Question

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

• A. \(\frac{1}{\sqrt3}\)
• B. 1
• C. \(\frac{1}{\sqrt2}\)
• D. \(\frac{1}{2}\)

Answer 1

The Correct Option is C

To find the maximum possible value of \(\frac{x}{\sqrt{1+x^4}}\), let's define it as a function: \(f(x) = \frac{x}{\sqrt{1+x^4}}\). We will use calculus to determine its maximum.

First, calculate the derivative, \(f'(x)\), using the quotient rule:

Given \(f(x) = \frac{u}{v}\), then \(f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}\) where \(u = x\) and \(v = \sqrt{1+x^4}\). 

Thus, \(u' = 1\) and \(v' = \frac{4x^3}{2\sqrt{1+x^4}} = \frac{2x^3}{\sqrt{1+x^4}}\).

Then \(f'(x) = \frac{\sqrt{1+x^4} \cdot 1 - x \cdot \frac{2x^3}{\sqrt{1+x^4}}}{1+x^4}\)

= \(\frac{\sqrt{1+x^4} - \frac{2x^4}{\sqrt{1+x^4}}}{1+x^4}\)

= \(\frac{\sqrt{1+x^4}^2 - 2x^4}{(1+x^4)\sqrt{1+x^4}}\)

= \(\frac{1+x^4 - 2x^4}{(1+x^4)\sqrt{1+x^4}}\)

= \(\frac{1-x^4}{(1+x^4)\sqrt{1+x^4}}\).

Setting \(f'(x) = 0\) gives:

\(1-x^4 = 0\) implies \(x^4 = 1\), thus \(x = \pm1\).

Evaluate \(f(x)\) at these critical points:

For \(x = 1\), \(\frac{1}{\sqrt{1+1^4}} = \frac{1}{\sqrt{2}}\).

For \(x = -1\), \(\frac{-1}{\sqrt{1+(-1)^4}} = -\frac{1}{\sqrt{2}}\).

The maximum value is \(\frac{1}{\sqrt{2}}\).

Thus, the maximum possible value of \(\frac{x}{\sqrt{1+x^4}}\) is \(\frac{1}{\sqrt{2}}\).

Answer 2

Step 1: Rewrite the Expression 

We start by rationalizing and simplifying the expression algebraically:

\[ \frac{x}{\sqrt{1 + x^4}} = \frac{1}{\sqrt{\frac{1 + x^4}{x^2}}} = \frac{1}{\sqrt{\frac{1}{x^2} + x^2}} \quad \text{…… (1)} \]

Step 2: Apply the AM ≥ GM Inequality

We use the inequality:

\[ \frac{a + b}{2} \geq \sqrt{ab} \quad \text{(with equality when } a = b \text{)} \]

Let: \[ a = \frac{1}{x^2}, \quad b = x^2 \] Then: \[ \frac{\frac{1}{x^2} + x^2}{2} \geq \sqrt{ \frac{1}{x^2} \cdot x^2 } = 1 \Rightarrow \frac{1}{x^2} + x^2 \geq 2 \]

Step 3: Plug Back into Equation (1)

\[ \frac{x}{\sqrt{1 + x^4}} = \frac{1}{\sqrt{\frac{1}{x^2} + x^2}} \leq \frac{1}{\sqrt{2}} \]

Equality occurs when: \[ \frac{1}{x^2} = x^2 \Rightarrow x^4 = 1 \Rightarrow x = \pm1 \]

✅ Final Answer:

The maximum value is: \[ \boxed{\frac{1}{\sqrt{2}}} \] and it occurs when \( x = \pm 1 \)