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

- $ \frac{1}{\sqrt{3}}$
- $1$
- $\frac{1}{\sqrt{2}}$
- $\frac{1}{2}$

## 1 Answer

1 vote

Given that, $x$ is a real number.

Now, $\dfrac{x}{\sqrt{1+x^{4}}} = \dfrac{1}{\frac{\sqrt{1+x^{4}}}{x}} = \dfrac{1}{\sqrt{\frac{{1+x^{4}}}{x^{2}}}} = \dfrac{1}{\sqrt{\frac{1}{x^{2}} + x^{2}}} \quad \longrightarrow (1) $

We know that, $ \boxed{\text{AM} \geq \text{GM}}$

$ \Rightarrow \dfrac{x^{2} + \dfrac{1}{x^{2}}}{2} \geq \sqrt{x^{2} \cdot \dfrac{1}{x^{2}}} $

$ \Rightarrow \boxed {x^{2} + \frac{1}{x^{2}} \geq 2} $

From equation $(1),$ we get

$ \dfrac{x}{\sqrt{1+x^{4}}} = \dfrac{1}{\sqrt{\underbrace{x^{2}+\frac{1}{x^{2}}}_{2}}}$

$ = \dfrac{1}{\sqrt{2}} \quad [ \because \text{For the minimum value}] $

$\therefore$ The minimum possible value of $\dfrac{x}{\sqrt{1+x^{4}}}$ is $\dfrac{1}{\sqrt{2}}.$

Correct Answer$: \text{C}$

Now, $\dfrac{x}{\sqrt{1+x^{4}}} = \dfrac{1}{\frac{\sqrt{1+x^{4}}}{x}} = \dfrac{1}{\sqrt{\frac{{1+x^{4}}}{x^{2}}}} = \dfrac{1}{\sqrt{\frac{1}{x^{2}} + x^{2}}} \quad \longrightarrow (1) $

We know that, $ \boxed{\text{AM} \geq \text{GM}}$

$ \Rightarrow \dfrac{x^{2} + \dfrac{1}{x^{2}}}{2} \geq \sqrt{x^{2} \cdot \dfrac{1}{x^{2}}} $

$ \Rightarrow \boxed {x^{2} + \frac{1}{x^{2}} \geq 2} $

From equation $(1),$ we get

$ \dfrac{x}{\sqrt{1+x^{4}}} = \dfrac{1}{\sqrt{\underbrace{x^{2}+\frac{1}{x^{2}}}_{2}}}$

$ = \dfrac{1}{\sqrt{2}} \quad [ \because \text{For the minimum value}] $

$\therefore$ The minimum possible value of $\dfrac{x}{\sqrt{1+x^{4}}}$ is $\dfrac{1}{\sqrt{2}}.$

Correct Answer$: \text{C}$