CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2007

  • question_answer
    \[\frac{3{{x}^{2}}+1}{{{x}^{2}}-6x+8}\]is equal to

    A)  \[3+\frac{49}{2(x-4)}-\frac{13}{2(x-2)}\]

    B)  \[\frac{49}{2(x-4)}-\frac{13}{2(x-2)}\]

    C)  \[\frac{-49}{2(x-4)}-\frac{13}{2(x-2)}\]

    D)  \[\frac{49}{2(x-4)}+\frac{13}{2(x-2)}\]

    Correct Answer: A

    Solution :

    Given,   \[\frac{3{{x}^{2}}+1}{{{x}^{2}}-6x+8}\] On dividing, we get \[\frac{3{{x}^{2}}+1}{{{x}^{2}}-6x+8}=3+\frac{18x-23}{{{x}^{2}}-6x+8}\]                                 ....(i) Now,  \[\frac{18x-23}{(x-2)(x-4)}=\frac{A}{x-2}+\frac{B}{x-4}\] \[\Rightarrow \]  \[18x-23=A(x-4)+B(x-2)\] \[\Rightarrow \] \[18x-23=(A+B)x-4A-2B\] Equating the coefficient of x and constant term, we get \[A+B=18\] \[-4A-2B=-23\] On solving these equations, we get \[A=-\frac{13}{2},\]   \[B=\frac{49}{2}\] \[\therefore \] \[\frac{18x-23}{(x-2)(x-4)}=-\frac{13}{2(x-2)}+\frac{49}{2(x-4)}\] Then, from Eq. (i), we get \[\frac{3{{x}^{2}}+1}{{{x}^{2}}-6x+8}=3-\frac{13}{2(x-2)}+\frac{49}{2(x-4)}\]


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