Difference between revisions of "2010 AIME I Problems/Problem 6"
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<math>x - 2 \le Q'(x) \le 2x - 3</math> | <math>x - 2 \le Q'(x) \le 2x - 3</math> | ||
− | For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However, <math>\lim_{x \to 1} x - 2 | + | For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However, |
+ | |||
+ | <math>\lim_{x \to 1} x - 2 =</math> <math>lim_{x \to 1}</math> <math>2x - 3 = -1</math>, and thus by the [[sandwich theorem]] <math>lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>. | ||
== See also == | == See also == |
Revision as of 17:51, 11 March 2012
Problem
Let be a quadratic polynomial with real coefficients satisfying for all real numbers , and suppose . Find .
Solution
Solution 1
Let , . Completing the square, we have , and , so it follows that for all (by the Trivial Inequality).
Also, , so , and obtains its minimum at the point . Then must be of the form for some constant ; substituting yields . Finally, .
Solution 2
It can be seen that the function must be in the form for some real and . This is because the derivative of is , and a global minimum occurs only at (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at ). Substituting and we obtain two equations:
Solving, we get and , so . Therefore, .
Solution 3
Let ; note that . Setting , we find that and therefore ; this is true iff , so .
Let ; clearly , so we can write , where is some linear function. Plug into the given inequality:
, and thus
For all ; note that the inequality signs are flipped if , and that the division is invalid for . However,
, and thus by the sandwich theorem ; by the definition of a continuous function, . Also, , so ; plugging in and solving, . Thus , and so .
See also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |