Let `f:R to R` be a function satisfying `f(x+y)=f(x)=lambdaxy+3x^(2)y^(2)"for all "x,y in R`
If f(3)=4 and f(5)=52, then f'(x) is equal to

Options

A. `10x`
B. `-10x`
C. `20x`
D. `128x`

Explanation

To solve the problem, we need to find the derivative \( f'(x) \) of the function \( f: \mathbb{R} \to \mathbb{R} \) that satisfies the functional equation:

\[ f(x+y) = f(x) + \lambda xy + 3x^2y^2 \]

for all \( x, y \in \mathbb{R} \). We also know that \( f(3) = 4 \) and \( f(5) = 52 \).

### Step 1: Find the value of \( \lambda \)

To find \( \lambda \), we can use the given values of \( f(3) \) and \( f(5) \).

Substituting \( x = 3 \) and \( y = 2 \) into the functional equation:

\[ f(3 + 2) = f(3) + \lambda \cdot 3 \cdot 2 + 3 \cdot 3^2 \cdot 2^2 \]

This simplifies to:

\[ f(5) = f(3) + 6\lambda + 3 \cdot 9 \cdot 4 \]

Substituting the known values \( f(5) = 52 \) and \( f(3) = 4 \):

\[ 52 = 4 + 6\lambda + 108 \]

Now, simplify the equation:

\[ 52 = 112 + 6\lambda \]

Rearranging gives:

\[ 6\lambda = 52 - 112 \]
\[ 6\lambda = -60 \]
\[ \lambda = -10 \]

### Step 2: Find \( f'(x) \)

Now that we have \( \lambda = -10 \), we can find \( f'(x) \) using the definition of the derivative:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

Using the functional equation, we can express \( f(x+h) \):

\[ f(x+h) = f(x) + \lambda xh + 3x^2h^2 \]

Substituting \( \lambda = -10 \):

\[ f(x+h) = f(x) - 10xh + 3x^2h^2 \]

Now, substituting this into the derivative formula:

\[ f'(x) = \lim_{h \to 0} \frac{(f(x) - 10xh + 3x^2h^2) - f(x)}{h} \]

This simplifies to:

\[ f'(x) = \lim_{h \to 0} \frac{-10xh + 3x^2h^2}{h} \]

Factoring \( h \) out of the numerator:

\[ f'(x) = \lim_{h \to 0} (-10x + 3x^2h) \]

As \( h \to 0 \), the term \( 3x^2h \) goes to 0:

\[ f'(x) = -10x \]

### Final Answer

Thus, the derivative \( f'(x) \) is:

\[ f'(x) = -10x \]