`f:R rarr R`, `f(x)=tan^(-1)(2^x) + k` is an odd function, then k=
(A) `-pi/4`(B) `(pi)/(4)` (C) `-(pi)/(2)` (D) `(pi)/(2)`

f:R rarr R,f(x)=tan^(-1)(2^(x))+k is an odd function, then k is equal to:

The function f(x)=tan^(-1)(sin x+cos x) is an increasing function in (-(pi)/(2),(pi)/(4))(b)(0,(pi)/(2))(-(pi)/(2),(pi)/(2))(d)((pi)/(4),(pi)/(2))

The function f(x)=tan^(-1)(sin x+cos x) is an increasing function in (1)((pi)/(4),(pi)/(2))(2)(-(pi)/(2),(pi)/(4))(3)(0,(pi)/(2))(4)(-(pi)/(2),(pi)/(2))^(((pi)/(4)),(pi)/(2))^(((pi)/(4)))(2)

Show that f(x)=tan^(-1)(sin x+cos x) is decreasing function on the interval ((pi)/(4),(pi)/(2))

For f(x)=sin^2x ,x in (0,pi) point of inflection is: pi/6 (b) (2pi)/4 (c) (3pi)/4 (d) (4pi)/3

Show that f(x)=tan^(-1)(sin x+cos x) is a decreasing function on the interval on (pi/4,pi/2).

The domain of the function f given by f(x)=sqrt(log_(0.3)((e^(x)-1)/(e^(x)-tan x))) contains (A) (0,(pi)/(4)](B)[(pi)/(4),(pi)/(2))(C)[-(3 pi)/(4),-(pi)/(2)) (D) ((pi)/(2),(5 pi)/(4)]

If f:F rarr R is defind by f(x)=7+cos(5x+3),AA x in R then period of f is (A)2 pi(B)pi(C)(pi)/(5) (D) 2(pi)/(5)

If f:R rarr [pi/6,pi/2], f(x)=sin^(-1)((x^(2)-a)/(x^(2)+1)) is an onto function, the set of values a is