Given function is continuous at x=2π
⇒f(2π)=f(2π+)=f(2π−)
⇒k+52=(54)tan5xtan4x
⇒k+52=(54)tan(2π)cot(25π)
⇒k+52=(54)0
⇒k+52=1
⇒k=53
JEE Main 2017 — Mathematics Calculus
The value of k which the function f(x)= {\begin{matrix}{(\frac{4}{5})}^{\frac{\mathrm{tan}4x}{\mathrm{tan}5x}}, & 0<x<\frac{\pi }{2} \\ k+\frac{2}{5}, & x=\frac{\pi }{2}\end{matrix} is continuous at x=2π, is
Held on 9 Apr 2017 · Verified 6 Jul 2026.
52
−52
2017
53
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