| If \[2\sin \left[ \frac{(2x+1)\pi }{2} \right]={{x}^{2}}+\frac{1}{{{x}^{2}}},\]then the value of\[\left( x-\frac{1}{x} \right)\]is [SSC (CGL) 2012] |
A) \[-\,\,1\]
B) \[2\]
C) \[1\]
D) \[0\]
Correct Answer: D
Solution :
| \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=2\sin \left( \frac{(2x+1)\pi }{2} \right)\] |
| \[\Rightarrow \] \[{{\left( x-\frac{1}{x} \right)}^{2}}+2=2\sin \left( \frac{(2x+1)\pi }{2} \right)\] |
| \[[\because \,{{a}^{2}}+{{b}^{2}}={{(a-b)}^{2}}+2ab]\] |
| \[\therefore \] \[x-\frac{1}{x}=0\] |
| [\[\sin \frac{(2x+1)\pi }{2}=1\]for all integer values of \[x\]] |
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