Solutions to the second test in Mathematics 2026: step-by-step solutions for high school students.

The second exam in Mathematics for the Scientific High School took place today, June 19, 2026. The ministerial exam contains two problems and eight questions, with the requirement to choose one problem and four questions. This article provides a detailed and complete outline of the official MIM exams, including all the analytical steps, tests, and numerical results.

The structure of the test

The time available is 6 hoursThe track is composed of 2 Problems (choose 1) and 8 Questions (choose 4). The maximum rating is 20/20The commission evaluates not only the numerical result but also the procedure, the clarity of the presentation, and the ability to argue.

Problem 1 – The level of Lake Bracciano

The problem models the level of Lake Bracciano from 2016 to 2026 with a piecewise-defined function. The data in the table show the difference in level from the hydrometric zero (optimal level), expressed in dm. The unit on the x-axis corresponds to one year, with January 1, 2016, set at x=0.

Point a) – Determination of parameters and definition of the model

First section (0 ≤ x < 3, from 2016 to 2019): the polynomial model is

y = a(x−2)⁴ + b(x−2)³ + c(x−2)² − 20

From the data in the table we have the points: (0;−6), (1;−16), (2;−20), (3;−18). We substitute into the equation:

• x=2 (2018): y = a 0⁴ + b 0³ + c 0² − 20 = −20. The given is automatically satisfied.
• x=0 (2016): 16a − 8b + 4c − 20 = −6 → 8a − 4b + 2c = 7 (dividing by 2)
• x=1 (2017): a − b + c − 20 = −16 → a − b + c = 4
• x=3 (2019): a + b + c − 20 = −18 → a + b + c = 2

From a − b + c = 4 and a + b + c = 2 we obtain, by subtracting member by member: −2b = 2 → b = −1. Adding up: 2a + 2c = 6 → a + c = 3.

We substitute b = −1 into 8a − 4b + 2c = 7: 8a + 4 + 2c = 7 → 8a + 2c = 3. With c = 3 − a: 8a + 6 − 2a = 3 → 6a = −3 → a = −1/2, from which c = 7/2.

Second section (3 ≤ x ≤ 7, from 2019 to 2023): y = mx − 24 + sin²(πx). At the boundary x=3 we must have y=−18 by continuity: 3m − 24 + sin²(3π) = 3m − 24 = −18 → m = 2Let's check that ax=7: 14 − 24 + sin²(7π) = −10, which corresponds to the data for 2023.

Third section (7 < x ≤ 10, from 2023 to 2026): y = 2cos(2πx) + k. By continuity at x=7: 2cos(14π) + k = 2 1 + k = −10 → k = −12Let's check that x=10: 2cos(20π) − 12 = 2 − 12 = −10, consistent with the 2026 data.

Complete model f(x):

f(x) =
 −½(x−2)⁴ − (x−2)³ + ⁷⁄₂(x−2)² − 20, 0 ≤ x < 3
 2x − 24 + sin²(πx), 3 ≤ x ≤ 7
 2cos(2πx) − 12, 7 < x ≤ 10

Point b) – Study of function and graph

Continuity: Let's check the connections.
x=3: left limit: −½(1)⁴ − (1)³ + ⁷⁄₂(1)² − 20 = −½ − 1 + ⁷⁄₂ − 20 = 3 − 20 = −18. Right limit: 2 3 − 24 + sin²(3π) = 6 − 24 = −18. f(3) = −18. Continued ✓
x=7: left limit: 14 − 24 + 0 = −10. Right limit: 2cos(14π) − 12 = 2 − 12 = −10. f(7) = −10. Continued ✓

Derivability: Let's calculate the derivatives at the boundaries.
f₁′(x) = −2(x−2)³ − 3(x−2)² + 7(x−2). At x=3: f₁′(3) = −2(1)³ − 3(1)² + 7(1) = 2.
f₂′(x) = 2 + 2π·sin(πx)·cos(πx) = 2 + π·sin(2πx). At x=3: f₂′(3) = 2 + π·sin(6π) = 2. At x=7: f₂′(7) = 2 + π·sin(14π) = 2.
f₃′(x) = −4π·sin(2πx). At x=7: f₃′(7) = −4π sin(14π) = 0.

