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)

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