Difference between revisions of "2013 MPFG Problem 17"
(Created page with "==Problem== Let f be the function defined by <math>f(x) = -2 sin(\pi x)</math> . How many values of x such that <math> -2 \leq x \leq 2</math> satisfy the equation <math>f(f(f...") |
(→Solution 1) |
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We first create the graph of <math>f(x)</math> by transforming a regular <math>sin</math> graph by scaling down <math>x</math> by <math>\pi</math> and scaling up <math>y</math> by <math>-2</math>. | We first create the graph of <math>f(x)</math> by transforming a regular <math>sin</math> graph by scaling down <math>x</math> by <math>\pi</math> and scaling up <math>y</math> by <math>-2</math>. | ||
| − | [ | + | [[File:Graph_of_f(x).jpg|400px|center]] |
We observe that the domain & range of <math>f(x)</math> is restricted in <math>[-2,2]</math> and the graph is restricted in a <math>2x2</math> square. Trying out <math>f(f(x))</math>, we see that every <math>\frac{1}{4}</math> of a cycle of <math>f(x)</math> ranges over <math>[0,\pm 2]</math>, which means a cycle for <math>(f(x))</math> will have <math>4</math> cycles for <math>f(f(x))</math>. <math>f(x)</math> and <math>f(f(x))</math> have two crossing points in each small cycle of <math>f(f(x))</math> | We observe that the domain & range of <math>f(x)</math> is restricted in <math>[-2,2]</math> and the graph is restricted in a <math>2x2</math> square. Trying out <math>f(f(x))</math>, we see that every <math>\frac{1}{4}</math> of a cycle of <math>f(x)</math> ranges over <math>[0,\pm 2]</math>, which means a cycle for <math>(f(x))</math> will have <math>4</math> cycles for <math>f(f(x))</math>. <math>f(x)</math> and <math>f(f(x))</math> have two crossing points in each small cycle of <math>f(f(x))</math> | ||
| − | [ | + | [[File:Graph_of_f(f(x)).jpg|400px|center]] |
As such, <math>f(f(f(x))</math> will have <math>2\cdot4^{2} = 32</math> cycles in the domain <math>[-2,2]</math>. Each has <math>2</math> points that cross with <math>f(x)</math>. However, at <math>3</math> special locations, the crossing points of two consecutive <math>f(f(f(x)))</math> cycles overlapped (at <math>x=-1, 0, 1</math>). | As such, <math>f(f(f(x))</math> will have <math>2\cdot4^{2} = 32</math> cycles in the domain <math>[-2,2]</math>. Each has <math>2</math> points that cross with <math>f(x)</math>. However, at <math>3</math> special locations, the crossing points of two consecutive <math>f(f(f(x)))</math> cycles overlapped (at <math>x=-1, 0, 1</math>). | ||
As such, number of total points is <math>2\cdot32-3 = \boxed{61}</math> | As such, number of total points is <math>2\cdot32-3 = \boxed{61}</math> | ||
Revision as of 01:34, 21 August 2025
Problem
Let f be the function defined by 
. How many values of x such that 
satisfy the equation 
?
Solution 1
We first create the graph of 
by transforming a regular 
graph by scaling down 
by 
and scaling up 
by 
.
.jpg)
We observe that the domain & range of is restricted in ![$[-2,2]$](http://latex.artofproblemsolving.com/f/6/0/f60010fd01c6801d1b857f1acefb200f1ee8c682.png)
and the graph is restricted in a 
square. Trying out 
, we see that every 
of a cycle of ranges over ![$[0,\pm 2]$](http://latex.artofproblemsolving.com/a/7/b/a7b7dad4fb35e4c2f4d7b9c767ddb8709ebf2446.png)
, which means a cycle for 
will have 
cycles for . and have two crossing points in each small cycle of
).jpg)
As such, 
will have 
cycles in the domain . Each has 
points that cross with . However, at 
special locations, the crossing points of two consecutive 
cycles overlapped (at 
).
As such, number of total points is 
