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KVPY Sample Paper KVPY Stream-SX Model Paper-13

  • question_answer
    If the expression \[{{z}^{5}}-32\] can be factorized into linear and quadratic factors over real coefficient as \[({{z}^{5}}-32)=(z-2)({{z}^{2}}-pz+4)({{z}^{2}}-qz+4),\] then\[({{p}^{2}}+2q)\] is equal to

    A) 2

    B) 3       

    C) 4

    D) 8

    Correct Answer: C

    Solution :

    We have
    \[{{z}^{5}}-32=(z-2)({{z}^{2}}-pz+4)({{z}^{2}}-qz+4)\]
    \[{{z}^{5}}-32=(z-{{z}_{0}})(z-{{z}_{1}})(z-{{z}_{2}})(z-{{z}_{3}})(z-{{z}_{4}})\]
    \[\Rightarrow \]\[{{z}_{i}}=2\left( \cos \frac{2k\pi }{5}+i\sin \frac{2k\pi }{5} \right)\]
    where  \[{{z}_{i}}=0,1,2,3,4\]
    \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{z}_{0}}=2,\,{{z}_{1}}=2\left( \cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5} \right)\]
    \[{{z}_{2}}=2\left( \cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5} \right)\]
    \[{{z}_{3}}=2\left( \cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5} \right)\]
    \[{{z}_{4}}=2\left( \cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5} \right)\]
    \[{{z}^{5}}-32=(z-2)\left( {{z}^{2}}-4\cos \frac{2\pi }{5}+4 \right)\left( {{z}^{2}}-4\cos \frac{2\pi }{5}+4 \right)\]\[[\because (z-\alpha ((z-\bar{\alpha })=({{z}^{2}}-(\alpha +\bar{\alpha })z+\alpha \bar{\alpha })]\]
    \[\therefore \]\[p=q=4\cos \frac{2\pi }{5}=4\cos 72{}^\circ =4\sin 18{}^\circ \]
    \[{{p}^{2}}+2q=16{{\sin }^{2}}18{}^\circ +8\sin 18{}^\circ \]
    \[=8(1-\cos 36{}^\circ +\sin 18{}^\circ )\]
    \[=8\left( 1+\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4} \right)=4\]


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