Simplify
(i) `(6 x-2 y)+(3 x+4 y)`
ii) `(x-2 y)-(2 x-y)`
iii) `(5+3 x)+(3-2 x)`
iv) `(4 x-2 y)-(6 x-7 y)`

If Q is the image of the point P (2,3,4) under the reflection in the plane x-2y+5z=6 then the equation of the line PQ is a) (x-2)/(-1)= (y-3)/2= (z-4)/5 b) (x-2)/(1)= (y-3)/(-2)= (z-4)/5 c) (x-2)/(-1)= (y-3)/(-2)= (z-4)/5 d) (x-2)/(1)= (y-3)/(2)= (z-4)/5

Prove that (i) 2 cos x cos y=cos (x+y)+cos (x-y) (ii) -2 . sin x sin y=cos (x+y)-cos (x-y) (iii) 2 sin x cos y=sin (x+y)+sin (x-y) (iv) 2 cos x sin y=sin (x+y)-sin (x-y)

If (3x, x + y) =( 6,3), find x and y

The solution of 2 (y + 3) - x y(dy)/(dx) = 0 with y =-2, when x =1 is :a) (y + 3) = x ^(2) b) x ^(2) (y + 3) = 1 c) x ^(4) (y + 3) = 1 d) x ^(2) ( y +3) ^(2) = e ^(y +2)

Show that (i) cos x+cos y=2 cos (x+y)/(2) cos (x-y)/(2) (ii) cos x-cos y=-2 sin (x+y)/(2) sin (x-y)/(2) (iii) sin x+sin y=2 sin (dot(x)+y)/(2) cos (x-y)/(2) (iv) sin x-sin y=2 cos (x+y)/(2) sin (x-y)/(2)

The locus of a point which moves so that the ratio of the length of the tangents to the circles x^(2)+ y^(2)+ 4x+3 =0 and x^(2)+ y^(2) -6x +5=0 is 2 : 3 is a) 5x^(2) +5y^(2) - 60x +7=0 b) 5x^(2) +5y^(2) +60x -7=0 c) 5x^(2) +5y^(2) -60x -7=0 d) 5x^(2) +5y^(2) +60x +7=0

Solve y^'= (y(x-2 y))/(x(x-3 y)), x ne 0, x ne 3 y^

Which of the following is a homogeneous differential equation? (4 x+6 y+5) d y-(3 y+2 x+4) d x=0 x y d x-(x^3+y^3) d y=0 (x^3+2 y^2) d x+2 x y d y=0 y^2 d x+(x^2-x y-y^2) d y=0

The equation of one of the diameters of the circle x^(2) - y^(2) - 6x + 2y = 0 is a)x - 3y = 0 b)x + 3y = 0 c)3x + y = 0 d)3x - y = 0