The Inverse of cosine function is defined in the intervals
(a) `(0,pi/2)`
(b) `[-pi,0]`
(c) `(-pi/2,0)`
(d) `[pi,pi/2]`

f(x) = cos x is strictly increasing in the interval : (a) ((pi)/(2),pi) (b) (pi,(3 pi)/(2)) (c) (0,(pi)/(2))

The equation (cos p-1)x^(2)+(cos p)x+sin p=0 in the variable x has real roots.The p can take any value in the interval (a) ( 0,2 pi)( b) (-pi,0) (c) (-(pi)/(2),(pi)/(2))( d )(0,pi)

One of the root equation cos x-x+(1)/(2)=0 lies in the interval (0,(pi)/(2))(b)(-(pi)/(2,0))(c)((pi)/(2),pi)(d)(pi,(3 pi)/(2))

One root of the equation cos x-x+(1)/(2)=0 lies in the interval (A)[0,(pi)/(2)](B)[-(pi)/(2),0] (C) [(pi)/(2),0] (D) none

If |veca+vecb|lt|vecavecb| then the angle between veca and vecb lies in the interval (A) (-pi/2, pi/2) (B) (0,pi0) (C) (pi/2,(3pi)/2) (D) (0,2pi)

Find the area under the curve y=cosx over the interval (a) [0,(pi)/(2)] (b) [0,pi]

Let f:(-1,1)rarr B be a function defined by f(x)=tan^(-1)[(2x)/(1-x^(2))]. Then f is both one- one and onto when B is the interval.(a) [0,(pi)/(2))(b)(0,(pi)/(2))(c)(-(pi)/(2),(pi)/(2))(d)[-(pi)/(2),(pi)/(2)]

If the unit vectors veca and vecb are inclined at an angle 2theta such that |veca-vecb|lt1 and 0lethetalepi then theta lies in the intervasl. (A) [0,pi/6] (B) (5pi/6,pi] (C) [pi/2,5pi/6] (D) [pi/6,pi/2]

The gratest value of the function f(x)=(sinx)/(sin(x+pi/4)) on thhe interval [0,pi//2] is

On the interval (0,(pi)/2) the function log sinx is