Question

If $a$ be any real number, then $\frac{a^2}{1+a^4}$

• A. $\in (0,1)$
• B. $\in (0,\frac{1}{2}]$
• C. $\in (0,1]$
• D. $[0,1]$

Answer 1

The correct option is B

$\in (0,\frac{1}{2}]$

Explanation for the correct option:

Find the domain of given function:

Given $\frac{a^2}{1+a^4}$.

If $a=0$, then $\frac{0}{[1+0]}\le \frac{1}{2},0\le \frac{1}{2}$ is true.

If $a\ne 0$, $a^2$ is positive.

$\left(\frac{1}{a^2}\right)$ is also positive.

$\frac{\left(a^2\right)+\left({\displaystyle \frac{1}{a^2}}\right)}{2}\ge \sqrt{\left(a^2\right)\times \left(\frac{1}{a^2}\right)}$ $\left[\mathbf{\because }\mathit{A}\mathit{M}\mathbf{}\mathbf{\ge }\mathbf{}\mathit{G}\mathit{M}\right]$

$\Rightarrow$$\left(a^2\right)+\left(\frac{1}{a^2}\right)\ge 2$

$\Rightarrow$ $\frac{a^4+1}{a^2}\ge 2$

$\mathbf{\therefore }\frac{{\mathbf{a}}^{\mathbf{2}}\mathbf{}}{\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{a}}^{\mathbf{4}}}\mathbf{}\mathbf{\le }\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}$

Hence, Option ‘B’ is Correct.