\[ 2x f(x) = 2x(ax + b) = 2ax^2 + 2bx \]

### Step 3: Set the two sides equal

Now we equate the two expressions:

\[ 2ax^2 - a + b = 2ax^2 + 2bx \]

### Step 4: Compare coefficients

To satisfy this equation for all \( x \), we compare coefficients of like terms:

1. Coefficient of \( x^2 \): \( 2a = 2a \) (This is satisfied for any \( a \))
2. Coefficient of \( x \): \( 0 = 2b \) implies \( b = 0 \)
3. Constant term: \( -a + b = 0 \) implies \( -a = 0 \) or \( a = 0 \)

### Step 5: Conclude the form of \( f(x) \)

From the above, we find that \( a = 0 \) and \( b = 0 \). Therefore:

\[ f(x) = 0 \]

### Step 6: Find the fourth derivative

Since \( f(x) = 0 \), all derivatives of \( f \) are also zero:

\[ f^{(iv)}(x) = 0 \]

Thus, we have:

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

### Final Answer

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

\[ \boxed{0} \]

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