A) \[[1,\ \sqrt{2}]\]
B) \[[1,\ \infty )\]
C) \[[-\sqrt{2},\ -1]\cup [1,\ \sqrt{2}]\]
D) \[(-\infty ,\ -1]\cup [1,\ \infty )\]
Correct Answer: A
Solution :
\[f(x)=\sec \left( \frac{\pi }{4}\,{{\cos }^{2}}x \right)\] We know that, \[0\le {{\cos }^{2}}x\le 1\] at \[\cos x=0,\,\]\[f(x)=1\] and at \[\cos x=1,\]\[=\alpha .1-\beta .1=\alpha -\beta .\]; \\[1\le x\le \sqrt{2}\]Þ\[x\in [1,\,\,\sqrt{2}]\].
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