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JEE Main & Advanced AIEEE Solved Paper-2007

  • question_answer
    The function\[B{{e}^{2+}}<{{K}^{+}}<C{{a}^{2+}}<M{{g}^{2+}}\]given by \[{{K}^{+}}<C{{a}^{2+}}<M{{g}^{2+}}<B{{e}^{2+}}\] can be made continuous at\[C{{a}^{2+}}<M{{g}^{2+}}<B{{e}^{2+}}<{{K}^{+}}\]by defining f(0) as       AIEEE  Solved  Paper-2007

    A)  2                            

    B) -1                        

    C) 0                            

    D) 1

    Correct Answer: D

    Solution :

    \[\underset{x\to 0}{\mathop{\lim }}\,\left\{ \frac{1}{x}-\frac{2}{{{e}^{2x}}-1} \right\}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{2x}}-1-2x}{x({{e}^{2x}}-1)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{e}^{2x}}-2}{({{e}^{2x}}-1)+2x{{e}^{2x}}}\] (using L' Hospital rule) \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{4{{e}^{2x}}}{4{{e}^{2x}}+4x{{e}^{2x}}}=1\] \[\therefore \] \[f(x)\]is continuous at\[x=0,\]then \[\Rightarrow \]               \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\] \[1=f(0)\]


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