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, we get
`(dy)/(dx)cosx+y(-sinx)+x(-siny)(dy)/(dx)+cosy=0`
`implies(dy)/(dx)=(ysinx-cosy)/(cosx-xsiny)`
Again differentiating both sides, we get
`(d^(2)y)/(dx^(2))=`

At x = 0,
`f''(0)=(1(y+siny)-(-1)(-siny))/((1-0)^(2))=y`
As `ycosx+xcosy=pi`
`:." At "x=0impliesy=pi`
Hence, `" "f''(0)=pi`