Question

The angle at which curve $y=k{e}^{kx}$ intersects the y axis is

• A. $ta{n}^{-1}\left(k^2\right)$
• B. $co{t}^{-1}\left(k^2\right)$
• C. $si{n}^{-1}\frac{1}{k^2}$
• D. None of these

Answer 1

The correct option is B $co{t}^{-1}\left(k^2\right)$

y = $k{e}^{kx}$ cuts the y-axis at x = 0
$\frac{dy}{dx}=k^2{e}^{kx}$
at x = 0
$\frac{dy}{dx}=k^2$
$\alpha =ta{n}^{-1}\left(k^2\right)$ this is angle made by x- axis,so angle made by y-axis is $\frac{\pi }{2}-ta{n}^{-1}\left(k^2\right)$
or $co{t}^{-1}\left(k^2\right)$