Question
For real x, the maximum possible value of \(\frac{x}{{\sqrt {1 + {x^4}} }}\) is
- \(\frac{1}{{\sqrt 3 }}\)
- 1
- \(\frac{1}{{\sqrt 2 }}\)
- \(\frac{1}{2}\)
Option: 3
Solution:
Solution:
\(\frac{x}{{\sqrt {1 + {x^4}} }} = \frac{1}{{\sqrt {\frac{1}{{{x^2}}} + {x^2}} }}\)
\({x^2} + \frac{1}{{{x^2}}} \ge 2\)
Hence the maximum value of \(\frac{1}{{\sqrt {\frac{1}{{{x^2}}} + {x^2}} }}\) is \(\frac{1}{{\sqrt 2 }}\)
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