Q. The equation of the tangent top the curve $y = 1 - e^{x/2}$ at the point on intersection with the $y$-axis is
A) $x + 2y = 0$
B) $2x + y = 0$
C) $x - y = 2$
D) None of these
Explanation
(a)
For $y$-axis, $x\text{=}0$
$\therefore\ \ y = 1 - e^{0} = 1 - 1 = 0$
$\Rightarrow \ \frac{dy}{dx} = 0 - \frac{1}{2}\ e^{x/2}\ \Rightarrow \ \left( \frac{dy}{dx} \right)_{(0,\ 0)} = - \frac{1}{2}$
$\therefore$ Equation of tangent is
$y - 0 = - \frac{1}{2}(x - 0)$
$\Rightarrow \ \ \ x + 2y = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.