A) \[-\frac{1}{{{(\pi +1)}^{2}}}\]
B) \[-\frac{2}{{{(\pi +1)}^{2}}}\]
C) \[-\frac{1}{\pi +1}\]
D) \[\frac{2}{\pi +1}\]
Correct Answer: B
Solution :
[b] \[x\text{ }sin\text{ }y+x=y\] Differentiating w.r.t. x, we get \[x(\cos y)y'+\sin y+1=y'\] ...(1) When \[x=\pi \] then \[y=\pi \]. Putting these values into (1), we get \[\pi (-1)y'+0+1=y'\] \[\Rightarrow \,\,\,y'(\pi )=\frac{1}{\pi +1}\] Differentiating (1) w.r.t. x, we get \[(\cos \,\,y)'-x(\sin y){{(y')}^{2}}+x(\cos \,y)y''+(\cos y)y'=y''\] ...(2) Putting \[x=\pi \] and \[y=\pi \] into (2), we get \[-\frac{1}{\pi +1}-0-\pi y''-\frac{1}{\pi +1}=y''\] \[\Rightarrow \,\,y''\left( 1+\pi \right)=-\frac{2}{{{\left( \pi +1 \right)}^{2}}}\]
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