RAJASTHAN ­ PET Rajasthan PET Solved Paper-2009

  • question_answer
    If\[{{x}_{r}}=\cos \left( \frac{\pi }{{{2}^{r}}} \right)+i\sin \left( \frac{\pi }{{{2}^{r}}} \right),\]then value of\[{{x}_{1}}\,{{x}_{2}}\,\,{{x}_{3}}....\]is

    A)  \[i\]

    B)  \[1\]

    C)  \[-1\]

    D)  \[-i\]

    Correct Answer: C

    Solution :

     \[{{x}_{1}}{{x}_{2}}{{x}_{3}}=\left[ \cos \left( \frac{\pi }{2} \right)+i\sin \left( \frac{\pi }{2} \right) \right]\] \[\left[ \cos \left( \frac{\pi }{{{2}^{2}}} \right)+i\sin \left( \frac{\pi }{{{2}^{2}}} \right) \right].....\] \[=\cos \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+\frac{\pi }{{{2}^{3}}}+.... \right)\] \[+i\sin \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+\frac{\pi }{{{2}^{3}}}+.... \right)\] \[=\cos \left( \frac{\pi /2}{1-\frac{1}{2}} \right)+i\sin \left( \frac{\pi /2}{1-\frac{1}{2}} \right)\] \[=\cos (\pi )+\sin (\pi )=-1\]


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