So f is differentiable at x=3 (left derivative = right derivative = 2), but it is not differentiable at x=7: f′(7⁻) = 2 while f′(7⁺) = 0. This is a corner point.

Stationary points – First section: Let f₁′(x) = 0 with t = x−2:

−2t³ − 3t² + 7t = t(−2t² − 3t + 7) = 0 → t = 0 (x=2) or 2t² + 3t − 7 = 0 → t = (−3 ± √65)/4. The two solutions t≈1,265 (x≈3,265, outside the first segment) and t≈−2,765 (x≈−0,765, outside the domain). The only stationary point in the first segment is x = 2, that is a relative minimum (f₁″(2) = 7 > 0). f(2) = −20 dm.

Second section: f₂′(x) = 2 + π·sin(2πx) = 0 → sin(2πx) = −2/π ≈ −0,637. In [3,7], the solutions of 2πx = arcsin(−2/π) + 2πk and 2πx = π − arcsin(−2/π) + 2πk give:

x ≈ 3,610 — local max (f₂ ≈ −16,96)
x ≈ 3,890 — local min (f₂ ≈ −18,45)
x ≈ 4,610 — local max (f₂ ≈ −14,96)
x ≈ 4,890 — local min (f₂ ≈ −16,45)
x ≈ 5,610 — local max (f₂ ≈ −12,96)
x ≈ 5,890 — local min (f₂ ≈ −14,45)
x ≈ 6,610 — local max (f₂ ≈ −10,96)
x ≈ 6,890 — local min (f₂ ≈ −12,45)

Third section: f₃′(x) = −4π sin(2πx) = 0 → sin(2πx) = 0 → x = k/2 for k integers. In (7,10]: x = 7,5; 8; 8,5; 9; 9,5; 10. Points with cos(2πx) = 1 (x integers) give maxima y = 2−12 = −10; those with cos(2πx) = −1 (x half-integers) give minima y = −2−12 = −14.

Values ​​at the extremes: f(0) = −6, f(10) = −10.

The chart shows an initial sharp decline to −20 dm in 2018, a slow, oscillating recovery from 2019 to 2023, and finally a stable oscillatory pattern between −14 and −10 dm from 2023 onwards.

Point c) – Lagrange's Theorem

Lagrange's theorem requires that f be continuous at [0,10] and differentiable at (0,10). We have just shown that f is not differentiable at x=7 (corner point), then it is not applicable in [0,10].

On the second question: we look for s ∈ (0,10) such that f′(s) = [f(10) − f(0)] / 10. We calculate: [−10 − (−6)] / 10 = −4/10 = −0,4Are there points where the derivative is −0,4? Yes, in the third section f₃′(x) = −4π sin(2πx). We solve −4π sin(2πx) = −0,4 → sin(2πx) = 0,4/(4π) ≈ 0,0318. This equation has periodic solutions in (7,10], e.g. x ≈ 7,005, x ≈ 7,495, and symmetric. Since f is continuous and differentiable at every single point, we can apply the intermediate value theorem to the function f₃′ in the interval (7,10] to guarantee the existence of such points. It is enough to note that f₃′(x) varies continuously between −4π and 4π, so it takes on all intermediate values, including −0,4.

Point d) – Integral Mean Theorem and volume estimation

Applicability: The Integral Mean Theorem only requires the continuity of f on the closed interval [0,10]. f is continuous in [0,10], as verified in point b). So it is applicable.

