y=2x

y=2-x^(2), y=x^(2)

The area bounded by the curves y^(2)=x^(3) and |y|=2x

The axis of a parabola is along the line y=x and the distance of its vertex and focus from the origin are sqrt(2) and 2sqrt(2), respectively.If vertex and focus both lie in the first quadrant, then the equation of the parabola is (x+y)^(2)=(x-y-2)(x-y)^(2)=4(x+y-2)(x-y)^(2)=4(x+y-2)(x-y)^(2)=4(x+y-2)(x-y)^(2)=8(x+y-2)

If 2^x-2^y=2^(x+y) then dy/dx= (a) (2^x+2^y)/(2^x-2^y) (b) (2^x+2^y)/(1+2^(x+y)) (c) (2^(x-y))(1-2^y)/(2^x+1) (d) (2^x-2^y)/(2^x+2^y)

A parabola has its vertex and focus in the first quadrant and axis along the line y=x . If the distances of the vertex and focus from the origin are respectively sqrt(2)&2sqrt(2) , then equation of the parabola is x^2+y^2-8x+8y+2x y=16 x^2+y^2-8x-8y+16=2x y (x-y)^2=8(x+y-2) (x+y)^2=(x-y+2)

Area bounded by the curve y=2+x, y=2-x and x = 2 is

Area bounded by the lines y=2+x, y=2-x and x=2 is

The axis of a parabola is along the line y=x and the distance of its vertex and focus from the origin are sqrt(2) and 2sqrt(2) , respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is (x+y)^2=(x-y-2) (x-y)^2=(x+y-2) (x-y)^2=4(x+y-2) (x-y)^2=8(x+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)