Comprehensive Math Exam Notes

Matura 2019 Task 5: Stochastics

• This task involves probability and combinatorics within a political context.
• A small country is divided into 12 electoral districts.
• Three political parties (A, B, C) each nominate one candidate per district.
• One person from each district enters the country's parliament.
• All candidates have different names.

a) Possible Name Lists Sorted by Electoral District

• Task: Determine the number of ways to form a name list of elected parliament members sorted by electoral district.
• Solution: Since there are three candidates in each of the 12 districts, there are 3 choices for each district.
• Total possible ways: $3^{12} = 531441$

b) Proportional Distribution of Parliamentary Seats

• Task: Determine the number of ways the parliamentary seats can be distributed among the three parties.
• Context: Each party nominates one candidate in each of the 12 districts. After the election, one person from each district will enter the country's parliament. Need to find number of ways the 12 seats can be distributed among the three parties.
• Solution: This is equivalent to distributing 12 identical items (seats) into 3 distinct boxes (parties).
• Stars and bars method: Using stars and bars, we have 12 seats (stars) and need 2 bars to divide them among the 3 parties.
• Total arrangements: $\binom{12 + 2}{2} = \binom{14}{2} = \frac{14 \cdot 13}{2} = 91$

c) Alphabetically Sorted Name List with Equal Wins

• Task: How many ways to form an alphabetically sorted name list if each party wins exactly four electoral districts?
• Condition: Each party wins exactly four electoral districts.
• Solution: First, choose 4 districts won by party A, then 4 districts from the remaining for party B, and the rest go to C.
• Number of ways to choose districts for A: $\binom{12}{4}$ ways.
• Number of ways to choose districts for B from the remaining 8: $\binom{8}{4}$ ways.
• Party C gets the remaining 4 districts: $\binom{4}{4} = 1$ way.
• Ways to choose districts: $\binom{12}{4} \cdot \binom{8}{4} \cdot \binom{4}{4} = \frac{12!}{4!8!} \cdot \frac{8!}{4!4!} \cdot 1 = \frac{12!}{4!4!4!} = 34650$
• Since the list needs to be alphabetically sorted, there's only one way to arrange the names within each set of districts won by a party.
• Total ways: 34650

Matura 2024 Task 4

a) Seating 7 People in 10 Numbered Seats

• Task: Find the number of ways 7 people can be seated in 10 numbered seats.
• Solution: This is a permutation problem since the order matters.
• Number of ways: $P(10, 7) = \frac{10!}{(10-7)!} = \frac{10!}{3!} = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 = 604800$

b) Distributing 13 People Among 3 Wagons (No Restrictions)

• Task: Find the number of ways 13 people can be distributed among 3 wagons without any restrictions.
• Solution: Each person has 3 choices of wagons.
• Number of ways: $3^{13} = 1594323$

c) Distributing 15 People Among 3 Wagons (5 in Each Wagon)

• Task: Find the number of ways 15 people can be distributed among 3 wagons such that each wagon has exactly 5 people.
• Solution: Choose 5 people for the first wagon, 5 for the second, and the remaining 5 for the third.
• Number of ways: $\binom{15}{5} \cdot \binom{10}{5} \cdot \binom{5}{5} = \frac{15!}{5!10!} \cdot \frac{10!}{5!5!} \cdot 1 = \frac{15!}{5!5!5!} = 3003 \cdot 252 \cdot 1 = 756756$

d) Seating 4 Married Couples Together

• Task: Find the number of different sequences possible if each of the 4 married couples wants to board together.
• Solution: Treat each couple as a single unit, so arrange 4 units, then each couple can switch places.
• Arrangements of couples: $4! = 24$
• Each couple can switch: $2^4 = 16$
• Total ways: $4! \cdot 2^4 = 24 \cdot 16 = 384$

e) Percentage of Women and Probability of Satisfied Male Passenger

• Given: 72% satisfied, 65% of satisfied are women, 40% of dissatisfied are women.

