Question

Let $f\left(x\right)=\log (1+x^2)$ and $A$ be a constant such that $\frac{|f\left(x\right)-f\left(y\right)|}{|x-y|}\le A$ for all $x,y$ real and $x\ne y$. Then the least possible value of $A$ is

• A. Equal to $1$
• B. Bigger than $1$ but less than $2$
• C. Bigger than $0$ but less than $1$
• D. Bigger than $2$

Answer 1

The correct option is A Equal to $1$

$\frac{|f\left(x\right)-f\left(y\right)|}{|x-y|}=f^{\prime }\left(x\right)=\frac{d\left(log\right(1+x^2\left)\right)}{dx}=\frac{2x}{1+x^2}$

If $f^{\prime }\left(x\right)$ is maximum, then $f^{\prime \prime }\left(x\right)$ is $0$.

Therefore, $\frac{4x^2-2x^2-2}{(1+x^2)}=0⟹x=1$

Maximum value of $f\left(x\right)=\frac{2\left(1\right)}{1+(1{)}^2}=1$

$A=1$