Value of the determinant `|(x,1,1),(0,1+x, 1),(-y, 1+x, 1+y)|` is (A) `xy` (B) `xy(x+2)` (C) `x(x+1)(y+1)` (D) `xy(x+1)`

|(1+x,1,1),(1,1+y,1),(1,1,1+z)|=xy+yz+zx+xyz

The normal at the point (1,1) on the curve 2y+x^2=3 is (A) x + y = 0 (B) x y = 0 (C) x + y +1 = 0 (D) x y = 0

The normal at the point (1,1) on the curve 2y+x^2=3 is (A) x - y = 0 (B) x y = 0 (C) x + y +1 = 0 (D) x y = 0

Factorise : (iv) xy^(2) + (x-1)y-1

The value of |{:(x,x^2-yz,1),(y,y^2-zx,1),(z,z^2-xy,1):}| is

If A(x_(1), y_(1)), B(x_(2), y_(2)) and C (x_(3), y_(3)) are the vertices of a Delta ABC and (x, y) be a point on the internal bisector of angle A, then prove that b|(x,y,1),(x_(1),y_(1),1),(x_(2),y_(2),1)|+c|(x,y,1),(x_(1),y_(1),1),(x_(3),y_(3),1)|=0 where, AC = b and AB = c.

If sin theta = x and sec theta = y , then tan theta is (a) xy (b) x/y (c) y/x (d) 1/(xy)

The solution of the differential equation (x^2+1)(dy)/(dx)+(y^2+1)=0 is a. y=2+x^2 b. y=(1+x)/(1-x) c. y=x(x-1) d. y=(1-x)/(1+x)

If (dy)/dx = (xy)/(x^2 + y^2), y(1) = 1 and y(x) = e then x =

If 1//x + 1//y = 1//3 ,then (xy)/(x + y) =