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2025 AMC 10A Problems/Problem 9: Difference between revisions

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~[[User:grogg007|grogg007]]
~[[User:grogg007|grogg007]]


===Alternate Solution===
===Alternative Solution===
Since we are trying to find the number of roots of <imath>-100a^3+100a-25</imath> (which is the same as the number of roots of <imath>4a^3-4a+1</imath> because we divided by a constant, namely <imath>-25</imath>), we can sketch a graph of this function, which we will call <imath>g(x)</imath>. We find that the function is strictly decreasing when <imath>a<-1</imath>, <imath>g(0)=g(1)=1</imath>, and the function is strictly increasing when <imath>a>1</imath>. We immediately see that since <imath>g(-1)</imath> is negative, there is a root between <imath>-1</imath> and <imath>0</imath> by the Intermediate Value Theorem. From here, we use the first derivative test as follows. Taking the derivative, we have <imath>g'(x)=12a^2-4</imath>. We solve for the roots of this equation, obtaining <imath>\pm \frac{\sqrt{3}}{3}</imath>. We know that at these two values, the function will either reach a local minimum or maximum if the second derivative isn't <imath>0</imath>. Proceeding by the second derivative rule, we have that <imath>g''(x)=24a</imath> is positive at <imath>a=\frac{\sqrt{3}}{3}</imath> and negative at <imath>a=-\frac{\sqrt{3}}{3}</imath>, making them local minimums and local maximums, respectively. We don't need to care about the latter case because the local maximum doesn't give any more roots besides the one we've already found. Considering the first case when <imath>a=\frac{\sqrt{3}}{3}</imath>, we must find the sign of <imath>g(a)</imath>. Using some approximations, we have that  
Since we are trying to find the number of roots of <imath>-100a^3+100a-25</imath> (which is the same as the number of roots of <imath>4a^3-4a+1</imath> because we divided by a constant, namely <imath>-25</imath>), we can sketch a graph of this function, which we will call <imath>g(x)</imath>. We find that the function is strictly decreasing when <imath>a<-1</imath>, <imath>g(0)=g(1)=1</imath>, and the function is strictly increasing when <imath>a>1</imath>. We immediately see that since <imath>g(-1)</imath> is negative, there is a root between <imath>-1</imath> and <imath>0</imath> by the Intermediate Value Theorem. From here, we use the first derivative test as follows. Taking the derivative, we have <imath>g'(x)=12a^2-4</imath>. We solve for the roots of this equation, obtaining <imath>\pm \frac{\sqrt{3}}{3}</imath>. We know that at these two values, the function will either reach a local minimum or maximum if the second derivative isn't <imath>0</imath>. Proceeding by the second derivative rule, we have that <imath>g''(x)=24a</imath> is positive at <imath>a=\frac{\sqrt{3}}{3}</imath> and negative at <imath>a=-\frac{\sqrt{3}}{3}</imath>, making them local minimums and local maximums, respectively. We don't need to care about the latter case because the local maximum doesn't give any more roots besides the one we've already found. Considering the first case when <imath>a=\frac{\sqrt{3}}{3}</imath>, we must find the sign of <imath>g(a)</imath>. Using some approximations, we have that  
<cmath>g(\frac{\sqrt{3}}{3})</cmath>
<cmath>g(\frac{\sqrt{3}}{3})</cmath>
Line 45: Line 45:


~minor LaTeX edits by i_am_not_suk_at_math
~minor LaTeX edits by i_am_not_suk_at_math


==Solution 2==
==Solution 2==

Revision as of 00:30, 8 November 2025

Let $f(x) = 100x^3 - 300x^2 + 200x$. For how many real numbers $a$ does the graph of $y = f(x - a)$ pass through the point $(1, 25)$?

(A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) more than $4$

Video Solution

https://youtu.be/l1RY_C20Q2M

Solution 1

Substitute $1 - a$ for $x$ and set this expression equal to $25.$ The problem boils down to finding how many real roots \[100(1-a)^3 - 300(1-a)^2 + 200(1-a) = 25\] has. We further simplify this expression and create a function $f(x):$

\[f(x) = -100a^3 + 100a - 25\]

Using Descarte's Rule of Signs we get:

Sign changes for $f(x)$ (possible number of positive roots): 2

\[f(-x) = +100a^3 - 100a - 25\]

Sign changes for $f(-x)$ (possible number of negative roots): 1


Possibilities for roots:

1) $2$ positive roots, $1$ negative root

2) $0$ positive roots, $1$ negative root, $2$ imaginary roots


So which one is it? We know if the function changes sign between an interval, then a root exists in that interval. From $a = 0$ to $\frac{1}{2},$ the function changes sign because $f(0) = -25$ while $f(\frac{1}{2}) = +\frac{25}{2}$, so a positive root exists. This eliminates the second possibility, implying that there must be $2$ positive and $1$ negative roots. So the answer is $2 + 1 = \boxed{\textbf{(C) } 3}.$


~grogg007

Alternative Solution

Since we are trying to find the number of roots of $-100a^3+100a-25$ (which is the same as the number of roots of $4a^3-4a+1$ because we divided by a constant, namely $-25$), we can sketch a graph of this function, which we will call $g(x)$. We find that the function is strictly decreasing when $a<-1$, $g(0)=g(1)=1$, and the function is strictly increasing when $a>1$. We immediately see that since $g(-1)$ is negative, there is a root between $-1$ and $0$ by the Intermediate Value Theorem. From here, we use the first derivative test as follows. Taking the derivative, we have $g'(x)=12a^2-4$. We solve for the roots of this equation, obtaining $\pm \frac{\sqrt{3}}{3}$. We know that at these two values, the function will either reach a local minimum or maximum if the second derivative isn't $0$. Proceeding by the second derivative rule, we have that $g''(x)=24a$ is positive at $a=\frac{\sqrt{3}}{3}$ and negative at $a=-\frac{\sqrt{3}}{3}$, making them local minimums and local maximums, respectively. We don't need to care about the latter case because the local maximum doesn't give any more roots besides the one we've already found. Considering the first case when $a=\frac{\sqrt{3}}{3}$, we must find the sign of $g(a)$. Using some approximations, we have that \[g(\frac{\sqrt{3}}{3})\] \[=\frac{4 \sqrt{3}}{9} - \frac{4 \sqrt{3}}{3} + 1\] \[\approx \frac{6.9}{9} - \frac{6.9}{3} +1\] \[= \frac{23}{30} - \frac{69}{30} + 1\] \[= -\frac{16}{30}\] \[<0.\] By the Intermediate Value Theorem again, we see that there is one root between $0$ and $\frac{\sqrt{3}}{3}$ and one root between $\frac{\sqrt{3}}{3}$ and $1$. Thus, we have a total of $1+2=3$ roots. Therefore, the answer is $\boxed{\textbf{(C) 3}}$

~scjh999999

~minor LaTeX edits by i_am_not_suk_at_math

Solution 2

This problem is essentially asking how many values of $x$ satisfy $f(x)=25,$ since the graph $f(x-a)$ is a shift $a$ units right from the original graph of $f(x).$ Now we just need to determine the amount of real solutions to $100x^3-300x^2+200x=25.$

Dividing all terms by 25, we get $4x^3-12x^2+8x-1=0.$ 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 $x$ $\to$ $+\infty,$ $f(x)$ $\to$ $+\infty.$ When $x$ $\to$ $-\infty,$ $f(x)$ $\to$ $-\infty.$ Substituting $x=0$ gives us $f(x)=-1.$ Substituting $x=1$ gives us $f(x)=-1$ 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 $1/2.$ Substituting $x=1/2,$ we get $1/2-3+4-1=1/2.$ Aha! Connecting the dots, we see there are two solutions from $0\leq x \leq 1.$ Also, $f(x)$ must continue increasing after $x>1,$ which tells us there is a third root when $x>1.$ Now we can conclude that the problem has 3 real roots, meaning that 3 values of a satisfy the problem. The answer is $\boxed {C}.$

~ eqb5000/Esteban Q.

Solution 4 (Graphing)

We can factor $f(x)$ as $100x(x-1)(x-2)$. We just need to check that if the local maximum between $0$ and $1$ is below or above 25. Observe that $f(1/2)>25$, so the graph will cross the line $y=25$ 3 times, $\boxed {C}.$

~gimmeaworkingusername

Somewhat cheese (not recommended)

Instead of plugging in a value between $0$ and $1,$ we can assume that the equation will reach a value above $25$ between those numbers. Thus, the graph crosses $y=25$ 3 times, $\boxed {C}.$

~dodobird150

Chinese Video Solution

https://www.bilibili.com/video/BV1bj2uBxEXo/

~metrixgo

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution (In 1 Min)

https://youtu.be/zvB7LuEBbww?si=OJ9H-Wkkx3WUxTU5 ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU?si=MPPNaoCB_Bqo918z&t=1132 ~MK

Video Solution by Daily Dose of Math

https://youtu.be/gPh9w3X3QSw?si=FhsZ3VSzFXKEO5pf&t=749 ~Thesmartgreekmathdude

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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