Let f be a differentiable function satisfying
`f(x)+f(y)+f(z)+f(x)f(y)f(z)=14" for all "x,y,z inR` Then,

Options

A. `f'(x)lt0" for all "x in R`
B. `f'(x)=0" for all "x in R`
C. `f'(x)gt0" for all "x in R`
D. none of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the given equation.
The function \( f \) satisfies the equation:
\[ f(x) + f(y) + f(z) + f(x)f(y)f(z) = 14 \]
for all \( x, y, z \in \mathbb{R} \).

### Step 2: Substitute \( x = y = z = 0 \).
Let’s substitute \( x = 0, y = 0, z = 0 \) into the equation:
\[ f(0) + f(0) + f(0) + f(0)f(0)f(0) = 14 \]
This simplifies to:
\[ 3f(0) + f(0)^3 = 14 \]
Let \( f(0) = a \). Then we have:
\[ 3a + a^3 = 14 \]

### Step 3: Rearrange the equation.
Rearranging gives us:
\[ a^3 + 3a - 14 = 0 \]

### Step 4: Find the roots of the cubic equation.
We can test possible rational roots. By substituting \( a = 2 \):
\[ 2^3 + 3(2) - 14 = 8 + 6 - 14 = 0 \]
Thus, \( a = 2 \) is a root, so \( f(0) = 2 \).

### Step 5: Substitute \( y = x \) and \( z = x \) into the original equation.
Now, substitute \( y = x \) and \( z = x \):
\[ f(x) + f(x) + f(x) + f(x)f(x)f(x) = 14 \]
This simplifies to:
\[ 3f(x) + f(x)^3 = 14 \]
Let \( f(x) = b \). Then we have:
\[ 3b + b^3 = 14 \]

### Step 6: Differentiate the equation.
Differentiating both sides with respect to \( x \):
\[ 3f'(x) + 3f(x)^2 f'(x) = 0 \]
Factoring out \( f'(x) \):
\[ f'(x)(3 + 3f(x)^2) = 0 \]

### Step 7: Analyze the factored equation.
This gives us two cases:
1. \( f'(x) = 0 \)
2. \( 3 + 3f(x)^2 = 0 \) (which cannot happen since \( f(x)^2 \) is always non-negative)

Thus, we conclude:
\[ f'(x) = 0 \quad \text{for all } x \in \mathbb{R} \]

### Step 8: Conclusion.
Since \( f'(x) = 0 \) for all \( x \), \( f(x) \) must be a constant function. Given \( f(0) = 2 \), we conclude:
\[ f(x) = 2 \quad \text{for all } x \in \mathbb{R} \]

### Final Answer:
The correct option is that \( f'(x) = 0 \) for all \( x \in \mathbb{R} \).

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