A) \[n=-1,0,3,5\]
B) \[n=1,2,4,5\]
C) \[n=0,2,4\]
D) \[n=-1.1,3,5\]
Correct Answer: B
Solution :
| \[\because -\frac{3\pi }{4}\le \frac{x}{2}-\frac{\pi }{2}\le \frac{3\pi }{4}\Rightarrow \,\,-\frac{\pi }{2}\le x\le \frac{5\pi }{2}\]\[\Rightarrow \] But given \[x=\frac{n\pi }{2}\Rightarrow n=-1,0,1,2,3,4,5\] |
| \[\sin \frac{x}{2}-\cos \frac{x}{2}=1-\sin x\]\[\Rightarrow \]\[\sin \frac{x}{2}-\cos \frac{x}{2}={{\left( \sin \frac{x}{2}-\cos \frac{x}{2} \right)}^{2}}\]\[\Rightarrow \left( \sin \frac{x}{2}-\cos \frac{x}{2} \right)\left[ \sin \frac{x}{2}-\cos \frac{x}{2}-1 \right]=0\] |
| \[\Rightarrow \]either \[\tan \frac{x}{2}=1\] or \[\sin \frac{x}{2}-\cos \frac{x}{2}=1.\] |
| If \[\tan \frac{x}{2}=1\Rightarrow x=\frac{\pi }{2},\frac{5\pi }{2}\Rightarrow n=1,5\,\,;\] |
| If \[\sin \frac{x}{2}-\cos \frac{x}{2}=1\Rightarrow \sin \left( \frac{x}{2}-\frac{\pi }{4} \right)\] |
| \[=\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}\Rightarrow \frac{x}{2}-\frac{\pi }{4}=k\pi +{{(-1)}^{k}}\frac{\pi }{4}\] for odd values of k, \[x=2k\pi \] |
| \[\Rightarrow x=0,\,\,2\pi ,\] \[\therefore n=0,4\] for even values of k, \[x=2k\pi +\pi \] |
| \[\Rightarrow x=\pi ,\] \[\therefore n=2\] Hence overall \[n=1,2,4,5\] |
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