If `y=f(x)" and "ycosx+cosy=pi`, then the value of f''(0) is

Options

A. `pi`
B. `-pi`
C. 0
D. `2pi`

Answer

A

Explanation

Given , `y cos x + x cos y = pi`
On differentiating both sides w.r.t.x, we get
`rArr (dy)/(dx) = (y sin x - cos y)/(cos x - x sin y)`
Again , differentiating both sides w.r.t.x , we get
`(d^(2)y)/(dx^(2)) = ([{:("(cos x - x sin y)(y cos x + sin x" (dy)/(dx) + sin y (dy)/(dx)),(- "(y sin x - cos y)(- sin x - siny - x cos y" (dy)/(dx)):}])/((cos x - x sin y)^(2))`
At ` x = 0 , f''(0) = (1( y + - siny cos y)- (-cos )(-sin y))/((1 - 0)^(2))`
` = y - 2 sin pi cos pi `
As ` y cos x + x cos y = pi `
` therefore "At "x = 0 rArr y = pi`
Hence , ` f''(0) = pi ` .