CAT 2020 Quant Question [Slot 2] with Solution 10

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

\({x^2} + \frac{1}{{{x^2}}} \ge 2\)

Hence the maximum value of \(\frac{1}{{\sqrt {\frac{1}{{{x^2}}} + {x^2}} }}\) is \(\frac{1}{{\sqrt 2 }}\)