The average variation Δh is:

Δh = 1/(b−a) ∫ₐᵇ f(x) dx = 1/10 ∫₀¹⁰ f(x) dx

Let's calculate the piecewise integral:

Section 1 [0,3]: ∫₀³ [−½(x−2)⁴ − (x−2)³ + ⁷⁄₂(x−2)² − 20] right. With t = x−2, t from −2 to 1:
∫₋₂¹ [−½t⁴ − t³ + ⁷⁄₂t² − 20] dt = [−t⁵/10 − t⁴/4 + ⁷⁄₆t³ − 20t]₋₂¹
At t=1: −1/10 − 1/4 + 7/6 − 20 = −0,1 − 0,25 + 1,167 − 20 = −19,183
At t=−2: −(−32)/10 − (16)/4 + ⁷⁄₆(−8) − 20(−2) = 3,2 − 4 − 9,333 + 40 = 29,867
Integral = −19,183 − 29,867 = −49,05 (rounded)

Section 2 [3,7]: ∫₃⁷ [2x − 24 + sin²(πx)] dx. Recall sin²(πx) = (1−cos(2πx))/2, whose antiderivative is x/2 − sin(2πx)/(4π).
∫₃⁷ (2x − 24) dx = [x² − 24x]₃⁷ = (49−168) − (9−72) = −119 + 63 = −56
∫₃⁷ sin²(πx) dx = [x/2 − sin(2πx)/(4π)]₃⁷ = (7/2 − 0) − (3/2 − 0) = 2
Total integral = −56 + 2 = −54

Section 3 [7,10]: ∫₇¹⁰ [2cos(2πx) − 12] dx = [sin(2πx)/π − 12x]₇¹⁰
At x=10: sin(20π)/π − 120 = 0 − 120 = −120
At x=7: sin(14π)/π − 84 = 0 − 84 = −84
Integral = −120 − (−84) = −36

Total integral: −49,05 + (−54) + (−36) = −139,05

Average variation: Δh = −139,05/10 ≈ −13,9 dm = -1,39 m

Volume estimate: Lake surface ≈ 57 km² = 57·10⁶ m². Volume = surface area × Δh = 57·10⁶·(−1,39) ≈ −79,2·10⁶ m³. In litres: 1 m³ = 10³ litres, therefore -79,2 billion liters approximately. The mean integral of f(x) indicates that in the period 2016-2026 the estimated average level was approximately 13,9 dm below freezing.

Problem 2 – Study of parametric functions

Point a) – Monotonicity of fₐ and common tangent

Study of fₐ(x) = ax²/(x−1), a ≠ 0:

Domain: ℝ{1}. First derivative: fₐ′(x) = [2ax(x−1) − ax²] / (x−1)² = [2ax² − 2ax − ax²] / (x−1)² = [ax² − 2ax] / (x−1)² = ax(x−2) / (x−1)².

The sign of fₐ′ is determined by the sign of a·x·(x−2) (the denominator is always positive).

If a > 0:
fₐ′ > 0 for x < 0 or x > 2 → increasing f
fₐ′ < 0 for 0 < x < 1 or 1 < x < 2 → f decreasing
x=0: max relative (f(0)=0)
x=2: relative min (f(2)=4a)

If a < 0: opposite monotony (it increases where it previously decreased and vice versa).

Study of g(x) = |x|/(x²+1):

Domain: ℝ. For x ≥ 0, g(x) = x/(x²+1); for x < 0, g(x) = −x/(x²+1).

For x ≥ 0: g′(x) = [1·(x²+1) − x·2x] / (x²+1)² = (1−x²) / (x²+1)². g′(x) > 0 for 0 < x < 1, g′(x) < 0 for x > 1. Maximum ax=1 → g(1)=½.

For x < 0: g(x) is even, symmetric. g(0)=0 is minimum. g(−1)=½ is maximum.

The stationary points are x=±1 with g(±1)=½. So the stationary point of γ with x_B > 0 is B(1;½).

Common Tangent: We are looking for a horizontal line y=k tangent to φₐ and γ. γ has a horizontal tangent to y=½ (at the points x=±1). k=½. For φₐ, f(0)=0 and f(2)=4a. With a such that f(2)=½: 4a=½ → a=⅛. With a such that f(0)=½: impossible. So a=⅛, k=½.

