Let `f(x)=-x^2 sin^2 theta-2x cos theta + 1, theta !=npi, nepsilonZ`. If (1, 0) lies between the point where parabola cuts x-axis, then `theta` belongs to: (A) `[-pi/2, pi/2]` (B) `[pi/2, pi]` (C) `[-pi, - pi/2]` (D) none of these

Let f(x)=-x^2 sin^2 theta-2x cos theta + 1, theta !=npi, nepsilonZ The parabola y=f(x) cuts x-axis at : (A) no point (B) at exactly one point (C) at two points (D) None of these

The range of f(x)=sin^(-1)((x^2+1)/(x^2+2)) is (a) [0,pi/2] (b) (0,pi/6) (c) [pi/6,pi/2] (d) none of these

cos theta + sin theta - sin 2 theta = (1)/(2), 0 lt theta lt (pi)/(2)

If tan^(-1)(cottheta)=2\ theta , then theta= (a) +-pi/3 (b) +-pi/4 (c) +-pi/6 (d) none of these

For 0 cos^(-1)(sintheta) is true when theta belongs to (a) (pi/4,pi) (b) (pi,(3pi)/2) (c) (pi/4,(3pi)/4) (d) ((3pi)/4,2pi)

The range of f(x)=sin^(-1)((x^2+1)/(x^2+2)) is (a)[0,pi/2] (b) (0,pi/6) (c) [pi/6,pi/2] (d) none of these

If 1 lies between the roots of the quadratic equation 3x^2-(3sintheta)x-2cos^2theta=0 , then : (A) -pi/3 lt theta lt(5pi)/3 (B) npilttheta lt 2npi (C) 2npi+pi/6 lt theta lt 2npi+(5pi)/6 (D) none of these

If the point (2costheta, 2sintheta), for theta in (0, 2pi) lies in the region between the lines x+y=1 and x-y=2 containing the origin then theta lies in (A) (0, pi/2)cup((3pi)/2, 2pi) (B) [0,pi] (C) (pi/2, (3pi)/2) (D) [pi/4, pi/2]

Period of sin^2 theta is(A) pi^2 (B) pi (C) 2pi (D) pi/2

Solution of the equation sin(sqrt(1+sin2theta))=sintheta+costhetai s(n in Z) npi-pi/4 (b) npi+pi/(12) (c) npi+pi/6 (d) none of these