All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
\mathrm{ln} e^{2-2x} = \mathrm{ln}2 + \mathrm{ln}e^{-x}
2 - 2x = \mathrm{ln} 2 - x
x = 2 - ln2

(ii)
\frac{dy}{dx} = -2 e^{2-2x} + 2 e ^{-x} = 0
e^{2-2x} = e^{-x}
x = 2
y = -e^{-2}
\therefore, \mathrm{required~coordinates~is~} (2, -e^{-2})

(iii)

Graph of 5(iii)
Graph of 5(iii)

(iv)
Area = \int_0^1 e^{2-2x} - 2e^{-x} dx = 1.93 units^{2} using graphing calculator.

KS Comments

Students must show all the requirements of the graph, and please use a GC to check it. Lastly, since (iv) did not require exact answers, we can easily use a GC

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