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JEE Main & Advanced Sample Paper JEE Main Sample Paper-27

  • question_answer
    If Rolle's theorem holds for the function \[f(x)=2{{x}^{3}}+a{{x}^{2}}+bx\] in \[x\in \] [-1,1] for the point \[c=\frac{1}{2}\], then the value of (2a + b) is equal to

    A)  2                                

    B)  1

    C)  -2                               

    D)  -1

    Correct Answer: D

    Solution :

    \[f(-1)\,=f(1)\,\Rightarrow \,b=-2\] (Using Rolle?s theorem) Also \[f'\left( \frac{1}{2} \right)=0\Rightarrow \,a=\frac{1}{2}\]. \[\therefore \,\,\,2a+b=2\left( \frac{1}{2} \right)\,+(-2)\,=-1\]


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