2025 AMC 10A Problems/Problem 9: Difference between revisions
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(A) <imath>1</imath> (B) <imath>2</imath> (C) <imath>3</imath> (D) <imath>4</imath> (E) more than <imath>4</imath> | (A) <imath>1</imath> (B) <imath>2</imath> (C) <imath>3</imath> (D) <imath>4</imath> (E) more than <imath>4</imath> | ||
Solution 1: | |||
This problem is essentially asking how many values of <imath>x</imath> satisfy <imath>f(x)=25,</imath> since the graph <imath>f(x-a)</imath> is a shift <imath>a</imath> units right from the original graph of <imath>f(x).</imath> Now we just need to determine the amount of real solutions to <imath>100x^3-300x^2+200x=25.</imath> | |||
Dividing all terms by 25, we get <imath>4x^3-12x^2+8x-1=0.</imath> Using the rational root theorem and testing, we find no rational roots. Our next method to determine the amount of roots is graphing. Notice how when <imath>x</imath> approaches infinity, <imath>f(x)</imath> also approaches infinity. When <imath>x</imath> approaches negative infinity, <imath>f(x)</imath> also approaches negative infinity. Substituting <imath>x=0</imath> gives us <imath>f(x)=-1.</imath> Substituting <imath>x=1</imath> gives us <imath>f(x)=-1</imath> also. This makes us think to see if there was any sudden bump or curve in the graph, so we test the midpoint which is <imath>1/2.</imath> Substituting <imath>x=1/2,</imath> we get <imath>1/2-3+4-1=1/2.</imath> Aha! So connecting the dots, we see there is two solutions from <imath>0\geq x \geq 1.</imath> Looking at when <imath>x<0,</imath> we see that if <imath>x</imath> is negative, the whole expression of <imath>f(x)</imath> will be negative also, since <imath>4x^3<0</imath> when <imath>x<0,</imath> <imath>-12x^2<0</imath> when <imath>x<0,</imath> <imath>8x<0</imath> when <imath>x<0,</imath> and <imath>-1<0.</imath> Adding all inequalities gives <imath>f(x)<0</imath> when <imath>x<0.</imath> Since there are only two roots satisfy the equation <imath>f(x)=25,</imath> we conclude that there only are two values of <imath>a</imath> that satisfy <imath>f(x-a) = 25.</imath> This means that the answer is <imath>\boxed {B}.</imath> | |||
~ eqb5000 | |||
==Solution== | ==Solution== | ||
Revision as of 13:49, 6 November 2025
Let 
. For how many real numbers 
does the graph of 
pass through the point 
?
(A) 
(B) 
(C) 
(D) 
(E) more than
Solution 1:
This problem is essentially asking how many values of 
satisfy 
since the graph 
is a shift units right from the original graph of 
Now we just need to determine the amount of real solutions to 
Dividing all terms by 25, we get 
Using the rational root theorem and testing, we find no rational roots. Our next method to determine the amount of roots is graphing. Notice how when approaches infinity, 
also approaches infinity. When approaches negative infinity, also approaches negative infinity. Substituting 
gives us 
Substituting 
gives us 
also. This makes us think to see if there was any sudden bump or curve in the graph, so we test the midpoint which is 
Substituting 
we get 
Aha! So connecting the dots, we see there is two solutions from 
Looking at when 
we see that if is negative, the whole expression of will be negative also, since 
when 
when 
when and 
Adding all inequalities gives 
when 
Since there are only two roots satisfy the equation we conclude that there only are two values of that satisfy 
This means that the answer is 
~ eqb5000
Solution
The problem boils down to how many real roots does the equation ![\[100(1-a)^3 - 300(1-a)^2 + 200(1-a) = 25\]](http://latex-new.aopstest.com/f/f/8/ff8338b0bd12c4ae1c49d59ea8496bc430e48d8d.png)
have? Using Descarte's Rule of Signs we find there shouldn't be any imaginary roots, so the answer is 
real roots.
