A) \[e(x+y)=1\]
B) \[y+ex=1\]
C) \[y+x=e\]
D) None of these
Correct Answer: D
Solution :
[d] We have equation of tangent to any curve f(x) at \[({{x}_{1}},\,\,{{y}_{1}})\] is \[(y-{{y}_{1}})={{\left. \frac{dy}{dx} \right|}_{({{x}_{1}},{{y}_{1}})}}(x-{{x}_{1}})\] Given curve is \[y={{e}^{-\left| x \right|}}\] Point of intersection is \[\left( 1,\,\,\frac{1}{e} \right)\,\,at\,\,x=1,\,\,\left| x \right|=x\] So, \[y={{e}^{-x}}\Rightarrow \frac{dy}{dx}=-{{e}^{-x}}\therefore {{\left( \frac{dy}{dx} \right)}_{x=1}}=-{{e}^{-1}}\] Therefore, equation of tangent is \[y-\frac{1}{e}=\frac{-1}{e}(x-1)\Rightarrow x+ey=2\]
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