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JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    If \[f(x)=\left\{ \begin{matrix}    x\ln x, & x>0  \\    0, & x=0  \\ \end{matrix} \right.\]and conclusion of LMVT holds at \[x=1\] in the interval [0, a] for \[f(x)\], then \[[{{a}^{2}}]\] is equal to [Note: [k] denotes the greatest integer less than or equal to k.]

    A)  2                                

    B)  4

    C)  7                                

    D)  9

    Correct Answer: C

    Solution :

    Given that, So, \[f(x)=\,\left\{ \begin{matrix}    x\ln \,x, & x>0  \\    0, & x=0  \\ \end{matrix} \right.\]                   So,\[f'(1)x=\frac{f(a-f)(0)}{a-0}\]                         ?(1) As \[f'(x)\,=\left( x.\frac{1}{x}\,+\ln \,x.1 \right)\,=1+\ln \,x\] \[f'(1)=1+\ln \,1=1\]                               ?(2) \[\therefore \] Using (2) in (1), we get \[1=\frac{a\ln a-0}{a}\] \[\Rightarrow \,\,\ln \,a=1\,\Rightarrow \,a=e\] \[\therefore \,\,[{{a}^{2}}]\,=[{{a}^{2}}]=7\]


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