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J & K CET Engineering J and K - CET Engineering Solved Paper-2004

  • question_answer
    A function \[y=f(x)\] has second order derivative \[f''(x)=6(x-1)\]. If its graph passes through the point \[(2,1)\] and at that point the tangent to the graph is \[y=3x-5,\] then the function is

    A)  \[{{(x-1)}^{2}}\]      

    B)  \[{{(x-1)}^{3}}\]

    C)  \[{{(x+1)}^{3}}\]

    D)  \[{{(x+1)}^{2}}\]

    Correct Answer: B

    Solution :

    We have, \[f''\,(x)=6x-6\] On integrating, we get \[f'(x)=3{{x}^{2}}-6x+c\] At \[x=2,\] \[f'(2)=3\] \[\Rightarrow \] \[12-12+c=3\] \[\Rightarrow \] \[c=3\] \[\therefore \] \[f'(x)=3{{x}^{2}}-6x+3\] Again integrating we get \[f(x)={{x}^{3}}-3{{x}^{2}}+3x+d\] when \[x=2,\,\,y=1\] \[f(2)=1={{(2)}^{3}}-3{{(2)}^{2}}+3(2)+d\] \[\Rightarrow \] \[d=-1\] \[\therefore \] \[f(x)={{x}^{3}}-3{{x}^{2}}+3x-1\] \[={{(x-1)}^{3}}\]


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