Let $f$ be a differentiable function satisfying $f(x)+f(y)+f(z)+f(x) f(y) f(z)=14$ for all $x, y, z \in R$ Then,
Answer & explanation
Correct answer: option 2
We have,
$f(x)+f(y)+f(z)+f(x) f(y) f(z)=14$ for all $x, y, z \in R$ ......(i)
Putting x = y = z = 0, we get
$3 f(0)+\{f(0)\}^3=14$
$\Rightarrow \{f(0)\}^3+3 f(0)-14=0$
$\Rightarrow f(0)=2$
Now, putting $y=z=x$ in (i), we get
$3 f(x)+\{f(x)\}^3=14$ for all $x \in R$
Differentiating with respect to $x$, we get
$3 f^{\prime}(x)+3\{f(x)\}^2 f^{\prime}(x)=0$ for all $x \in R$
$\Rightarrow \left\{\{f(x)\}^2+1\right\} f^{\prime}(x)=0$ for all $x \in R$
$\Rightarrow f^{\prime}(x)=0$ for all $x \in R$