Point b) – Minimum distance AB

The stationary points are: for γ, B(1;½); for φₐ, the points with x_A ≠ 0 are x=2 (the minimum for a>0). Hence A(2;4a).

AB² = (2−1)² + (4a−½)² = 1 + (4a−½)². The minimum occurs when |4a−½| is minimum, that is, when 4a−½=0 → a=⅛. Therefore, the minimum distance is AB = √1 = 1, and corresponds to a=⅛.

Point c) – Graphs and inequalities

With a=⅛: f(x) = x²/[8(x−1)] = x²/(8x−8).

Domain: ℝ{1}. Vertical asymptote: x=1 (left limit: −∞, right: +∞). Oblique asymptote: degree of numerator = degree of denominator + 1, so it exists. Dividend: x²/(8x−8) = (x+1)/8 + 1/[8(x−1)]. For x→∞, f(x) ∼ (x+1)/8. Asymptote: y = x/8 + 1/8.

Derivative: f′(x) = x(x−2)/[8(x−1)²]. f increases for x<0 and x>2, decreases for 0

g(x) = |x|/(x²+1): as already studied, it has max at x=±1: y=½, minimum at x=0: y=0.

Inequality f(x) > g(x): Let's study the cases. For x < 0: g(x) = −x/(x²+1) > 0 while f(x) < 0 (positive numerator, negative denominator), so g > f. For 0 < x < 1: g(x) > 0 while f(x) ≤ 0 (f(0)=0 and f negative immediately after), so g > f. For x = 1: f is not defined. For x > 1: both f(x) and g(x) are positive. f(x) − g(x) = x²/(8x−8) − x/(x²+1). For x→1⁺: f→+∞, so f > g. For x > 1 the equation f(x) = g(x) translates to x²/(8x−8) = x/(x²+1). Simplifying x > 0: x/(8x−8) = 1/(x²+1) → x(x²+1) = 8x−8 → x³ + x − 8x + 8 = 0 → x³ − 7x + 8 = 0. The positive roots of this cubic do not belong to the interval (1,+∞) except for values ​​greater than the point where f definitely exceeds g. From the asymptotic comparison: for x→+∞, f(x) ∼ x/8 → +∞ while g(x) → 0, so f > g definitely. The inequality is therefore verified for x ∈ (1, +∞).

Point d) – Area of ​​the region bounded by γ

Let's find the inflection points of g(x). For x≥0: g(x)=x/(x²+1). g″(x) = 2x(x²−3)/(x²+1)³. g″(x)=0 → x=0 (no inflection, it's a minimum) and x=√3. For x<0: by symmetry, x=−√3. The inflection points are x=±√3. g(√3)=√3/4 ≈ 0,433.

The finite region bounded by γ, the x-axis and the lines x=−√3 and x=√3 has area:

A = 2∫₀^√3 x/(x²+1) dx (by symmetry). ∫ x/(x²+1) dx = ½ ln(x²+1). So:

A = 2·[½·ln(x²+1)]₀^√3 = [ln(x²+1)]₀^√3 = ln(3+1) − ln(1) = ln(4) = 2ln(2).

Questions

Question 1 – Cover the Spot

Square ABCD of side √2 dm, area = (√2)² = 2 dm²Cecilia places a circular card with radius ⅔ dm with its center on diagonal AC and its edge passing through A. The center is ⅔ dm from A and is on diagonal AC. Let's calculate the area of ​​the circle that falls inside the square.

The center of the circle is located on the diagonal at a distance r = ⅔ from vertex A. With respect to a system of axes with A at the origin, the center C₀ has coordinates (r/√2, r/√2) = (⅔√2/2, ⅔√2/2) = (√2/3, √2/3). The circle is internally tangent at A and intersects the sides of the square. The area of ​​the portion of the circle internal to the square is calculated by decomposing it into circular sectors and triangles.

