CAT 2020 Slot 2 QA Question & Solution

AlgebraMedium

Question

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

Options

$\frac{1}{2}$

$1$

$\frac{1}{\sqrt{3}}$

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

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