The equation of a curve is $y = \mathrm{e}^{-2x}\tan x$. Show that \(\frac{\mathrm{d}y}{\math

Question

The equation of a curve is $y = \mathrm{e}^{-2x}\tan x$. Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-2x}(1 - \tan x)^2$.[3] Find, in terms of $\pi$, the $x$-coordinate of the stationary point for $0 < x < \frac{\pi}{2}$.[2] Explain why $y$ is never decreasing.[2] What does your answer to part (c) imply about the stationary point found in part (b)? Explain your answer. [2] For parts (c) and (d), consider the interval $0<x<\pi/2$.

Accepted Answer

(a) $\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{e}^{-2x}(1-\tan x)^2$ (shown). (b) $x=\frac\pi4$ (c) $\frac{\mathrm{d}y}{\mathrm{d}x}\ge0$ on the domain. (d) Stationary point of inflexion; no gradient sign change.

2025 PHS P2 Q8

2025 PHS P2 Q8

Solution:

Solution locked