A) 1
B) 2
C) 10
D) not defined
Correct Answer: A
Solution :
| \[f(x)=\frac{10}{{{x}^{12}}+2+3{{x}^{4}}+\frac{3}{{{x}^{4}}}+\frac{1}{{{x}^{12}}}}\]\[=\frac{10}{{{x}^{12}}+3{{x}^{4}}+\frac{3}{{{x}^{4}}}+\frac{1}{{{x}^{12}}}+2}=\frac{10}{{{\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)}^{3}}+2}\] |
| \[\because \,\,\,{{x}^{4}}+\frac{1}{{{x}^{4}}}\ge 2\] |
| \[\Rightarrow {{\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)}^{3}}+2\ge 10\] |
| \[\therefore \,\,\,f(x)\le \frac{10}{10}=1\] |
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