Difference between revisions of "2010 AIME I Problems/Problem 6"
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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>. | 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>. | ||
+ | |||
+ | === Solution 4 === | ||
+ | |||
+ | Let <math>Q(x) = P(x) - (x^2-2x+2)</math>, then <math>0\le Q(x) \le (x-1)^2</math>. Therefore, <math>0\le Q(x+1) \le x^2 \Rightarrow Q(x) = Ax^2</math> for some real value A. | ||
+ | |||
+ | <math>Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}</math>. | ||
+ | |||
+ | <math>Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}</math> | ||
== See Also == | == See Also == |
Revision as of 22:12, 10 August 2015
Contents
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 equality holds when and therefore when ; 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 .
Solution 4
Let , then . Therefore, for some real value A.
.
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 |
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