The equation of the tangent to the curve $y = e^{- | x |}$ at the point where the curve cuts the line $x = 1$ is

x + y = e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e (x + y) = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y + e x = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + e y = 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $x + e y = 2$
$y = e^{- x}$
$y^{'} = \frac{- 1 e^{- x}}{e^{- x}}$
$y^{'} = - e^{- 1} a t x = 1$
$y = e^{- 1} a t x = 1$
$(y - e^{- 1}) = - e^{- 1} (x - 1)$
$x + e y = 2$

Suggest Corrections

Similar questions

y = e^{- | x |}

y = e^{- x}

y = e^{- x}

y = e^{- 2 | x |}

x = \frac{1}{2}

y = e x log x

y = \frac{log x}{e x}

log e = 1

Join BYJU'S Learning Program