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JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    The value of K for which the function\[f(x)=\left\{ \begin{matrix}    {{\left( \frac{4}{5} \right)}^{\frac{\tan 4x}{\tan 5x}}}, & 0<x<\frac{\pi }{2}  \\    k+\frac{2}{5} & x=\frac{\pi }{2}  \\ \end{matrix} \right.\] is continuous at \[x=\frac{\pi }{2},\] is        [JEE Online 09-04-2017]

    A)  \[\frac{2}{5}\]                                  

    B)  \[-\frac{2}{5}\]

    C)  \[\frac{17}{20}\]                                             

    D)  \[\frac{3}{5}\]

    Correct Answer: D

    Solution :

                    \[k+2/5\,=(4/5)\] \[=k+\frac{2}{4}=1\] \[K=\frac{3}{5}\]


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