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Explanation:

$\frac{x}{\sqrt{1+x^4}}$ = $\frac{1}{\sqrt{\frac{1+x^4}{x^2} }}$ = $\frac{1}{\sqrt{\frac{1}{x^2}+x^2 }}$

We know x2 is positive and sum of a positive number and its reciprocal is always ≥ 2, i.e., 

$\frac{1}{x^2}+x^2$ ≥ 2.

⇒ $\frac{1}{\sqrt{\frac{1}{x^2}+x^2 }}$ will be maximum when 1$\frac{1}{x^2}+x^2$ is minimum i.e., 2.

∴ Maximum value of $\frac{1}{\sqrt{\frac{1}{x^2}+x^2 }}=\frac{1}{\sqrt{2}}$

Hence, option (c).