If $$f(x) = \int\limits_0^x {t\left( {\sin x - \sin t} \right)dt\,\,\,} $$ then :

If $$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}} $$ $$\left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,$$ is a constant of integration) then the ordered pair (K, A) is equal to :

If $$x = \sqrt {{2^{\cos e{c^{ - 1}}}}} $$ and $$y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),$$ then $${{dy} \over {dx}}$$ is equal to :

If the function f defined as

$$f\left( x \right) = {1 \over x} - {{k - 1} \over {{e^{2x}} - 1}},x \ne 0,$$ is continuous at

x = 0, then the ordered pair (k, f(0)) is equal to :