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2000 AMC 12 Problems/Problem 11

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Problem

Two non-zero real numbers, a and b, satisfy ab = a - b. Which of the following is a possible value of \frac {a}{b} + \frac {b}{a} - ab?

\text{(A)} \ - 2 \qquad \text{(B)} \ \frac { - 1}{2} \qquad \text{(C)} \ \frac {1}{3} \qquad \text{(D)} \ \frac {1}{2} \qquad...

Solution

\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab....

Alternatively, we could test simple values, like (a,b)=\left(1, \frac{1}{2}\right), which would yield \frac {a}{b} + \frac {b}{a} - ab=2.

See also

2000 AMC 12 (ProblemsResources)
Preceded by
Problem 10
Followed by
Problem 12
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
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