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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    If \[f\left( x \right)=g\left( x \right)|(x-1)\,(x-2)...(x-10)|-2\]is differentiable \[\forall x\in R,\] where \[g(x)\] is a polynomial of degree 9, then \[\frac{d}{dx}\,\left( f\left( {{x}^{2}}+g\left( x \right) \right) \right)\]at \[x=0\]is

    A) \[0\]          

    B)                    \[-2\]                     

    C) \[-4\]                     

    D)        \[4\]

    Correct Answer: A

    Solution :

    [a] Since \[f\left( x \right)=g\left( x \right)|(x-1)(x-2)...(x-10)|-2\]is differentiable \[\forall \,\,x\in R,\] we have \[g\left( x \right)=0\]for \[x=1,2,3,.....,10\] So, \[g\left( x \right)=0\]for all real x as degree of \[g(x)\] is 9. \[\therefore \,\,\,\,f(x)=-2\] \[\therefore \,\,\,\frac{d}{dx}\left( f\left( {{x}^{2}}+g(x) \right) \right)\] at \[x=0\] is zero.


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