Question

If $x\in R$ and $\alpha =\frac{x^2}{1+x^4}$, then $\alpha$ lies in the interval

• A. $[0,2]$
• B. $[0,1]$
• C. $[0,\frac{1}{2}]$
• D. None of the above.

Answer 1

The correct option is C $[0,\frac{1}{2}]$

We know that,

$A.M.\ge G.M.⟹\frac{a+b}{2}\ge \sqrt{ab}$

Let $a=1$ and $b=x^4$

$⟹\frac{1+x^4}{2}\ge \sqrt{1\ast x^4}$

$⟹\frac{1+x^4}{2}\ge x^2$

$⟹\frac{1+x^4}{2x^2}\ge 1$

$⟹\frac{2x^2}{1+x^4}\le 1$

$⟹\frac{x^2}{1+x^4}\le \frac{1}{2}$

$⟹\alpha \le \frac{1}{2}$

since given that $\alpha =\frac{x^2}{1+x^4}$

Therefore, $\alpha$ lies between $[0,\frac{1}{2}]$. Since, $\alpha$ is always positive because $x^2$ and $1+x^4$ are always positive.