Euler's Eyes
2026-09-26
FUVEST 2024 – 2nd phase: Mathematics solved
Here are the resolutions for M01 to M06, with the reasoning for each item and verification of the results. Use the official notebook, 2nd day to consult the statements and drawings; Below we only return the data necessary for the calculations.
The explanations are copyright. The checking of answers uses the official FUVEST guide, pages 40 to 48, available in the collection of 2024. M02 requires attention: the printed data is inconsistent and the guide accepts different approaches.
Template
| Question | Theme | Answers |
|---|---|---|
| M01 | Mobiles and systems | a) top 64; empty circles: 4, 11, 8, 8, 16, from left to right; b) 27; c) $5/4$, $35/12$, $7/4$, $7/6$ |
| M02 | Conditional probability | Incompatible data: preferences add up to 110%; see the discussion and calculations conditioned on an explicit hypothesis |
| M03 | Lines and parabola | a) $(-1,1)$ e $(3,9)$; b) $(28/25,21/25)$; c) $B=(5/2,25/4)$, $C=(3/2,9/4)$ |
| M04 | Exponentials | a) $2$ m; b) $9\sqrt2/4$ m; c) $2\log_2(2+\sqrt3)$ m |
| M05 | Solids of revolution | a) $16$ area units; b) $128\pi/3$ volume units; c) $32\sqrt2\pi/3$ volume units |
| M06 | Floor function | a) $10/3$; b) $[5/2,3)$; c) $[1,\sqrt2)\cup[\sqrt5,\sqrt6)\cup[3,\sqrt{10})$ |
M01: How to read a mathematical mobile
See the three drawings on page 3 of the test PDF. The rule is $ac=bd$ and $x=ac+bd$; a branch without a number has coefficient 1. In calculations, the value of a substructure includes the circles it supports.
a) Mobile I
Start at the bottom left end. The blank circle balances the 4 circle, so 4 is also valid. The set hanging below the circle 3 is worth $4+4=8$. Including 3 itself, this branch is worth $3+8=11$; therefore the circle to its right receives 11.
Now go up a level. The structure below the circle 10 is worth $11+11=22$. The left side of the main support totals
$$10+22=32.$$The right side must also be worth 32. Its first division has two equal branches, with value 16: the circle on the right receives 16, and the pair on the left adds 16. Each circle in this pair receives 8.
Thus, the triangle at the top receives $32+32=\boxed{64}$. A direct conference consists of adding up all the circles:
$$10+3+4+4+11+8+8+16=64.$$Caution: the circle 3 does not alone represent the entire branch; there are still two circles hanging on it.
b) Mobile II
Each fork divides the value into two equal parts. Starting from 912, the values of the successive branches are
$$912\longrightarrow456\longrightarrow228\longrightarrow114\longrightarrow57.$$Circle $2n+3$ is on the last level. Therefore,
$$2n+3=57\quad\Rightarrow\quad n=\boxed{27}.$$Replacing: $2\cdot27+3=57$, and four duplications recovers 912.
c) Mobile III
Name the four circles, from left to right, $x,y,z,w$. The equilibrium conditions are
$$7x=3y,\qquad2z=3w.$$Set the values of the lower sets by $L=7x+3y$ and $R=2z+3w$. On the top support,
$$2L=5R,\qquad2L+5R=70.$$As the installments are equal, each one is worth 35. Then $L=35/2$ and $R=7$. Again dividing each set into two equal contributions,
$$7x=3y=\frac{35}{4},\qquad2z=3w=\frac72.$$Like this,
$$\boxed{x=\frac54,\quad y=\frac{35}{12},\quad z=\frac74,\quad w=\frac76}.$$The coefficients are part of the mobile rule: adding just the four circles should not produce 70.
M02: Check the data before calculating
Kits A, B and C have indicated preferences of 40%, 30% and 40%. The chances of service, depending on the kit chosen, are 80%, 90% and 70%.
The inconsistency of the statement
Each customer indicates a single preference. Therefore, the events of preferring A, B, or C are mutually exclusive and cover all customers. Your probabilities should add up to 1, but
$$0{,}4+0{,}3+0{,}4=1{,}1.$$There is no probability distribution that simultaneously satisfies these data. This justification is one of the accepted answers in the official guide to the question.
It is possible to see the problem in two ways for item a. The direct sum of non-attendance contributions produces
$$0{,}4\cdot0{,}2+0{,}3\cdot0{,}1+0{,}4\cdot0{,}3=0{,}23.$$The sum of the service contributions produces $0{,}87$, whose complement would be $0{,}13$. The disagreement does not come from the complement rule: it comes from using weights that add up to 1,1 as if they were probabilities.
