let `f:[-pi/3,pi/6]rarr B` defined by `f(x)=2cos^2x+sqrt3sin2x+1`. Find B such that `f^-1` exists. Also find `f^-1`

Let f:[-pi/3,(2pi)/3] rarr[0,4] be a function defined as f(x) = sqrt3 sin x - cos x +2 . Then f^(-1) (x) is given by a) sin^(-1)((x-2)/2)-pi/6 B) sin^(-1)((x-2)/2)+pi/6 C) (2pi)/3 + cos^(-1)((x-2)/2) D)None of these

Let f:[-pi/3, 2pi/3] rarr [0,4] be a function defined as f(x)=sqrt3 sinx-cosx+2 . Then f^-1(x) is given by

Let f : [-(pi)/(2), (pi)/(2)] rarr [3, 11] defined as f(x) = sin^(2)x + 4 sin x + 6 . Show that f is bijective function.

Let f : [-(pi)/(2), (pi)/(2)] rarr [3, 11] defined as f(x) = sin^(2)x + 4 sin x + 6 . Show that f is bijective function.

Let f:[-(pi)/(3),(2pi)/(3)]rarr[0,4] be a function defined as f(x) = sqrt(3)sin x -cosx +2 . Then f^(-1)(x) is given by

Let f:[-(pi)/(3),(2 pi)/(3)]rarr[0,4] be a function defined as f(x)=sqrt(3)sin x-cos x+2. Then f^(-1)(x) is given by sin^(-1)((x-2)/(2))-(pi)/(6)sin^(-1)((x-2)/(2))+(pi)/(6)(2 pi)/(3)+cos^(-1)((x-2)/(2))(d) none of these

Let f: [-pi/2, pi/2]rarr[-1, 1] , where f(x)=sinx. Find whether f(x) is one-one or not.

Let A=RR - {3} and B =RR-{1} . Prove that the function f: A rarr B defined by , f(x)=(x-2)/(x-3) is one-one and onto. Find a formula that defines f^(-1)

Let f:[-(pi)/(3),(2pi)/(3)]rarr[0,4] be a function defined as f(x) as f(x) = sqrt(3)sin x -cos +2 . Then f^(-1)(x) is given by

Let f:[-pi/3,(2pi)/3] rarr [0,4] be a function defined as f(x)=sqrt(3)sinx-cosx+2. Then f^(-1)(x) is given by (a) sin^(-1)((x-2)/2)-pi/6 (b) sin^(-1)((x-2)/2)+pi/6 (c) (2pi)/3+cos^(-1)((x-2)/2) (d) none of these