If `x^2-4x+5-siny=0, yepsilon [0,2pi]` then (A) `x=1,y=0` (B) `x=1,y=pi/2` (C) `x=2,y=0` (D) `x=2,y=pi/2`

If tan^-1 (x+h)=tan^-1 (x)+(h sin y)(sin y)-(h sin y)^2. (sin^2y)/2+ (h sin y)^3. (sin^3y)/3+…, where xepsilon (0,1), yepsilon (pi/4, pi/2) then (A) y=tan^-1x (B) y=sin^-1x (C) y=cot^-1x (D) y=cos^-1x

If s in^(-1)x = y , then (A) 0lt=ylt=pi (B) -pi/2lt=ylt=pi/2 (C) 0ltyltpi (d) -pi/ 2ltyltpi/ 2

sin^2 theta= (4xy)/(x+y)^2 is true if and only if (A) x-y!=0 (B) x=-y (C) x+y!=0 (D) x!=0,y!=0

If sinx+cosx=sqrt(y+1/y) for x in [0,pi] , then (a) x=pi/4 (b) y=0 (c) y=1 (d) x=(3pi)/4

The equation of the plane containing the line 2x+z-4=0 nd 2y+z=0 and passing through the point (2,1,-1) is (A) x+y-z=4 (B) x-y-z=2 (C) x+y+z+2=0 (D) x+y+z=2

x=2, y=−1 is a solution of the linear equation (a) x+2y=0 (b) x+2y=4 (c) 2x+y=0 (d) 2x+y=5

The image of line 2x+y=1 in line x+y+2=0 is : (A) x+2y-7=0 (B) 2x+y-7=0 (C) x+2y+7=0 (D) 2x+y+7=0

The equation of the line inclined at an angle of pi/4 to the x-axis ,such that the two circles x^2+y^2=4 and x^2+y^2 10 x-14 y+65=0 intercept equal length on it, is (A) 2x-2y-3=0 (B) 2x-2y+3=0 (C) x-y+6=0 (D) x-y-6=0

If sinx+cosx=sqrt(y+1/y) for x in [0,pi] , then x=pi/4 (b) y=0 y=1 (d) x=(3pi)/4

One of the bisector of the angle between the lines a(x-1)^2 + 2h(x-1)(y-2) + b (y-2)^2 = 0 is x + 2y - 5 = 0 . Then other bisector is (A) 2x-y=0 (B) 2x+y=0 (C) 2x+y-4=0 (D) x-2y+3=0