A coherent model to study the three items
To practice, we can adopt an additional hypothesis: treat $40:30:40$ as proportions and normalize them. This changes the printed model and does not constitute a one-time official correction. In this hypothesis,
$$P(A)=\frac4{11},\qquad P(B)=\frac3{11},\qquad P(C)=\frac4{11}.$$Let $T$ be the event of having the preference met.
a) No service. Separate customers into three groups and multiply the proportion of each group by the chance of failure in it:
$$P(T^c)=\frac4{11}\cdot\frac2{10}+\frac3{11}\cdot\frac1{10}+\frac4{11}\cdot\frac3{10}=\boxed{\frac{23}{110}}.$$The chance of service is $87/110$; now the two probabilities add up to exactly 1.
b) Two friends with different preferences. Assume independent choices and independent assistance given preferences. Before knowing the order of friends, pairs AB, AC and BC have weights proportional to
$$4\cdot3:4\cdot4:3\cdot4=12:16:12.$$Conditioning on different preferences, these weights become $3/10$, $4/10$ and $3/10$. The chance of both being met is
$$\frac3{10}(0{,}8\cdot0{,}9)+\frac4{10}(0{,}8\cdot0{,}7)+\frac3{10}(0{,}9\cdot0{,}7)=\boxed{0{,}629}.$$A simple average of the three products would be wrong: the AC pair is more frequent in this model.
c) Preference for B among those served. The joint probability of choosing B and being served is $(3/11)(9/10)=27/110$. Divide it by the probability of service:
$$P(B\mid T)=\frac{27/110}{87/110}=\boxed{\frac9{29}}\approx31{,}03\%.$$These three results belong to the normalized hypothesis. In the test as printed, the fundamental conclusion is the incompatibility of the data; the official guide details the acceptance criteria for other approaches.
M03: Lines and parabola
We work with $P:y=x^2$, $r:ax+by+c=0$ and $O=(0,0)$.
a) Intersection for $(a,b,c)=(2,-1,3)$
On the straight line, $y=2x+3$. A common point must satisfy both expressions of $y$:
$$x^2=2x+3\quad\Rightarrow\quad(x-3)(x+1)=0.$$For $x=-1$, we have $y=1$; for $x=3$, we have $y=9$.
Response: $\boxed{(-1,1)\text{ e }(3,9)}$.
b) Point of $4x+3y=7$ closest to the origin
The shortest distance from a point to a straight line is measured by the perpendicular segment. The given straight line has angular coefficient $-4/3$; the perpendicular that passes through $O$ has equation $y=3x/4$.
Replacing,
$$4x+3\cdot\frac{3x}{4}=7\quad\Rightarrow\quad25x=28.$$Then,
$$\boxed{\left(\frac{28}{25},\frac{21}{25}\right)}.$$Conference: $4(28/25)+3(21/25)=7$. The vector up to this point is proportional to $(4,3)$, normal to the straight line, confirming perpendicularity.
c) Parallelism with $A=(-1,1)$ and distance $BC=\sqrt{17}$
Write $B=(u,u^2)$ and $C=(v,v^2)$. Since B is in the first quadrant, $u>0$. The OC segment must be non-degenerate, therefore $v\ne0$.
The slopes of AB and OC are
$$\frac{u^2-1}{u+1}=u-1,\qquad\frac{v^2}{v}=v.$$Parallelism gives $v=u-1$. Using distance squared,
$$17=(u-v)^2+(u^2-v^2)^2=1+(u+v)^2.$$Thus, $u+v=\pm4$. Combining with $u-v=1$, we obtain $u=5/2$ or $u=-3/2$. The first quadrant condition selects
$$\boxed{B=\left(\frac52,\frac{25}{4}\right),\qquad C=\left(\frac32,\frac94\right)}.$$The differences between the coordinates are 1 and 4, so $BC=\sqrt{1+16}=\sqrt{17}$.
M04: The height of the string
The rope model is $f(x)=2^x+2^{-x}$, with measurements in meters.
a) Minimum height
Do $t=2^x>0$. As $(t-1)^2\ge0$, dividing by $t$ we have
$$t+\frac1t\ge2.$$Equality requires $t=1$, that is, $x=0$. Therefore, the minimum height is $\boxed{2\text{ m}}$. The inequality proves that it is a global minimum, without depending only on the appearance of the graph.
b) Rods separated by 3 m
The function is even: $f(-x)=f(x)$. It is strictly increasing for $x\ge0$: if $1\le s
The stems are at $x=-3/2$ and $x=3/2$. The height is
$$f\left(\frac32\right)=2\sqrt2+\frac1{2\sqrt2}=2\sqrt2+\frac{\sqrt2}{4}=\boxed{\frac{9\sqrt2}{4}\text{ m}}.$$c) Distance between points of height 4 m
Solving $t+1/t=4$ is equivalent to solving
$$t^2-4t+1=0\quad\Rightarrow\quad t=2\pm\sqrt3.$$The values are positive and reciprocal, as $(2+\sqrt3)(2-\sqrt3)=1$. Therefore, the abscissa are opposite:
$$x=\pm\log_2(2+\sqrt3).$$The points have the same ordinate. Their distance is the difference between the abscissas, and not the arc length of the string:
$$\boxed{d=2\log_2(2+\sqrt3)\text{ m}}.$$M05: One triangle, three axes of rotation
The region is the triangle of vertices $O=(0,0)$, $A=(4,0)$ and $B=(0,4)$, bounded by $x+y=4$ and the coordinate axes.
a) Rotation around Oy: section area
Rotation produces a cone with radius 4 and height 4. The section through the Oxy plane crosses its axis and contains both the original triangle and its reflection to the left of Oy. It is a triangle with base 8 and height 4:
$$\boxed{A=\frac{8\cdot4}{2}=16}.$$Note what is asked for: the area of the flat section of the solid, not the lateral area of the cone.
b) Rotation around $x=4$: volume
At a height $y\in[0,4]$, the triangle occupies $0\le x\le4-y$. The distances from these ends to the $x=4$ axis are 4 and $y$. When rotating, the band generates a circular crown with outer radius 4 and inner radius $y$.
This identifies the solid as a cylinder of radius 4 and height 4 with a conical cavity of the same dimensions. Therefore,
$$V=\pi\cdot4^2\cdot4-\frac13\pi\cdot4^2\cdot4=\boxed{\frac{128\pi}{3}}.$$c) Rotation around $x+y=4$: volume
The axis is the hypotenuse AB. The foot of the perpendicular from O to AB is $H=(2,2)$. The relevant distances are
$$OH=AH=BH=\sqrt{2^2+2^2}=2\sqrt2.$$Divide the triangle by the segment OH. Rotating, OAH and OBH generate two cones that share the circular base and lie on opposite sides of it. Each cone has radius $OH$ and height $AH$ or $BH$.
$$V=2\cdot\frac13\pi(2\sqrt2)^2(2\sqrt2)=\boxed{\frac{32\sqrt2\pi}{3}}.$$The height of each cone is half the original hypotenuse; Using 4 as the height would confuse the sides with the axis of rotation.
M06: Equations with the floor function
The definition we will use is
$$\lfloor x\rfloor=n\quad\Longleftrightarrow\quad n\le xFirst, check the negative numbers. If $-3\le x<0$, then $-\lfloor x\rfloor\le3$ and $-x\le3$, then $x\lfloor x\rfloor\le9$. If $x<-3$, we have $\lfloor x\rfloor\le-4$ and the product is greater than 12. There is no negative solution.
For $0\le x<3$, the product is less than 6; for $x\ge4$, it is at least 16. There remains $3\le x<4$, the range in which the floor is 3:
$$3x=10\quad\Rightarrow\quad\boxed{x=\frac{10}{3}}.$$As $3\le10/3<4$, the solution respects the range used.
b) $\lfloor2x\rfloor+\lfloor x\rfloor=7$
Write $x=n+r$, with $n\in\mathbb Z$ and $0\le r<1$. Then
$$\lfloor2x\rfloor+\lfloor x\rfloor=3n+\lfloor2r\rfloor.$$The top floor can only be 0 or 1. If it is 0, we would have $3n=7$, impossible for an integer. If it is 1, we have $3n+1=7$, therefore $n=2$ and $1/2\le r<1$.
Then,
$$\boxed{x\in\left[\frac52,3\right)}.$$The extreme $5/2$ enters because $5+2=7$. The extreme 3 is left out because $6+3=9$.
c) $\lfloor x^2\rfloor-4\lfloor x\rfloor+3=0$
Rearrange the equation:
$$\lfloor x^2\rfloor=4\lfloor x\rfloor-3.$$If $x<1$, the right side is negative and the left side is non-negative. Therefore $x\ge1$.
Set $n=\lfloor x\rfloor\ge1$. As $x\ge n$, we have $\lfloor x^2\rfloor\ge n^2$. A necessary condition is
$$n^2\le4n-3\quad\Rightarrow\quad(n-1)(n-3)\le0.$$So, just examine $n=1,2,3$. In each case, we need to impose both the $x$ and $x^2$ floor:
| $n$ | Condition in $x^2$ | Intersection with $n\le x |
|---|---|---|
| 1 | $1\le x^2<2$ | $[1,\sqrt2)$ |
| 2 | $5\le x^2<6$ | $[\sqrt5,\sqrt6)$ |
| 3 | $9\le x^2<10$ | $[3,\sqrt{10})$ |
Therefore,
$$\boxed{[1,\sqrt2)\cup[\sqrt5,\sqrt6)\cup[3,\sqrt{10})}.$$The upper extremes are excluded because, in them, the floor of $x^2$ increases by one unit. You cannot replace $\lfloor x^2\rfloor$ with $\lfloor x\rfloor^2$: for example, in $x=\sqrt5$, these values are 5 and 4.