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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    The real number \[x\] when added to its inverse gives the minimum sum at \[x\] equals to     AIEEE  Solved  Paper-2003

    A)                         2         

    B)                       1             

    C)       -1                           

    D) -2

    Correct Answer: B

    Solution :

    Let              \[f(x)=x+\frac{1}{x}\]                     \[f'(x)=1-\frac{1}{{{x}^{2}}}\] Put \[\,f'(x)=0\] for maxima and minima. \[\Rightarrow \]   \[1-\frac{1}{{{x}^{2}}}=0\] \[\Rightarrow \]   \[x=\pm 1\]                     \[f''(x)\frac{2}{{{x}^{3}}}\] At               \[x=1,\,f''(x)=+ve\]         (minima) and at \[x=-1\],     \[f''(x)=-ve\]            (maxima)


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