JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    The greatest value of \[f(x)={{(x+1)}^{1/3}}-{{(x-1)}^{1/3}}\] on [0, 1] is   AIEEE  Solved  Paper-2002

    A) 1                

    B)                           2  

    C)           3                                                

    D) 1/3

    Correct Answer: B

    Solution :

                       We have, \[f(x)={{(x+1)}^{1/3}}-{{(x-1)}^{1/3}}\] On differentiating w.r.t. x, we get                    \[f'(x)=\frac{1}{3}\left[ \frac{1}{{{(x+1)}^{2/3}}}-\frac{1}{{{(x-1)}^{2/3}}} \right]\]                    \[=\frac{{{(x-1)}^{2/3}}-{{(x+1)}^{2/3}}}{3{{({{x}^{2}}-1)}^{2/3}}}\] Clearly, \[f'(x)\] does not exist at \[x=\pm 1\]. Now, \[f'(x)=0,\] then \[{{(x-1)}^{2/3}}={{(x+1)}^{2/3}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=0\] Clearly, \[f'(x)\ne 0\]for any other value of \[x\in [0.1]\]. The value of \[f(x)\] at \[x=0\] is 2. Hence, the greatest value of \[f(x)\] is 2.

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