The slope of the tangent at `(x , y)` to a curve passing through `(1,pi/4)` is given by `y/x-cos^2(y/x),` then the equation of the curve is

Options

A. ` y = tan^(-1)[ log (e/x)]`
B. ` y = x tan ^(-1)[ log (x/e)]`
C. ` y = x tan ^(-1)[ log (e/x)]`
D. None of these

Answer

c

Explanation

According to the given condition
` (dy)/(dx) = y/x - cos^(2) (y/x)`
On putting y = vx
` rArr (dy)/(dx) = v+ x (dv)/(dx) , ` we get
` v + x (dv)/(dx) = v - cos^(2) v`
` rArr (dv) /(cos^(2)v) = - (dx)/x rArr sec^(2) v " dv" = (-1)/x dx`
On integrating both sides , we get
`tan v = - log x + log C`
` rArr tan (y/x) = - log x + log C " " [ :' v = y //x ] `
Since , this curve is passing through `(1,pi/4)`
` tan(pi/4) = - log 1 + log C rArr log C = 1 `
` :. tan (y/x) =- log x +1`
` rArr tan(y/x) = - log x + log e`
` rArr y = tan^(-1) [ log (e/x)]`