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JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
    The equation of the tangent to the curve\[y=x+\frac{4}{{{x}^{2}}},\]that is parallel to the x ? axis, is -       AIEEE  Solved  Paper-2010

    A) y = 0                      

    B) y = 1                      

    C)        y = 2      

    D)        y = 3

    Correct Answer: D

    Solution :

    \[y=x+\frac{4}{{{x}^{2}}}\] \[\frac{dy}{dx}=1-\frac{8}{{{x}^{3}}}=0\] \[1=\frac{8}{{{x}^{3}}}\] \[{{x}^{3}}=8\] \[x=2\] at \[x=2,\] \[y=x+\frac{4}{{{x}^{2}}}\] \[=2+\frac{4}{4}=3\] tangent \[y3=0(x2)\] \[y=3\]


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