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JEE Main & Advanced Mathematics Functions Question Bank Continuity

  • question_answer
    If \[f(x)=\frac{{{x}^{2}}-10x+25}{{{x}^{2}}-7x+10}\] for \[x\]\[\ne \,\]5 and f is continuous at \[x=5,\] then \[f(5)=\] [EAMCET 2001]

    A)            0

    B)            5

    C)            10

    D)            25

    Correct Answer: A

    Solution :

               \[f(5)=\underset{x\to 5}{\mathop{\text{lim}}}\,f(x)\]\[=\underset{x\to 5}{\mathop{\text{lim}}}\,\frac{{{x}^{2}}-10x+25}{{{x}^{2}}-7x+10}\]                           \[=\underset{x\to 5}{\mathop{\text{lim}}}\,\frac{{{(x-5)}^{2}}}{(x-2)(x-5)}=\frac{5-5}{5-2}=0\].


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