1982 AHSME Problems/Problem 19

Revision as of 20:17, 12 September 2021 by MRENTHUSIASM (talk | contribs) (Solution)

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 simplify 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(220);  import TrigMacros;  rr_cartesian_axes(-2,10,-2,4,useticks=true);  pair A[]; A[0] = (0,0); A[1] = (2,0); A[2] = (3,2); A[3] = (4,0); A[4] = (8,0);  draw(A[0]--A[1]--A[2]--A[3]--A[4],red+linewidth(1.5));  for(int i = 0; i <= 4; ++i) { dot(A[i],red+linewidth(4.5));  }  label("$(0,0)$",A[0],(0,-1.5),UnFill); label("$(2,0)$",A[1],(0,-1.5),UnFill); label("$(3,2)$",A[2],(0,1.5),UnFill); label("$(4,0)$",A[3],(0,-1.5),UnFill); label("$(8,0)$",A[4],(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 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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