Difference between revisions of "1994 AHSME Problems/Problem 12"
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==Solution== | ==Solution== | ||
We simplify step by step as follows: <cmath>\begin{align*}(i-i^{-1})^{-1}&=\frac{1}{i-i^{-1}}\\&=\frac{1}{i-\frac{1}{i}}\\&=\frac{1}{\left(\frac{i^2-1}{i}\right)}\\&=\frac{i}{i^2-1}\\&=\boxed{\textbf{(D) }-\frac{i}{2}.}\end{align*}</cmath> | We simplify step by step as follows: <cmath>\begin{align*}(i-i^{-1})^{-1}&=\frac{1}{i-i^{-1}}\\&=\frac{1}{i-\frac{1}{i}}\\&=\frac{1}{\left(\frac{i^2-1}{i}\right)}\\&=\frac{i}{i^2-1}\\&=\boxed{\textbf{(D) }-\frac{i}{2}.}\end{align*}</cmath> | ||
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| + | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
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| + | ==See Also== | ||
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| + | {{AHSME box|year=1994|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:34, 9 January 2021
Problem
If 
, then 
Solution
We simplify step by step as follows: 
--Solution by TheMaskedMagician
See Also
| 1994 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
