Question

Let $f$ be a differentiable function satisfying $f\left(x\right)+f\left(y\right)+f\left(z\right)+f\left(x\right)f\left(y\right)f\left(z\right)=14$ for all $x,y,z\in R$
Then,

• A. $f^{\prime }\left(x\right)<0$ for all $x\in R$
• B. $f^{\prime }\left(x\right)=0$ for all $x\in R$
• C. $f^{\prime }\left(x\right)>0$ for all $x\in R$
• D. none of these

Answer 1

The correct option is B $f^{\prime }\left(x\right)=0$ for all $x\in R$

We have,
$f\left(x\right)+f\left(y\right)+f\left(z\right)+f\left(x\right)f\left(y\right)f\left(z\right)=14$ for all $x,y,z\in R...\left(i\right)$
Putting $x=y=z=0$, we get,
$3f\left(0\right)+{\left\{f\right(0\left)\right\}}^3=14$
${\left\{f\right(0\left)\right\}}^3+3f\left(0\right)-14=0$
$f\left(0\right)=2.$
Now, putting $y=z=x$ in $\left(i\right)$, we get
$3f^{\prime }\left(x\right)+3{\left\{f\right(x\left)\right\}}^2{f}^{\prime }\left(x\right)=0$ for all $x\in R$
$\{{\left\{f\right(x\left)\right\}}^2+1\}{f}^{\prime }\left(x\right)=0$ for all $x\in R$
$f^{\prime }\left(x\right)=0$ for all $x\in R$