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JEE Main & Advanced Sample Paper JEE Main Sample Paper-7

  • question_answer
    Let \[f(x)={{x}^{2}}+xg'(1)+g''(2)\] and \[g(x)=f(1){{x}^{2}}+xf'(x)+f''(x)\] then the value of \[f'(1)+g'(1)\] is

    A)  - 4                                         

    B)  0

    C)  2                                            

    D)  4

    Correct Answer: A

    Solution :

    \[\because \]\[f(x)={{x}^{2}}+ax+b,\] where \[g'(1)=\alpha \] and \[g''(2)=b\] \[\because \] \[g(x)=(3+a+b){{x}^{2}}+ax+2\] \[\because \] \[g'(x)=2(3+a+b)x+a\] and \[g''(x)\] \[=2(3+a+b)\] \[=g''(2)\] \[\because \] \[g'(1)=2(3+a+b)+a=a\] \[\Rightarrow \,\] \[3+a+b=0\] \[\Rightarrow \,\]\[g''(2)=0=b\] and \[3+a+b=0\] \[\Rightarrow \,\]\[a=-3\]. \[\therefore \] \[f(x)={{x}^{2}}-3x\,\text{ang}\,g(x)=-3x+2\] \[\Rightarrow \,\] \[f'(x)=2x-3\] and \[g'(x)=-3\]. \[\therefore \,\,f'(1)+g'(1)=-1-3=-4\].


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