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junior college 1 | H2 Maths
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How to do the hence part?
Date Posted: 4 years ago
Views: 287
From i) you know that ½ < x < 12
For ii),
Divide both numerator and denominator by e^x
You get (3e^(-x) + 10) / (2e^(-x) - 1) > 2
Using the result from i),
½ < e^(-x) < 12
ln ½ < -x < ln12
-ln½ > x > - ln12
ln2 > x > -ln12
∴ -ln12 < x < ln2
For ii),
Divide both numerator and denominator by e^x
You get (3e^(-x) + 10) / (2e^(-x) - 1) > 2
Using the result from i),
½ < e^(-x) < 12
ln ½ < -x < ln12
-ln½ > x > - ln12
ln2 > x > -ln12
∴ -ln12 < x < ln2
See 1 Answer
done
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clear
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Here is the link between the 2 parts. After establishing how x is replaced by e^-x, use solution from first part (x>... Etc, to e^-x>...) Then make x the subject.
Date Posted:
4 years ago
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