Difference between revisions of "1993 AHSME Problems/Problem 26"
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The first term <math>\sqrt{p(x)}</math> is defined only when <math>p(x)\geq 0</math> which is the interval <math>[0,8]</math> and the second term <math>\sqrt{q(x)}</math> is defined only when <math>q(x)\geq 0</math> which is on the interval <math>[6,8]</math>, so the domain of <math>f(x)</math> is <math>[0,8] \cap [6,8] = [6,8]</math>. | The first term <math>\sqrt{p(x)}</math> is defined only when <math>p(x)\geq 0</math> which is the interval <math>[0,8]</math> and the second term <math>\sqrt{q(x)}</math> is defined only when <math>q(x)\geq 0</math> which is on the interval <math>[6,8]</math>, so the domain of <math>f(x)</math> is <math>[0,8] \cap [6,8] = [6,8]</math>. | ||
− | Now <math>p(x)</math> peaks at the midpoint of its roots at <math>x=4</math> and it decreases to 0 at <math>x=8</math>. Thus, <math>p(x)</math> is decreasing over the entire domain of <math>f(x)</math> and it obtains its maximum value | + | Now <math>p(x)</math> peaks at the midpoint of its roots at <math>x=4</math> and it decreases to 0 at <math>x=8</math>. Thus, <math>p(x)</math> is decreasing over the entire domain of <math>f(x)</math> and it obtains its maximum value at the left boundary <math>x=6</math> and <math>\sqrt{p(x)}</math> does as well. On the other hand <math>q(x)</math> obtains its minimum value of <math>q(x)=0</math> at the left boundary <math>x=6</math> and <math>\sqrt{q(x)}</math> does as well. Therefore <math>\sqrt{p(x)}-\sqrt{q(x)}</math> is maximized at <math>x=6</math>. (If this seems a little unmotivated, a quick sketch of the two parabolic-like curves makes it clear where the distance between them is greatest). |
The value at <math>x=6</math> is <math>\sqrt{ 6\cdot 2 } = 2\sqrt{3}</math> and the answer is <math>\fbox{D}</math>. | The value at <math>x=6</math> is <math>\sqrt{ 6\cdot 2 } = 2\sqrt{3}</math> and the answer is <math>\fbox{D}</math>. |
Revision as of 23:08, 27 May 2021
Contents
Problem
Find the largest positive value attained by the function , a real number.
Solution
We can rewrite the function as and then factor it to get . From the expressions under the square roots, it is clear that is only defined on the interval .
The factor is decreasing on the interval. The behavior of the factor is not immediately clear. But rationalizing the numerator, we find that , which is monotonically decreasing. Since both factors are always positive, is also positive. Therefore, is decreasing on , and the maximum value occurs at . Plugging in, we find that the maximum value is .
Solution 2
Note the form of the function is where and each describe a parabola. Factoring we have and .
The first term is defined only when which is the interval and the second term is defined only when which is on the interval , so the domain of is .
Now peaks at the midpoint of its roots at and it decreases to 0 at . Thus, is decreasing over the entire domain of and it obtains its maximum value at the left boundary and does as well. On the other hand obtains its minimum value of at the left boundary and does as well. Therefore is maximized at . (If this seems a little unmotivated, a quick sketch of the two parabolic-like curves makes it clear where the distance between them is greatest).
The value at is and the answer is .
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 |
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