A curve passes through the point (0,1) and the gradient at (x,y) on it is `y(xy-1)`. The equation of the curve isA. `y(x-1)=1`B. `y(x+1)=1`C. `x(y+1)=1`D. `x(y-1)=1`

A curve passes through the point (0,1) and the gradient at (x,y) on it

1.	A curve passes through the point (0,1) and the gradient at (x,y) on it is `y(xy-1)`. The equation of the curve isA. `y(x-1)=1`B. `y(x+1)=1`C. `x(y+1)=1`D. `x(y-1)=1`
Answer» Correct Answer - a We have , `(dy)/(dx) = y (xy -1)` ` rArr dy = xy^(2)dx - y dx` ` rArr y dx + dy = xy^(2) dx rArr (y" "dx+dy)/(y^(2)) = xdx` ` rArr (ye^(-x) dx + e^(-x) dy)/(y^(2)) = xe^(-x) dx` ` :. -d ((e^(-x))/y) = xe^(-x) dx` On integrating , we get ` - (e^(-x)/y ) = - xe^(-x) - e^(-x)-C` ` rArr 1/y = x+1 + Ce^(x) " " ` ...(i) Since , it passes through (0,1) Therefore , `1 = 1+C rArr C = 0 ` On putting C = 0 in Eq . (i) , we get ` 1/y = x+1 rArr y (x+1) =1`

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A curve passes through the point (0,1) and the gradient at (x,y) on it is `y(xy-1)`. The equation of the curve isA. `y(x-1)=1`B. `y(x+1)=1`C. `x(y+1)=1`D. `x(y-1)=1`

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