A) \[{{n}_{2}}>{{n}_{3}}>{{n}_{4}}>{{n}_{1}}\]
B) \[{{n}_{3}}>{{n}_{4}}>{{n}_{3}}>{{n}_{1}}\]
C) \[{{n}_{3}}={{n}_{1}}>{{n}_{4}}>{{n}_{2}}\]
D) \[{{n}_{2}}>{{n}_{1}}={{n}_{3}}>{{n}_{4}}\]
Correct Answer: D
Solution :
Given, \[{{n}_{1}}=\sin 7+\cos 7\] |
\[{{n}_{2}}=\sqrt{\sin 7}+\sqrt{\cos 7}\] |
\[{{n}_{2}}=\sqrt{1+\sin 14}\] |
\[{{n}_{4}}=1\] |
\[n_{1}^{2}={{\sin }^{2}}7+{{\cos }^{2}}7+2\sin 7\cos 7\] |
\[n_{1}^{2}=1+\sin 14\] |
\[{{n}_{1}}=\sqrt{1+\sin 14}={{n}_{3}}\] |
\[\sqrt{1+\sin 14}>1\] |
\[\therefore \]\[{{n}_{3}}>{{n}_{4}}\] |
Hence, \[{{n}_{2}}>{{n}_{1}}={{n}_{3}}>{{n}_{4}}\] |
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