Equation of the tangent to the curve $y=e^{-|x|}$ at the point where it cuts the line x = 1
Answer & explanation
Correct answer: option 1
$y=e^{-|x|}$ cut the line x = 1 at (1, 1/e)
$\left(\frac{d y}{d x}\right)_{(1,1 / e)}=\frac{-1}{e}$
Tangent $ y-\frac{1}{e}=-\frac{1}{e}(x-1)$
⇒ ey + x = 2