JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer If\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{4}}-1}{x-1}=\underset{x\to k}{\mathop{\lim }}\,\frac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},\]then k is: [JEE Main 10-4-2019 Morning]

    A) \[\frac{3}{8}\]            

    B)   \[\frac{3}{2}\]

    C) \[\frac{4}{3}\]                        

    D) \[\frac{8}{3}\]

    Correct Answer: D

    Solution :

    \[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{4}}-1}{x-1}=\underset{x\to k}{\mathop{\lim }}\,\frac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},\]           \[\Rightarrow \underset{x\to 1}{\mathop{\lim }}\,\left( x+1 \right)\left( {{x}^{2}}+1 \right)=\frac{{{k}^{2}}+{{k}^{2}}+{{k}^{2}}}{2k}\]           \[\Rightarrow k=8/3\]


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