Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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''Note'': We could also have used the result that minimum/maximum point of a parabola <math>y = ax^2 + bx + c</math> occurs at <math>x=-\frac{b}{2a}</math>. | ''Note'': We could also have used the result that minimum/maximum point of a parabola <math>y = ax^2 + bx + c</math> occurs at <math>x=-\frac{b}{2a}</math>. | ||
− | ==Solution 4== | + | ==Solution 4(guess with answer choices)== |
The expression is negative when an odd number of the factors are negative. This happens when <math>-2 < x < -1</math> or <math>-4 < x < -3</math>. Plugging in <math>x = -\frac32</math> or <math>x = -\frac72</math> yields <math>-\frac{15}{16}</math>, which is very close to <math>-1</math>. Thus the answer is <math>-1 + 2019 = \boxed{\textbf{(B) }2018}</math>. | The expression is negative when an odd number of the factors are negative. This happens when <math>-2 < x < -1</math> or <math>-4 < x < -3</math>. Plugging in <math>x = -\frac32</math> or <math>x = -\frac72</math> yields <math>-\frac{15}{16}</math>, which is very close to <math>-1</math>. Thus the answer is <math>-1 + 2019 = \boxed{\textbf{(B) }2018}</math>. |
Revision as of 22:05, 5 November 2021
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 the expression becomes .
We can now 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 (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: We could also have used the result that minimum/maximum point of a parabola occurs at .
Solution 4(guess with answer choices)
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 . Thus the answer is .
Solution 5 (using the answer choices)
Answer choices , , and are impossible, since can be negative (as seen when e.g. ). Plug in to see that it becomes , so round this to .
We can also see that the limit of the function is at least -1 since at the minimum, two of the numbers are less than 1, but two are between 1 and 2.
Video Solutions
https://www.youtube.com/watch?v=Lis8yKT9WXc (less than 2 minutes)
- https://youtu.be/NRa3VnjNVbw - Education, the Study of Everything
- https://www.youtube.com/watch?v=Mfa7j2BoNjI
- https://youtu.be/tIzJtgJbHGc - savannahsolver
- https://youtu.be/3dfbWzOfJAI?t=3319 - pi_is_3.14
- https://youtu.be/GmUWIXXf_uk?t=1134 ~ pi_is_3.14
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.