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Difference between revisions of "1980 AHSME Problems/Problem 6"
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<math>\text{(A)} \ x > \frac{1}{4} \qquad \text{(B)} \ x > 2 \qquad \text{(C)} \ x > 4 \qquad \text{(D)} \ x < \frac{1}{4}\qquad \text{(E)} \ x < 4</math>
 
<math>\text{(A)} \ x > \frac{1}{4} \qquad \text{(B)} \ x > 2 \qquad \text{(C)} \ x > 4 \qquad \text{(D)} \ x < \frac{1}{4}\qquad \text{(E)} \ x < 4</math>
  
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== Solution ==
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== Solution 1 ==
  
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<math> \sqrt{x}< 2x \\ x < 4x^2 \\ 0 < x(4x-1) \\ 0 < 4x-1 \\ 1 < 4x \\ x >\frac{1}{4} \\ \boxed{(A)}</math>
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Since both sides of this inequality are positive, we may square both sides to get <math>x < 4x^2</math>.  Since <math>x</math> is positive, we can divide both sides by <math>x</math> to get <math>1 < 4x</math>.  Dividing both sides by <math>4</math> yields <math>\frac{1}{4} < x</math>, or <math>\boxed{(\textbf{A})\ x > \frac{1}{4}}</math>.
  
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== Solution 2 (Graphing) ==
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Graphing both <math>\sqrt{x}</math> and <math>2x</math> in Quadrant I of the coordinate plane, we see that the two curves intersect at <math>\left(\frac{1}{4}, \frac{1}{2}\right)</math>, and <math>2x > \sqrt{x}</math> whenever <math>\boxed{(\textbf{A})\ x > \frac{1}{4}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1980|num-b=5|num-a=7}}
 
{{AHSME box|year=1980|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:14, 23 July 2025

Problem

A positive number $x$ satisfies the inequality $\sqrt{x} < 2x$ if and only if

$\text{(A)} \ x > \frac{1}{4} \qquad \text{(B)} \ x > 2 \qquad \text{(C)} \ x > 4 \qquad \text{(D)} \ x < \frac{1}{4}\qquad \text{(E)} \ x < 4$

Solution 1

Since both sides of this inequality are positive, we may square both sides to get $x < 4x^2$. Since $x$ is positive, we can divide both sides by $x$ to get $1 < 4x$. Dividing both sides by $4$ yields $\frac{1}{4} < x$, or $\boxed{(\textbf{A})\ x > \frac{1}{4}}$.

Solution 2 (Graphing)

Graphing both $\sqrt{x}$ and $2x$ in Quadrant I of the coordinate plane, we see that the two curves intersect at $\left(\frac{1}{4}, \frac{1}{2}\right)$, and $2x > \sqrt{x}$ whenever $\boxed{(\textbf{A})\ x > \frac{1}{4}}$.

See also

1980 AHSME (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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