Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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We can use the difference of two squares to get <math>(a^2-\tfrac{9}{4})(a^2-\tfrac{1}{4})</math>, and expand this to get <math>a^4-\tfrac{5}{2}a^2+\tfrac{9}{16}</math>. | We can use the difference of two squares to get <math>(a^2-\tfrac{9}{4})(a^2-\tfrac{1}{4})</math>, and expand this to get <math>a^4-\tfrac{5}{2}a^2+\tfrac{9}{16}</math>. | ||
− | Refactor this by completing the square to get <math>(a^2-\tfrac{5}{4})^2-1</math>, which has a minimum value of <math>-1</math>. The answer is thus <math>2019-1=\boxed{2018}</math>. | + | Refactor this by completing the square to get <math>(a^2-\tfrac{5}{4})^2-1</math>, which has a minimum value of <math>-1</math>. The answer is thus <math>2019-1=\boxed{\textbf{(B) }2018}</math>. |
==Solution 3 (using calculus)== | ==Solution 3 (using calculus)== |
Revision as of 15:33, 18 February 2019
Contents
Problem
What is the least possible value of
where
is a real number?
Solution 1
Grouping the first and last terms and two middle terms gives , which can be simplified to
. Noting that squares are nonnegative, and verifying that
for some real
, the answer is
.
Solution 2
Let . Then
becomes
.
We can use the difference of two squares to get , and expand this to get
.
Refactor this by completing the square to get , which has a minimum value of
. The answer is thus
.
Solution 3 (using calculus)
Similar to Solution 1, grouping the first and last terms and the middle terms, we get .
Letting , we get the expression
. Now, we can find the critical points of
to minimize the function:
To minimize the result, we use . Hence, the minimum is
, so
.
Note: The minimum/maximum point of a parabola occurs at
.
Solution 4
The expression is negative when an odd number of the factors are negative. This happens when or
. Plugging in
or
yields
, which is very close to
.
.
Solution 5 (using the answer choices)
Using the answer choices, we know that ,
, and
are impossible since
can be negative (as seen when
). Plug in
to see that it becomes
, so round this to
.
Video Solution
For those who want a video solution: https://www.youtube.com/watch?v=Mfa7j2BoNjI
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.