A) \[-34\]
B) \[-32\]
C) \[-2\]
D) \[4\]
Correct Answer: A
Solution :
Given, \[{{x}^{2}}+{{y}^{2}}+\sin y=4\] On differentiating the equation the above equation w.r.t , we get \[2x+2y\frac{dy}{dx}+\cos y\frac{dy}{dx}=0....(i)\] \[\Rightarrow 2x+(2y+\cos y)\frac{dy}{dx}=0\] \[\Rightarrow \frac{dy}{dx}=\frac{-2x}{2y+\cos y}\] At \[(-2,0),\frac{dy}{dx}=\frac{-2\times -2}{2\times 0+\cos 0}\] \[\Rightarrow \frac{dy}{dx}=\frac{4}{0+1}\] \[\Rightarrow \frac{dy}{dx}=4....(ii)\] Again differentiating equation (i) w.r.t to x, we get \[2+2{{(\frac{dy}{dx})}^{2}}+2y\frac{{{d}^{2}}y}{d{{x}^{2}}}-\sin y{{(\frac{dy}{dx})}^{2}}+\cos y\frac{{{d}^{2}}y}{d{{x}^{2}}}=0\] \[\Rightarrow 2+(2-\sin y){{(\frac{dy}{dx})}^{2}}+(2y+\cos y)\frac{{{d}^{2}}y}{d{{x}^{2}}}=0\] \[\Rightarrow (2y+\cos y)\frac{{{d}^{2}}y}{d{{x}^{2}}}=-2-2(2-\sin y){{(\frac{dy}{dx})}^{2}}\] \[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-2-(2-\sin y){{(\frac{dy}{dx})}^{2}}}{2y+\cos y}\] Therefore, at (-2,0), \[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-2-(2-0)\times {{4}^{2}}}{2\times 0+1}\] \[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-2-2\times 16}{1}\] \[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=-34\] Hence, correct option is\[(A)-34\]
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