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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    Let \[f\] be a positive function. Let \[{{I}_{1}}=\int_{1-k}^{k}{x\,f\left\{ x(1-x) \right\}}\,dx\],  \[{{I}_{2}}=\int_{1-k}^{k}{\,f\left\{ x(1-x) \right\}}\,dx\] when \[2k-1>0.\] Then \[{{I}_{1}}/{{I}_{2}}\] is [IIT 1997 Cancelled]

    A) 2    

    B) \[k\]

    C) \[1/2\]                                    

    D) 1

    Correct Answer: C

    Solution :

    • \[{{I}_{1}}=\int_{1-k}^{k}{xf\{x(1-x)\}dx}\]                   
    • \[=\int_{1-k}^{k}{(1-k+k-x)f[(1-k+k-x)\{1-(1-k+k-x)\}]dx}\] \[(\because \int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(a+b-x)dx)}}\]                   
    • \[=\int_{1-k}^{k}{\,\,(1-x)f\{x(1-x)\}}\,dx\]                   
    • \[=\int_{1-k}^{k}{f\{x(1-x)\}}\,dx-\int_{1-k}^{k}{xf\{x(1-x)\}}\,dx={{I}_{2}}-{{I}_{1}}\]                   
    • \[\therefore 2{{I}_{1}}={{I}_{2}}\Rightarrow \frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{2}\]       .


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