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Let $\mathrm{f}(\mathrm{x})=\log \left(1+\mathrm{x}^{2}\right)$ and $\mathrm{A}$ be a constant such that $\frac{|\mathrm{f}(\mathrm{x})-\mathrm{f}(\mathrm{y})|}{|\mathrm{x}-\mathrm{y}|} \leq \mathrm{A}$ for all $\mathrm{x}$, y real and $\mathrm{x} \neq \mathrm{y}$. Then the least possible value of $\mathrm{A}$ is
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The correct answer is:
equal to 1
$$
f(x)=\frac{2 x}{1+x^{2}}
$$
Range of $\frac{2 x}{1+x^{2}} \in[-1,1]$ $\frac{|f(x)-f(y)|}{|x-y|} \leq A$ means
maximum value of $\frac{|\mathrm{f}(\mathrm{x})-\mathrm{f}(\mathrm{y})|}{|\mathrm{x}-\mathrm{y}|}$ is always less than or equal to $\mathrm{A}$.
So, least value of $\mathrm{A}$ is 1
f(x)=\frac{2 x}{1+x^{2}}
$$
Range of $\frac{2 x}{1+x^{2}} \in[-1,1]$ $\frac{|f(x)-f(y)|}{|x-y|} \leq A$ means
maximum value of $\frac{|\mathrm{f}(\mathrm{x})-\mathrm{f}(\mathrm{y})|}{|\mathrm{x}-\mathrm{y}|}$ is always less than or equal to $\mathrm{A}$.
So, least value of $\mathrm{A}$ is 1
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