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JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    Let M and m be respectively the absolute maximum and the absolute minimum values of the function,\[f(x)=2{{x}^{3}}-9{{x}^{2}}+12x+5\]in the interval \[[0,3]\]. Then\[M-m\] is equal to. [JEE Main 16-4-2018]

    A)  1                                

    B)  5

    C)  4                                

    D)  9

    Correct Answer: A

    Solution :

     \[f(x)=2{{x}^{3}}-9{{x}^{2}}+12x+5\]                 \[f'(x)=6{{x}^{2}}-18x+12=0\]                 \[\Rightarrow \]\[{{x}^{2}}-3x+2=0\]                 \[x=1\]or \[x=2\]                 \[f''(x)=12x-18\]                 \[f''(1)=12(1)-18=-6<0\]                 Hence, function has maxima at \[x=1\]                 \[\Rightarrow \]\[M=f(1)=2-9+12+5=10\]                 \[f''(2)=12(2)-18=6>0\]                 Hence, function has minima at \[x=2\]                 \[\Rightarrow \]\[m=f(2)=2(8)-9(4)+12(2)+5=9\]                 Therefore, \[M-m=10-9=1\]                 Answer is option A.


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