A) \[f(x)<1\]
B) \[f(x)=1\]
C) \[1<f(x)<2\]
D) \[f(x)\ge 2\]
Correct Answer: D
Solution :
Since \[{{\left( x-\frac{1}{x} \right)}^{2}}\ge 0,\,\,\text{}\,\,x\in R,\] we have \[{{x}^{2}}+\frac{1}{{{x}^{2}}}\ge 2\] and Hence, \[f(x)={{\cos }^{2}}x+\frac{1}{{{\cos }^{2}}x}\ge 2\].
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