A) \[1\]
B) \[-1\]
C) \[\pi \]
D) \[-\pi \]
Correct Answer: C
Solution :
| Given that \[y=y(x)\] and \[x\cos y+y\cos x=\pi \] ...(i) |
| For \[x=0\]in (i) we get \[y=\pi \] |
| Differentiating (i) with respect to x, we get, \[-x\sin y.y'+\cos y+y'\cos x-y\sin x=0\] |
| \[\Rightarrow \,\,y'=\frac{y\sin x-\cos y}{\cos x-x\sin y}\] ...(ii) |
| \[\Rightarrow \,\,y'(0)=1\] (Using \[y(0)=\pi \]) |
| Differentiating (ii) with respect to x, we get, |
| \[\left( y'\sin x+y\cos x+\sin y.y' \right)\,\,\left( \cos x-x\sin y \right)\] |
| \[y''=\frac{-\left( -\sin x-\sin y-x\cos y.y' \right)\left( y\sin x-\cos y \right)}{{{\left( \cos x-x\sin y \right)}^{2}}}\] |
| \[\Rightarrow \,\,y''(0)=\frac{\pi \left( 1 \right)-0}{1}=\pi \]. |
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