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Question 74
For real x, the maximum possible value of $$\frac{x}{\sqrt{1+x^{4}}}$$ is
Solution
Now $$\frac{x}{\sqrt{\ 1+x^4}}=\ \frac{\ 1}{\sqrt{\ \ \frac{\ 1+x^4}{x^2}}}=\frac{1}{\sqrt{\ \frac{1}{x^2}+x^2}}$$
Applying A.M>= G.M.
$$\frac{\left(\frac{1}{x^2}+x^2\right)}{2}\ge\ 1\ or\ \ \frac{1}{x^2}+x^2\ge\ 2$$ Substituting we get the maximum possible value of the equation as $$\frac{1}{\sqrt{\ 2}}$$
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