2013 A-level H1 Mathematics (8864) Question 5 Suggested Solutions

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

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

\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})


Graph of 5(iii)

Graph of 5(iii)

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

Leave a Comment

Contact Us

CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

Not readable? Change text. captcha txt

Start typing and press Enter to search