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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    \[\sqrt{2+\sqrt{2+\sqrt{2}}}+...\infty \]is equal to:

    A)  \[-1\]                                   

    B) \[\sqrt{2}\]

    C)  2                                            

    D) \[1/2\]

    Correct Answer: C

    Solution :

    \[y=\sqrt{2+\sqrt{2+\sqrt{2}}}+...\] \[y=\sqrt{2+y}\] Squaring of both sides \[{{y}^{2}}=2+y\Rightarrow \] \[{{y}^{2}}-y-2=0\] Since \[y>0\,{{y}^{2}}-2y+y-2=0\] \[y(y-2)+1(y-2)=0\] \[y=2,-1\]


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