The required area is given by the sum of ¾ of the circle (angle of 270° swept inside the square) plus the area of ​​the triangle formed by A and the two points of intersection of the circle with the diagonal, which leads to:

A_portion = (3/4)·πr² + (1/2)·r² = (3π/4 + 1/2)·(4/9) = (3π/4)·(4/9) + (1/2)·(4/9) = π/3 + 2/9

Calculating: π/3 ≈ 1,047 and 2/9 ≈ 0,222, for a total of approximately 1,269 dm².

But be careful: the 270° sector is not exactly the internal area. A more rigorous decomposition shows that the area of ​​the portion of the circle that falls within the square is given by the sum of ¾ of the circle (which is equal to ¾ π (4/9) = π/3) plus the isosceles right-angled triangle with side r, which has area r²/2 = (4/9)/2 = 2/9. However, the precise calculation must be revised because part of the remaining quarter of the circle (the one outside the square) also cuts off the triangle.

Analyzing with analytical geometry: the circle has equation (x−√2/3)² + (y−√2/3)² = 4/9. The square has vertices A(0,0), B(√2,0), C(√2,√2), D(0,√2). For x≥0 and y≥0 (we are in the first quadrant). The part of the circle outside the square is the lower left quarter circle cut by vertex A. The area of ​​the circle outside the square is a 90° sector minus the triangle A-center-vertex A: therefore external area = πr²/4 − r²/2 = π/9 − 2/9 = (π−2)/9 ≈ 0,127 dm².

Internal area of ​​the square = total area of ​​the circle − external area = 4π/9 − (π−2)/9 = (3π+2)/9 ≈ (9,425+2)/9 ≈ 1,269 dm².

Since half of the square is 1 dm², the first circle actually covers more than half of the square (about 63,5%). Cecilia is right.

Question 2 – Regular tetrahedron and tangent plane

a) Check regular tetrahedron: Let's calculate all the distances between pairs of points.

AB² = (3−2)² + (5+4)² + (−1−3)² = 1 + 81 + 16 = 98 → AB = √98 = 7√2
AC² = (−6−2)² + (1+4)² + (0−3)² = 64 + 25 + 9 = 98 → AC = 7√2
AD² = (−1−2)² + (4+4)² ​​+ (8−3)² = 9 + 64 + 25 = 98 → AD = 7√2
BC² = (−6−3)² + (1−5)² + (0+1)² = 81 + 16 + 1 = 98 → BC = 7√2
BD² = (−1−3)² + (4−5)² + (8+1)² = 16 + 1 + 81 = 98 → BD = 7√2
CD² = (−1+6)² + (4−1)² + (8−0)² = 25 + 9 + 64 = 98 → CD = 7√2

All edges measure 7, so the tetrahedron is regular. ✓

b) Tangent plane: In a regular tetrahedron, the center of the circumscribed sphere coincides with the centroid. Coordinates of the centroid G: ((2+3−6−1)/4; (−4+5+1+4)/4; (3−1+0+8)/4) = (−2/4; 6/4; 10/4) = (−½; ³⁄₂; ⁵⁄₂).

Radius R = distance GA. For a regular tetrahedron with edge L=7√2, the radius of the circumscribed sphere is R = L √6/4 = 7√2 √6/4 = 7√12/4 = 7 2√3/4 = 7√3/2. Check: GA = √[(2+½)² + (−4−³⁄₂)² + (3−⁵⁄₂)²] = √[(⁵⁄₂)² + (−¹¹⁄₂)² + (½)²] = √[25/4 + 121/4 + 1/4] = √(147/4) = √147/2 = 7√3/2. ✓

The tangent plane at A is perpendicular to the vector GA (radius vector). GA = (⁵⁄₂; −¹¹⁄₂; ½). Equation of the plane: (⁵⁄₂)(x−2) + (−¹¹⁄₂)(y+4) + (½)(z−3) = 0. Multiplying by 2: 5(x−2) − 11(y+4) + (z−3) = 0 → 5x − 10 − 11y − 44 + z − 3 = 0 → 5x − 11y + z − 57 = 0.

