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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    If\[\int_{0}^{\infty }{\frac{{{x}^{2}}dx}{({{x}^{2}}+{{a}^{2}})({{x}^{2}}+{{b}^{2}})({{x}^{2}}+{{c}^{2}})}}\]\[=\frac{\pi }{2(a+b)(b+c)(c+a)},\]then the value of\[\int_{0}^{\infty }{\frac{1}{({{x}^{2}}+4)({{x}^{2}}+9)}}dx\]is:

    A)  \[\frac{\pi }{60}\]                           

    B)         \[\frac{\pi }{20}\]

    C)  \[\frac{\pi }{40}\]                           

    D)         \[\frac{\pi }{80}\]

    E)  \[\frac{\pi }{10}\]

    Correct Answer: A

    Solution :

    Given that, \[\int_{0}^{\infty }{\frac{{{x}^{2}}}{({{x}^{2}}+{{a}^{2}})({{x}^{2}}+{{b}^{2}})({{x}^{2}}+{{c}^{2}})}}dx\]                                 \[=\frac{\pi }{2(a+b)(b+c)(c+a)}\] On putting\[a=0,b=2,c=3\]in the above equation \[\therefore \]\[\int_{0}^{\infty }{\frac{dx}{({{x}^{2}}+4)({{x}^{2}}+9)}}\]                                 \[=\frac{\pi }{2(2+3)(3+0)(0+2)}\] \[=\frac{\pi }{2\times 5\times 3\times 2}=\frac{\pi }{60}\]           


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