The equation of the curve passing through the point `(1,pi/4)` and having a slope of tangent at any point (x,y) as `y/x - cos^2(y/x)` isA. ` x = e^(1+tan(y/x))`B. ` x = e^(1-tan(y/x))`C. ` x = e^(1+tan(x/y))`D. ` x = e^(1-tan(x/y))`

The equation of the curve passing through the point `(1,pi/4)` and hav

1.	The equation of the curve passing through the point `(1,pi/4)` and having a slope of tangent at any point (x,y) as `y/x - cos^2(y/x)` isA. ` x = e^(1+tan(y/x))`B. ` x = e^(1-tan(y/x))`C. ` x = e^(1+tan(x/y))`D. ` x = e^(1-tan(x/y))`
Answer» Correct Answer - c Given, `(dy)/(dx) = y/x - (cos^(2)) (y/x)` ` rArr (x-dy-y dx)/x = - (-cos^(2). y/x)dx` ` sec^(2) (y/x) ((x-dy-ydx)/(x^(2))) = -(dx)/2` `rArr sec^(2). y/2 .d(y/x) = -(dx)/x` On integrating both sides we get ltbRgt ` rArr tan. y/x = - log x+C` when x = 1, `y = pi/4 rArr C = 1` ` :. tan.(y/x) = 1- log x rArr x = e^(1-tan.(y/x))`

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The equation of the curve passing through the point `(1,pi/4)` and having a slope of tangent at any point (x,y) as `y/x - cos^2(y/x)` isA. ` x = e^(1+tan(y/x))`B. ` x = e^(1-tan(y/x))`C. ` x = e^(1+tan(x/y))`D. ` x = e^(1-tan(x/y))`

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