Question 3 – Friuli earthquake (1976)

The track mentions two shocks: one of magnitude M₁ = 6,5 (that of May 6, 1976, the main shock) and one of M₂ = 6,0 (one of the subsequent September tremors). From the formula M = log₁₀(A/A₀):

A₁/A₀ = 10⁶·⁵; A₂/A₀ = 10⁶·⁰.
A₁/A₂ = 10⁶·⁵ / 10⁶·⁰ = 10⁰·⁵ = √10 ≈ 3,16.

The ratio between the amplitudes is approximately 3,16: the main shock (M=6,5) produced seismic amplitudes more than 3 times greater than the subsequent shock (M=6,0).

From the Gutenberg-Richter law: log₁₀(E/E₀) = 1,5M + 4,8.

For M=6,5: log₁₀(E₁/E₀) = 1,5 6,5 + 4,8 = 9,75 + 4,8 = 14,55 → E₁ = E₀ 10¹⁴ ⁵⁵
For M=6,0: log₁₀(E₂/E₀) = 1,5 6,0 + 4,8 = 9,0 + 4,8 = 13,80 → E₂ = E₀ 10¹³ ⁸⁰

E₁/E₂ = 10¹⁴ ⁵⁵ / 10¹³ ⁸⁰ = 10⁰ ⁷⁵ ≈ 5,62.

Percentage change: (E₁−E₂)/E₂ · 100 = (5,62−1)·100 ≈ + 462 %The May earthquake released about 462% more energy than the September earthquake.

Question 4 – Constant function

F(x) = ∫₀ˣ dt/(1+t²) + ∫₀^(¹⁄x) dt/(1+t²), x > 0.

∫ dt/(1+t²) = arctan(t) + C. Therefore:

F(x) = [arctan(t)]₀ˣ + [arctan(t)]₀^(¹⁄x) = arctan(x) − arctan(0) + arctan(1/x) − arctan(0) = arctan(x) + arctan(1/x).

For x > 0, the identity arctan(x) + arctan(1/x) = π/2 holds (derivatable from: tan(π/2 − α) = cot(α) = 1/tan(α)).

Therefore F(x) = π/2, constant for every x > 0. ✓

Question 5 – Parameters of the logarithmic curve

y = h·ln(x²+k)⁵ = 5h·ln(x²+k), with h≠0.

Vertical asymptotes x=±√3: To have vertical asymptotes, the logarithm must tend to −∞ when x→±√3, so x²+k → 0⁺. We have (±√3)² + k = 0 → 3 + k = 0 → k = −3.

The function becomes y = 5h ln(x²−3).

Intersections with x-axis: y=0 → ln(x²−3)=0 → x²−3=1 → x²=4 → x=±2. The points are A(−2;0) and B(2;0).

Tangents at A and B that meet at C(0;−4):

y′ = 5h·2x/(x²−3) = 10hx/(x²−3).

At x=−2: y′(−2) = 10h (−2)/(4−3) = −20h. Tangent line: y − 0 = −20h(x+2), i.e. y = −20hx − 40h. Passing through C(0;−4): −4 = 0 − 40h → h = 4/40 = ⅒.

Check with point B: y′(2) = 10h·2/(4−3) = 20h. Tangent: y = 20h(x−2), for h=⅒: y = 2(x−2). At x=0: y=−4 ✓.

Solution: h=⅒, k=−3.

Question 6 – Polynomial and oblique asymptote

f(x) = p(x)/(2x+1). For oblique asymptote y = 3x−2, we write p(x) = (2x+1)(3x−2) + R, where R is the constant remainder. Expanding: (2x+1)(3x−2) = 6x² − 4x + 3x − 2 = 6x² − x − 2. So p(x) = 6x² − x − 2 + R.

Furthermore f(1)=0 for the passage through P(1;0): f(1) = p(1)/(3) = 0 → p(1) = 0.

p(1) = 6 − 1 − 2 + R = 3 + R = 0 → R = −3.

Therefore p(x) = 6x² − x − 5.

