Question

Consider a continuous function and differentiable function ${}^{\prime }{f}^{\prime }$ satisfies the functional equation $f\left(x\right)+f\left(y\right)=f(x+y+xy)+xy$ for $x,y>-1$ and $f^{\prime }\left(0\right)=0$,then

The value of $\frac{1}{f^{\prime \prime }\left(2\right)}$ is

• A. $9$
• B. $-9$
• C. $4$
• D. $-4$

Answer 1

Answer: (B)

$-9$

$f\left(x\right)+f\left(y\right)=f(x+y+xy)+xy$
$f\left(x\right)+f\left(y\right)=f(x+y(1+x\left)\right)+xy$
Put x=0 gives f(0) = 0
also given $f^{\prime }\left(0\right)=0$
hence ${lim}_{x\to 0}\frac{f\left(x\right)}{x}=0$
$\frac{f(x+y(1+x\left)\right)-f\left(x\right)}{y(1+x)}=\frac{f\left(y\right)-xy}{y(1+x)}$
Putting limit $y\to 0$
$li{m}_{y\to 0}\frac{f(x+y(1+x\left)\right)-f\left(x\right)}{y(1+x)}=\frac{{lim}_{y\to 0}\frac{f\left(y\right)}{y}-x}{(1+x)}$
$f^{\prime }\left(x\right)=\frac{f^{\prime }\left(0\right)-x}{(1+x)}$
$f^{\prime }\left(x\right)=\frac{-x}{(1+x)}$

Differentiate using product rule
$f^{\prime \prime }\left(x\right)=-\frac{1}{(1+x{)}^2}$
$f^{\prime \prime }\left(2\right)=-\frac{1}{9}$