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

The correct is (C): \(\frac{1}{\sqrt2}\)

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

\(x^2+\frac{1}{x^2}≥2\)

Hence the maximum value of is \(\frac{1}{\sqrt2}\)

Answer 2

Given : \(\frac{x}{\sqrt{1+x^4}}\)

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

= \(\frac{1}{\sqrt{\frac 1{x^2}+x^2}}\)         \(…….. (1)\)
We know that: \(AM ≥GM\)
\(\frac {\frac 1{x^2}+x^2}{2} ≥ 1\)

\(\frac {1}{x^2}+x^2≥2\)

From eq \((1)\),

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

So, the correct option is (C): \(\frac {1}{\sqrt 2}\)