The equation of a curve is $y=e^{-2x}\tan x$. Show that \(\dfrac{dy}{dx}=e^{-2x}(1-\tan x)^2\

Question

The equation of a curve is $y=e^{-2x}\tan x$. Show that $\dfrac{dy}{dx}=e^{-2x}(1-\tan x)^2$.[3] Find, in terms of $\pi$, the $x$-coordinate of the stationary point for $0<x<\dfrac{\pi}{2}$.[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) $\dfrac{dy}{dx}=e^{-2x}(1-\tan x)^2$ (b) $x=\dfrac{\pi}{4}$ (c) $y$ is never decreasing because $\dfrac{dy}{dx}\geq0$ wherever the curve is defined. (d) The stationary point is a stationary point of inflexion, not a maximum or minimum.

2025 PHS PRELIMS P2 Q8

2025 PHS PRELIMS P2 Q8

Solution:

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