Question
If $$y = y\left( x \right)$$ and it follows the relation $$x\,\cos \,y + y\,\cos \,x = \pi $$ then $$y''\left( 0 \right) = $$
A.
$$1$$
B.
$$-1$$
C.
$$\pi - 1$$
D.
$$ - \pi $$
Answer :
$$\pi - 1$$
Solution :
Given that $$y = y\left( x \right)$$
and $$x\,\cos \,y + y\,\cos \,x = \pi \,.....(1)$$
For $$x=0$$ in (1) we get $$y = \pi $$
Differentiating (1) with respect to $$x,$$ we get
$$\eqalign{
& - x\,\sin \,y.y' + \cos \,y + y'\,\cos \,x - y\,\sin \,x = 0 \cr
& \Rightarrow y' = \frac{{y\,\sin \,x - \cos \,y}}{{\cos \,x - x\,\sin \,y}}\,.....(2) \cr
& \Rightarrow y'\left( 0 \right) = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Using }}y\left( 0 \right) = \pi } \right) \cr} $$
Differentiating (2) with respect to $$x,$$ we get
$$\eqalign{
& y'' = \frac{{\left( {y'\sin \,x + y\,\cos \,x + \sin \,y.y'} \right)\left( {\cos \,x - x\,\sin \,y} \right) - \left( { - \sin \,x - \sin \,y - x\,\cos \,yy'} \right)\left( {y\,\sin \,x - \cos \,y} \right)}}{{{{\left( {\cos \,x - x\,\sin \,y} \right)}^2}}} \cr
& \Rightarrow y''\left( 0 \right) = \frac{{\pi \left( 1 \right) - 1}}{1} = \pi - 1 \cr} $$