JEE Main 20179 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002The value of k for which the function f(x)={(45)tan 4xtan 5x ,0<x<π2k+25 ,x=π2f\left( x \right) = \left\{ {\matrix{ {{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr {k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \cr } } \right.f(x)={(54)tan5xtan4x,k+52,0<x<2πx=2π is continuous at x = π2,{\pi \over 2},2π, is :AOption A. 1720{{17} \over {20}}2017BOption B. 25{{2} \over {5}}52COption C. 35{{3} \over {5}}53DOption D. −-− 25{{2} \over {5}}52CheckSkip · show solution
