KVPY Sample Paper KVPY Stream-SX Model Paper-25

  • question_answer
    If the complex number is \[{{\left( 1+ri \right)}^{3}}=\lambda \left( 1+i \right),\] when  \[i=\sqrt{-1},\] for some real \[\lambda \], the value of \[r\] can be

    A) \[\cos \frac{\pi }{5}\]

    B) \[\operatorname{cosec}\frac{3\pi }{2}\]

    C) \[\cos \frac{\pi }{12}\]

    D) \[\operatorname{cosec}\frac{\pi }{12}\]

    Correct Answer: B

    Solution :

    \[\therefore \]\[{{(1+ri)}^{3}}=\lambda (1+i)\]\[\Rightarrow \]\[1-{{r}^{3}}i+3ri-3{{r}^{2}}=\lambda +i\lambda \]
    On comparing real and imaginary parts. We get \[1-3{{r}^{2}}=\lambda \]and \[-{{r}^{3}}+3r=\lambda \] Then \[-{{r}^{3}}-3r=1-3{{r}^{2}}\]
    \[\Rightarrow \]\[{{r}^{3}}-3{{r}^{2}}-3r+1=0\] \[\Rightarrow \]\[({{r}^{3}}+1)-3r(r+1)=0\] \[\Rightarrow \]\[(r+1)({{r}^{2}}-r+1-3r)=0\]
    \[\Rightarrow \]\[(r+1)({{r}^{2}}-4r+1)=0\therefore r=-1,2\pm \sqrt{3}\]\[\Rightarrow \]\[r=\operatorname{cosec}\frac{3\pi }{2}\]

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