Question

The equation of the tangent to the curve $y=1-e^{x/2}$ at the point of intersection with the y- axis is

• A. x +2y =0
• B. 2x +y =0
• C. x - y =2
• D. 2x+y=0

Answer 1

The correct option is A

x +2y =0

For y -axis, x =0
$\therefore y=1-e^{\circ }=1-1=0\therefore \frac{dy}{dx}=0-\frac{1}{2}{e}^{x/2},{\left(\frac{dy}{dx}\right)}_{(0,0)}=-\frac{1}{2}$
$\therefore$ Equation of tangent
$y-0=-\frac{1}{2}(x-0)\Rightarrow x+2y=0$