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JEE Main & Advanced Sample Paper JEE Main Sample Paper-27

  • question_answer
    If \[\underset{x\to 2}{\mathop{\lim }}\,\frac{\tan (x-2).\,\left( {{x}^{2}}+(k-2)x-2k \right)}{({{x}^{2}}-4x+4)}=5\] , then k is equal to

    A)  2                                

    B)  1

    C)  0                                

    D)  3

    Correct Answer: D

    Solution :

    \[\underset{x\to 2}{\mathop{Lim}}\,\,\frac{\tan (x-2)\,.({{x}^{2}}+(k-2)(x-2k)}{({{x}^{2}}-4x+4)}=4\] \[\left( \frac{0}{0} \right)\] \[\Rightarrow \,\,k+2=5\] \[\Rightarrow \,\,k=3\]


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