A) \[2\cos \left( \frac{\pi }{{{2}^{n+1}}} \right)\]
B) \[2\sin \left( \frac{\pi }{{{2}^{n}}^{+1}} \right)\]
C) \[\sqrt{2}\cos \left( {{2}^{\operatorname{n}+1}}\pi \right)\]
D) none of these
Correct Answer: A
Solution :
| Let \[f(n)=\sqrt{2+\sqrt{2+.....+\sqrt{2}}}\] (number of roots is n) |
| Then \[f(1)=\sqrt{2}=2\cos \frac{\pi }{4}\] or \[2\sin \frac{\pi }{4}\] |
| \[\therefore \] f (1) may be true for [a] as well as for [b] again f(2) \[=\sqrt{2+\sqrt{2}}=2\times \frac{\sqrt{2+\sqrt{2}}}{2}=2\cos \frac{\pi }{4}\] |
| \[\therefore \]f (2) is true for [a]. |
| We check it for any integer. |
| Let \[\sqrt{2+\sqrt{2+....\sqrt{2}}}=2\cos \left( \frac{\pi }{{{2}^{\operatorname{k}+1}}} \right)\]for some \[k\ge 1...\left( \operatorname{i} \right)\] |
| Now, \[\underset{(k+1)\,terms}{\mathop{\sqrt{2+\sqrt{2+\sqrt{2+....\sqrt{2}}}}}}\,\]\[=\sqrt{2+\underbrace{\sqrt{2+\sqrt{2+.....\sqrt{2}}}}_{k\,\,terms}}\] |
| \[=\sqrt{2+2\cos \left( \frac{\pi }{{{2}^{k+1}}} \right)}\] [From (i)] |
| \[=\sqrt{2\left[ 1+\cos \frac{\pi }{{{2}^{k+1}}} \right]}\] |
| \[\sqrt{2.2{{\cos }^{2}}\frac{\pi }{{{2}^{k+2}}}}=2\cos \frac{\pi }{{{2}^{k+2}}}\] |
| \[\therefore \]The result is true for \[\operatorname{n}=k+1\]. |
| Hence, by the principle of mathematical induction \[\sqrt{2+\sqrt{2+....+\sqrt{2}}}=2\cos \left( \frac{\pi }{{{2}^{n+1}}} \right)\] for all \[n\in N\] |
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