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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    If \[\int_{a}^{b}{{{x}^{3}}dx}=0\] and \[\int_{a}^{b}{{{x}^{2}}}dx=\frac{2}{3}\], then the value of a and b will be respectively [AMU 2005]

    A) 1, 1

    B) \[-1,-1\]

    C) \[1,-1\]                                   

    D) \[-1,1\]

    Correct Answer: D

    Solution :

    • \[\int_{a}^{b}{{{x}^{3}}dx=0}\], (Given)                   
    • \[\left. \left| \frac{1}{4}{{x}^{4}} \right. \right|_{a}^{b}=0\] Þ \[\frac{1}{4}({{b}^{4}}-{{a}^{4}})=0\]Þ \[{{b}^{4}}-{{a}^{4}}=0\]                   
    • \[\int_{a}^{b}{{{x}^{2}}dx=\frac{2}{3}}\] Þ \[\left. \left| \frac{{{x}^{3}}}{3} \right. \right|_{a}^{b}=\frac{2}{3}\]\[\Rightarrow \] \[{{b}^{3}}-{{a}^{3}}=2\]            \[\Rightarrow \]\[{{b}^{4}}-{{a}^{4}}=0\] Þ \[({{b}^{2}}-{{a}^{2}})({{b}^{2}}+{{a}^{2}})=0\]                   
    • or  \[(b-a)(b+a)=0\]\[\Rightarrow \] \[b=\pm \,a\]                          
    • but \[b=a\]does not satisfy the equation, \[\therefore b=-a\]                   
    • Now, \[{{b}^{3}}-{{a}^{3}}=2\] Þ \[{{(-a)}^{3}}-{{a}^{3}}=2\]                   
    • \[-2{{a}^{3}}=2\] or \[{{a}^{3}}=-1\] Þ \[a=-1\] Þ \[b=-a\]           
    • Hence\[b=-(-1)=1\] Þ \[a=-1\], \[b=1\]Þ \[(a,\,\,b)=(-1,\ 1)\].


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