Difference between revisions of "1957 AHSME Problems/Problem 29"
(created solution page) |
m (typo fix) |
||
(One intermediate revision by the same user not shown) | |||
Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
− | <math>\fbox{D}</math> | + | |
+ | To solve this problem, think about the graph of <math>f(x)=x^2(x^2-1)=x^2(x-1)(x+1)</math>. The function equals zero only at the values <math>-1</math>, <math>0</math>, and <math>1</math>. Because the function is a quartic polynomial with a positive leading coefficient, it will go to positive infinity as <math>x</math> tends to either positive or negative infinity. Thus, when <math>x</math> is greatly negative, the function will be positive, and so the given inequality will hold. As <math>x</math> gradually becomes more positive, it will eventually equal <math>-1</math>, when the function will equal zero. Thus, the <math>(x+1)</math> term will switch to being positive, and so the whole function will become negative, where the inequality does not hold. <math>f(x)</math> again reaches <math>0</math> when <math>x=0</math>, but here the <math>x^2</math> term does not change sign, so the function stays negative afterwards. Finally, when <math>x=1</math>, the function crosses the <math>x</math>-axis as the <math>(x-1)</math> term changes sign, and the function goes off to positive infinity. Thus, the function is positive (and thus the given inequality will hold) when <math>x \leq -1</math>, <math>x=0</math>, and <math>x \geq 1</math>, which is answer choice <math>\fbox{\textbf{(D)}}</math>. | ||
==See Also== | ==See Also== | ||
− | {{AHSME box|year=1957|num-b=28|num-a=30}} | + | {{AHSME 50p box|year=1957|num-b=28|num-a=30}} |
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:AHSME]][[Category:AHSME Problems]] | [[Category:AHSME]][[Category:AHSME Problems]] | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Latest revision as of 16:56, 25 July 2024
Problem
The relation is true only for:
Solution
To solve this problem, think about the graph of . The function equals zero only at the values , , and . Because the function is a quartic polynomial with a positive leading coefficient, it will go to positive infinity as tends to either positive or negative infinity. Thus, when is greatly negative, the function will be positive, and so the given inequality will hold. As gradually becomes more positive, it will eventually equal , when the function will equal zero. Thus, the term will switch to being positive, and so the whole function will become negative, where the inequality does not hold. again reaches when , but here the term does not change sign, so the function stays negative afterwards. Finally, when , the function crosses the -axis as the term changes sign, and the function goes off to positive infinity. Thus, the function is positive (and thus the given inequality will hold) when , , and , which is answer choice .
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.