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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    The function\[f(x)=2{{x}^{3}}-15{{x}^{2}}+36x+4\] maximum at:

    A) \[x=2\]                                

    B) \[x=4\]

    C) \[~x=0\]                              

    D) \[x=3\]

    Correct Answer: A

    Solution :

    \[f(x)=2{{x}^{3}}-15{{x}^{2}}+36x+4\]put \[f'(x)=0\] \[f'(x)=6{{x}^{2}}-30x+36\]\[6{{x}^{2}}-30x+36=0\] \[f''=12x-30\]   \[{{x}^{2}}-5x+6=0\] put the value of \[x\]in \[f''(x)\] \[{{x}^{2}}-3x-2x+6=0\]                 \[=12\times 2-30\]                 \[=-6\] \[x(x-3)-2(x-3)=0\]     \[(x-2)(x-3)=0\]                 \[=12\times 03-30=+\,6\]is not maximum


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