Question

If $f:R\to R$ satisfies $|f\left(x\right)-f\left(y\right)|\le |x-y{|}^3$ and $g\left(x\right)=f(-x+f\left(x\right)),$ then the value of $g^{\prime }\left(1\right)$ is

• A. Cannot be determined
• B. $1$
• C. $-1$
• D. $0$

Answer 1

The correct option is D $0$

$|f\left(x\right)-f\left(y\right)|\le |x-y{|}^3$
$\Rightarrow \frac{|f\left(x\right)-f\left(y\right)|}{|x-y|}\le |x-y{|}^2$
Taking $\underset{x\to y}{lim}$on both sides
$\underset{x\to y}{lim}\frac{|f\left(x\right)-f\left(y\right)|}{|x-y|}\le \underset{x\to y}{lim}|x-y{|}^2$
$|f^{\prime }\left(x\right)|\le 0$
$\therefore f^{\prime }\left(x\right)=0\forall x\in R$
$g\left(x\right)=f(-x+f(x\left)\right)\Rightarrow g^{\prime }\left(x\right)=f^{\prime }(-x+f(x\left)\right).(-1+f^{\prime }(x\left)\right)$
$\therefore f^{\prime }\left(x\right)=0$
$\Rightarrow g^{\prime }\left(x\right)=0\forall x\in R$
$\therefore g^{\prime }\left(1\right)=0$