" 11."(5)/(x+1)-(2)/(y-1)=(1)/(2),(2)/(y-1)+(10)/(x+1)=2(1)/(2)

Solve the following system of equations: (5)/(x+1)-(2)/(y-1)=(1)/(2),quad (10)/(x+1)+(2)/(y-1)=(5)/(2) where x!=-1 and y!=1

( 5 ) /( x+ 1 ) - (2)/(y - 1 ) = (1)/(2) , (10)/(x + 1) + ( 2 ) /( y - 1) = ( 5)/(2), x ne - 1 and y ne 1

{:((5)/(x - 1) + (1)/(y - 2) = 2),((6)/(x - 1) - (3)/(y - 2) = 1):}

Solve the following system of equations: 5/(x+1)-2/(y-1)=1/2,\ \ \ (10)/(x+1)+2/(y-1)=5/2 , where x!=-1 and y!=1

The solution of (dy)/(dx)=(1)/(2x-y^(2)) is given by , (A) y=Ce^(-2x)+(1)/(4)x^(2)+(1)/(2)x+(1)/(4)," 4"(1)/(2)," (B) "x=Ce^(-y)+(1)/(4)y^(2)+(1)/(4)y+(1)/(2)," (C) "x=Ce^(y)+(1)/(4)y^(2)+y+(1)/(2)," (D) "x=Ce^(2y)+(1)/(2)y^(2)+(1)/(2)y+(1)/(4)

(5) / (x-1) + (1) / (y-2) = 2 (6) / (x-1) - (3) / (y-2) = 1

If (5)/(x-1)+(1)/(y-2)=2,(6)/(x-1)+(-3)/(y-2)=1 then x=……….

If range of y=(x)/(1+x^(2)),AA x varepsilon R is equal to (A)[-1,1](B)[-(1)/(2),(1)/(2)](C)R-[-(1)/(2),(1)/(2)] (D) R-[-1,1]

The x-coordinates of the vertices of a square of unit area are the roots of the equation x^2-3|x|+2=0 . The y-coordinates of the vertices are the roots of the equation y^2-3y+2=0. Then the possible vertices of the square is/are (a)(1,1),(2,1),(2,2),(1,2) (b)(-1,1),(-2,1),(-2,2),(-1,2) (c)(2,1),(1,-1),(1,2),(2,2) (d)(-2,1),(-1,-1),(-1,2),(-2,2)

The x-coordinates of the vertices of a square of unit area are the roots of the equation x^2-3|x|+2=0 . The y-coordinates of the vertices are the roots of the equation y^2-3y+2=0. Then the possible vertices of the square is/are (a)(1,1),(2,1),(2,2),(1,2) (b)(-1,1),(-2,1),(-2,2),(-1,2) (c)(2,1),(1,-1),(1,2),(2,2) (d)(-2,1),(-1,-1),(-1,2),(-2,2)