Question

If $ycosx+xcosy=\pi ,then{y}^{\prime \prime }\left(0\right)$ is

• A. 1
• B. $\pi$
• C. 0.0
• D. $-\pi$

Answer 1

The correct option is B $\pi$

$ycosx+xcosy=\pi$
Differentiate both sides with respect to x, we get
$-ysinx+cosx.y^{\prime }+x(-siny)y^{\prime }+cosy$
Again differentiate with respect to x
$-y"sinx-ycosx+cosx.y"+sinx.y^{\prime }-siny.y^{\prime }-x[cosy.(y^{\prime }{)}^2+siny.y"]-siny.y^{\prime }$
Putting $x=0,weget-y+y"-2siny{y}^{\prime }=0$
$y"=y+2y^{\prime }siny$
Since at $x=0,y=\pi ;(y"{)}_0=\pi$