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JEE Main & Advanced Sample Paper JEE Main Sample Paper-46

  • question_answer
    Let \[f(x)=\left\{ \begin{matrix}    |{{x}^{3}}+{{x}^{2}}+3x+\sin \,x| & x\ne 0  \\    0 & x=0  \\ \end{matrix}, \right.\] then number of points [where \[f(x)\] attains its minimum value] is

    A)  1                                

    B)  2                 

    C)  3                                

    D)  infinite many

    Correct Answer: A

    Solution :

     \[f(x)=\,\left\{ \begin{matrix}    |{{x}^{3}}+{{x}^{2}}+3x+\sin \,x\,|\,\left( 3+\sin \,\left( \frac{1}{x} \right) \right), & x\ne 0  \\    0 & ,x=0  \\ \end{matrix}\,\, \right.\] Let \[g(x)={{x}^{3}}+{{x}^{2}}+3x+\sin \,x\] \[\therefore \] \[g'(x)=3{{x}^{2}}+2x+3+\cos \,x\] \[=3\,\left( {{x}^{2}}+\frac{2x}{3}+1 \right)+\cos \,x\] \[=3\,\left\{ {{\left( x+\frac{1}{3} \right)}^{2}}+\frac{8}{9} \right\}+\cos \,x>0\] and      \[2<3+\sin \,\left( \frac{1}{x} \right)<4\] Hence, minimum value of \[f(x)\] is 0 at \[x=0\]. Hence, number of point is 1.


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