cat 2020 Question 10

Question

For real x, the maximum possible value of $\frac{x}{\sqrt(1 + x^{4})}$ is

Accepted Answer

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 $\frac{1}{\sqrt{2}}$

Question 10.