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Difference between revisions of "1982 AHSME Problems/Problem 19"

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int yMin = -2;
 
int yMin = -2;
 
int yMax = 4;
 
int yMax = 4;
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int numRays = 24;
 
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//Draws a polar grid that goes out to a number of circles
 
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//equal to big, with numRays specifying the number of rays:
 
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void polarGrid(int big, int numRays)
 
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{
 
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  for (int i = 1; i < big+1; ++i)
 
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  {
 
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    draw(Circle((0,0),i), gray+linewidth(0.4));
 
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  }
 
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  for(int i=0;i<numRays;++i)
 
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  draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
 
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}
 
  
 
//Draws the horizontal gridlines
 
//Draws the horizontal gridlines

Revision as of 06:50, 13 September 2021

Problem

Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$. The sum of the largest and smallest values of $f(x)$ is

$\textbf {(A)}\ 1 \qquad \textbf {(B)}\ 2 \qquad \textbf {(C)}\ 4 \qquad \textbf {(D)}\ 6 \qquad \textbf {(E)}\ \text{none of these}$

Solution

Note that at $x=2,3,4,$ one of the three absolute values is equal to $0.$

Without using absolute values, we rewrite $f(x)$ as a piecewise function: \[f(x) = \begin{cases} (x-2)+(4-x)-(6-2x) & \mathrm{if} \ 2\leq x<3 \\ (x-2)+(4-x)-(2x-6) & \mathrm{if} \ 3\leq x<4 \\ (x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 \end{cases},\] which simplifies to \[f(x) = \begin{cases} 2x-4 & \mathrm{if} \ 2\leq x<3 \\ -2x+8 & \mathrm{if} \ 3\leq x<4 \\ 0 & \mathrm{if} \ 4\leq x\leq8 \end{cases}.\] The graph of $y=f(x)$ is shown below. [asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 10; int yMin = -2; int yMax = 4; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/16,i)--(3/16,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/16)--(i,3/16), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),E); label("$y$",(0,yMax),N); pair A[]; A[0] = (2,0); A[1] = (3,2); A[2] = (4,0); A[3] = (8,0); draw(A[0]--A[1]--A[2]--A[3],red+linewidth(1.5)); for(int i = 0; i <= 3; ++i) { dot(A[i],red+linewidth(4.5)); } label("$(2,0)$",A[0],(0,-1.5),UnFill); label("$(3,2)$",A[1],(0,1.5),UnFill); label("$(4,0)$",A[2],(0,-1.5),UnFill); label("$(8,0)$",A[3],(0,-1.5),UnFill); [/asy] The largest value of $f(x)$ is $2,$ and the smallest value of $f(x)$ is $0.$ So, their sum is $\boxed{\textbf {(B)}\ 2}.$

~MRENTHUSIASM

See Also

1982 AHSME (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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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