Question

The tangent to the curve $y=e^{kx}$ at the point $(0,1)$ meets the $x-$axis at $(a,0)$ where $a\in [-2,-1]$, then $k\in$

• A. $[-\frac{1}{2},0]$
• B. $[-1,-\frac{1}{2}]$
• C. $[0,1]$
• D. $[\frac{1}{2},1]$

Answer 1

The correct option is D $[\frac{1}{2},1]$

$\frac{dy}{dx}=k{e}^{kx}$
At $(0,1)$ slope$=k$
$\therefore$ At $(0,1)$ Equation of the tangent is $y-1=kx$.
Point of intersection with $x-$ axis is $x=-\frac{1}{k}$
$\Rightarrow a=\frac{-1}{k}$, where $-2\le -\frac{1}{k}\le -1$
$\therefore \frac{1}{2}\le k\le 1$