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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    The value of \[p\] for which the function \[f(x)=\left\{ \begin{align}   & \frac{{{({{4}^{x}}-1)}^{3}}}{\sin \frac{x}{p}\log \left[ 1+\frac{{{x}^{2}}}{3} \right]},\,x\ne 0 \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,12{{(\log 4)}^{3}},\,\,x=0 \\ \end{align} \right.\]may be continuous at \[x=0\], is              [Orissa JEE 2004]

    A) 1

    B) 2

    C) 3

    D) None of these

    Correct Answer: D

    Solution :

    • For \[f(x)\] to be continuous at \[x=0,\] we should have \[\underset{x\to 0}{\mathop{\lim }}\,\,\,f(x)=f(0)=12\,{{(\log \,4)}^{3}}\] \[\underset{x\to 0}{\mathop{\lim }}\,\,\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\,{{\left( \frac{{{4}^{x}}-1}{x} \right)}^{3}}\times \frac{\left( \frac{x}{p} \right)}{\left( \sin \frac{x}{p} \right)}.\frac{p{{x}^{2}}}{\log \,\left( 1+\frac{1}{3}{{x}^{2}} \right)}\]                     
    • \[={{(\log 4)}^{3}}\,.\,1\,.\,p\,.\underset{x\to 0}{\mathop{\lim }}\,\,\left( \frac{{{x}^{2}}}{\frac{1}{3}{{x}^{2}}-\frac{1}{18}{{x}^{4}}+.........} \right)\]
    • \[=3p\,\,{{(\log 4)}^{3}}.\] Hence \[p=4.\]


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