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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    If \[l(m,\,n)=\int_{0}^{1}{{{t}^{m}}{{(1+t)}^{n}}dt,}\] then the expression for \[l(m,\,n)\] in terms of \[l(m+1,\,\,n-1)\] is [IIT Screening 2003]

    A) \[\frac{{{2}^{n}}}{m+1}-\frac{n}{m+1}l(m+1,\,n-1)\]           

    B) \[\frac{n}{m+1}l(m+1,\,n-1)\]

    C) \[\frac{{{2}^{n}}}{m+1}+\frac{n}{m+1}l(m+1,\,n-1)\]

    D) \[\frac{m}{n+1}l(m+1,\,n-1)\]

    Correct Answer: A

    Solution :

    • \[l(m,n)=\int_{0}^{1}{{{t}^{m}}(1+t}{{)}^{n}}dt\]                   
    • \[\left[ {{(1+t)}^{n}}\frac{{{t}^{m+1}}}{m+1} \right]_{0}^{1}-\int_{0}^{1}{n{{(1+t)}^{n-1}}\frac{{{t}^{m+1}}}{m+1}}\,dt\]                   
    • \[=\frac{{{2}^{n}}}{m+1}-\frac{n}{m+1}l(m+1,n-1)\].


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