A) \[\left( -\frac{1}{e},\frac{1}{{{e}^{2}}} \right)\]
B) \[\left( \frac{1}{e},\frac{1}{{{e}^{2}}} \right)\]
C) \[\left( \frac{1}{e},-\frac{1}{{{e}^{2}}} \right)\]
D) \[\left( -\frac{1}{e},-\frac{1}{{{e}^{2}}} \right)\]
Correct Answer: A
Solution :
\[{{e}^{y}}=xy=e\]differentiate w.r.t. x \[{{e}^{y}}\frac{dy}{dx}+x\frac{dy}{dx}+y=0\] \[\frac{dy}{dx}(x+{{e}^{y}})=-y,{{\left. \frac{dy}{dx} \right|}_{(0,1)}}=-\frac{1}{e}\] again differentiate w.r.t. x \[{{e}^{y}}.\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}.{{e}^{y}}.\frac{dy}{dx}+x.\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}+\frac{dy}{dx}=0\] \[(x+{{e}^{y}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{2}}.{{e}^{y}}+2\frac{dy}{dx}=0\] \[e\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{1}{{{e}^{2}}}e+2\left( -\frac{1}{e} \right)=0\] \[\therefore \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{1}{{{e}^{2}}}\]
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