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JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    Let\[a=\operatorname{Im}.\left( \frac{1+{{z}^{2}}}{2iz} \right)\], where z is any non-zero complex number and \[i=\sqrt{-1}\]. The set \[A=\{a:\left| z \right|=1\,and\,\ne \pm 1\}\] is equal to

    A)  \[[-1,\,1]\]                                        

    B)  \[(-1,\,0]\]

    C)  \[[0,\,1)\]                                          

    D)  \[(-1,\,1)\]

    Correct Answer: D

    Solution :

    \[a=\operatorname{Im}.\left( \frac{1+{{e}^{i2\theta }}}{2i{{e}^{i\theta }}} \right)=\operatorname{Im}.\left( \frac{{{e}^{-i\theta }}+{{e}^{i\theta }}}{2i} \right)\] \[=\operatorname{Im}.\left( \frac{2\cos \theta }{2i} \right)=-\cos \theta \]              As\[z\ne \pm 1,\,\,\text{so}\,\,a\in (-1,\,\,1)\]


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