Verification: f(1) = (6−1−5)/(2+1) = 0/3 = 0 ✓. Asymptote: dividing 6x² − x − 5 by 2x+1 we obtain quotient 3x − 2 and remainder −3/(2x+1) → as x→∞ the remainder tends to 0, confirmed. ✓

Question 7 – Probability in Scopone

40-card deck, 4 suits (wands, cups, coins, and swords), 10 cards per suit. 4 players, 10 cards each.

a) Probability that Massimo's first 3 cards are all cups:

Total cards: 40. Available cups: 10. The first 3 cards are dealt without replacement:

P = (10/40) · (9/39) · (8/38) = (10·9·8)/(40·39·38) = 720/59.280 = 3/247 ≈ 0,0121 ≈ 1,21%.

b) Probability that among Lorenzo's 10 cards there are the 3 aces of clubs, swords and coins:

Total number of ways to deal 10 cards to Lorenzo: C(40,10).
Favorable cases: Lorenzo receives the 3 specific aces + any 7 cards from the remaining 37: C(37,7).

P = C(37,7) / C(40,10) = 10.295.472 / 847.660.528 ≈ 0,01215 ≈ 1,21%.

Question 8 – Volleyball Tournament

16 teams divided into 4 groups (A, B, C, D) of 4 teams each. The 4 teams in Pot 1 have already been assigned (one per group, in ranking order, without a draw).

There remain: 4 teams from the 2nd tier to be distributed (one per group) and 8 teams from the 3rd tier to be distributed (two per group).

Phase 1: Assignment of 2nd tier teams. The 4 teams in the 2nd pot can be sorted in 4 ways to assign them to the 4 groups (one each): 24 ways.

Phase 2: Assignment of 3nd tier teams. You have to choose which two teams from the third tier go to each group. The groups are distinct (A,B,C,D). The number of ways is: C(8,2) · C(6,2) · C(4,2) · C(2,2) = 28 · 15 · 6 · 1 = 2.520 ways.

Total: 24 · 2.520 = 60.480 possible compositions of the groups.

FAQ

What is the value of question 3 on earthquake energy? Is it all there is to it, or is the formula enough?
The question requires the correct application of the two formulas (Richter and Gutenberg-Richter). The transition from magnitude to the amplitude ratio (10^0,5) must be shown, followed by the calculation of the energy ratio. The percentage change is the icing on the cake: the ratio alone isn't enough; it must be converted into a percentage.

In problem 1, could the integral be calculated with a graphing calculator?
Yes, a graphing calculator is permitted, as long as it doesn't have symbolic calculation capabilities. You can use it to verify the integral numerical value, but in the written test, you must still demonstrate the analytical procedure for each segment.

Point c) of problem 1 asks whether there exist s such that f′(s) = [f(10)−f(0)]/10. How do you answer?
We calculate the incremental ratio (−4/10 = −0,4). We then observe that the derivative in the third section is −4π sin(2πx), which varies continuously between −4π and 4π: by the intermediate value theorem it assumes all values ​​in this range. Since |−0,4| < 4π, there exist points where the derivative is −0,4. These can be found explicitly by solving sin(2πx) = 0,1/π.

In question 2, for the regular tetrahedron, was it enough to check 3 edges or all of them?
Technically, to prove regularity, five equal edges are enough (for a tetrahedron, five edges constrain the sixth). But during the exam, it's safer to show them all. Calculating the six distances with the formula is quick and leaves no doubt.

Question 8: Why isn't pure drawing used?
Because the four teams in the first tier have already been assigned without a draw (in ranking order: first in A, second in B, etc.). For the others, the text says "drawn," but the combinatorial scheme is: they must be positioned so that each group has one 2nd-tier team and two 3rd-tier teams. The total number of possible assignments is obtained by counting the possible assignments using combinatorics.

In problem 2 part d, is the area calculated only between x=−√3 and x=√3? What if there are other finite regions?
The text speaks of a "finite region of the plane bounded by γ, the x-axis, and the lines parallel to the y-axis passing through the inflection points." The inflection points of g are at x = ±√3, and the region is the part under the graph of g above the x-axis between these two points. There are no other finite regions.