Applications-Of-Derivatives Question 440
Question: The equation of the tangent to the curve $ y={e^{-| x |}} $ at the point where the curve cuts the line $ x=1 $ is
Options:
A) $ e(x+y)=1 $
B) $ y+ex=1 $
C) $ y+x=e $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] We have equation of tangent to any curve f(x) at $ (x_1,y_1) $ is $ (y-y_1)={{. \frac{dy}{dx} |}{(x_1,y_1)}}(x-x_1) $ Given curve is $ y={e^{-| x |}} $ Point of intersection is $ ( 1,\frac{1}{e} )atx=1,| x |=x $ So, $ y={e^{-x}}\Rightarrow \frac{dy}{dx}=-{e^{-x}}\therefore {{( \frac{dy}{dx} )}{x=1}}=-{e^{-1}} $ Therefore, equation of tangent is $ y-\frac{1}{e}=\frac{-1}{e}(x-1)\Rightarrow x+ey=2 $
BETA
