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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 13

  • question_answer
    Let \[f(x)=\int{\frac{({{x}^{2}}+1)}{\sqrt[3]{{{x}^{3}}+3x+6}}}dx\]and \[f(-1)=\frac{1}{\sqrt[3]{2}},\] then the value of \[f(-2)\] is

    A) \[-4\]                     

    B)        \[-2\]                      

    C) \[2\]            

    D)        \[4\]

    Correct Answer: C

    Solution :

          [c] \[f(x)=\int{\frac{({{x}^{2}}+1)}{{{({{x}^{3}}+3x+6)}^{1/3}}}}dx\] Put  \[{{x}^{3}}+3x+6={{t}^{3}}\] \[\Rightarrow \,\,\,3({{x}^{2}}+1)dx=3{{t}^{2}}dt\] \[\therefore \,\,\,f(x)=\int{\frac{{{t}^{2}}\,dt}{t}}=\frac{{{t}^{2}}}{2}+C\] \[\therefore \,\,\,f(x)=\frac{1}{2}{{({{x}^{3}}+3x+6)}^{2/3}}+C\] \[f(-1)=\frac{1}{2}{{(2)}^{2/3}}+C\] \[\Rightarrow \,\,C=0\] \[\therefore \,\,f(x)=\frac{1}{2}{{({{x}^{3}}+3x+6)}^{2/3}}\] \[\therefore \,\,f(-2)=\frac{1}{2}{{(-8)}^{2/3}}=\frac{1}{2}{{[{{(-2)}^{3}}]}^{2/3}}=2\]   


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