(i) Percentage of Women Among All Passengers

• Let S = satisfied, D = dissatisfied, W = women, M = men.
• $P(S) = 0.72, P(D) = 0.28$
• $P(W|S) = 0.65, P(W|D) = 0.40$
• $P(W) = P(W|S)P(S) + P(W|D)P(D) = 0.65 \cdot 0.72 + 0.40 \cdot 0.28 = 0.468 + 0.112 = 0.58$
• Percentage of women: 58%

(ii) Probability of Satisfied Male Passenger

• $P(M) = 1 - P(W) = 1 - 0.58 = 0.42$
• $P(S|M) = \frac{P(M|S)P(S)}{P(M)}$
• $P(M|S) = 1 - P(W|S) = 1 - 0.65 = 0.35$
• $P(S|M) = \frac{0.35 \cdot 0.72}{0.42} = \frac{0.252}{0.42} = 0.6$
• Probability a male passenger is satisfied: 60%

f) Fare Evasion

• Fare evasion rate: 15%, random check of 250 passengers.

(i) Probability of Exactly 30 Fare Evaders

• Using binomial distribution: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$
• $P(X=30) = \binom{250}{30} (0.15)^{30} (0.85)^{220}$

(ii) Expected Number of Fare Evaders

• Expected value: $E(X) = np = 250 \cdot 0.15 = 37.5$

(iii) 95% Confidence Interval

• Standard deviation: $\sigma = \sqrt{np(1-p)} = \sqrt{250 \cdot 0.15 \cdot 0.85} = \sqrt{31.875} \approx 5.646$
• 95% confidence interval: $E(X) \pm 1.96\sigma = 37.5 \pm 1.96 \cdot 5.646 = 37.5 \pm 11.066$
• Interval: $(26.434, 48.566)$
• Smallest possible 95% confidence interval: approximately (26, 49) since the number of fare evaders must be an integer.

g) Hypothesis Test for Delayed Trains

• Initial assumption: 18% of trains are delayed. Significant investment aims to decrease this proportion.

(i) Hypothesis Test

• Null hypothesis ($H_0$): $p = 0.18$
• Alternative hypothesis ($H_1$): p < 0.18
• Significance level: $\alpha = 0.05$

(ii) Inference

• Sample: 300 trains, 44 delayed.
• Sample proportion: $\hat{p} = \frac{44}{300} \approx 0.1467$
• Test statistic (z): $z = \frac{\hat{p} - p<em>0}{\sqrt{\frac{p</em>0(1-p_0)}{n}}} = \frac{0.1467 - 0.18}{\sqrt{\frac{0.18(0.82)}{300}}} \approx \frac{-0.0333}{0.0221} \approx -1.507$
• Critical value for a one-tailed test at $\alpha = 0.05$: -1.645
• Since z = -1.507 > -1.645, we fail to reject the null hypothesis. The result is not statistically significant at the 5% level.
• Interpretation: There is not enough evidence to conclude that the proportion of delayed trains has decreased after the infrastructure investments.

h) Glacier Express Overbooking

• 300 seats available. 10% cancellation rate. Maximize occupancy with at least 95% certainty that all travelers will have a seat.
• Let X be the number of travelers who show up. We want $P(X \le R) \ge 0.95$, where R is the number of reservations accepted.
• X follows a binomial distribution with $n = R$ and $p = 0.9$.
• Approximate using normal distribution: $\mu = Rp = 0.9R$ and $\sigma = \sqrt{Rp(1-p)} = \sqrt{R \cdot 0.9 \cdot 0.1} = \sqrt{0.09R} = 0.3\sqrt{R}$
• We want $P(X \le 300) \ge 0.95$, so we solve for R such that: $0.9R + 1.645 \cdot 0.3\sqrt{R} = 300$
• $0.9R + 0.4935\sqrt{R} - 300 = 0$; Let $y = \sqrt{R}$, so $0.9y^2 + 0.4935y - 300 = 0$
• Using the quadratic formula: $y = \frac{-0.4935 \pm \sqrt{0.4935^2 - 4 \cdot 0.9 \cdot (-300)}}{2 \cdot 0.9}$
• $y = \frac{-0.4935 \pm \sqrt{0.2435 + 1080}}{1.8} = \frac{-0.4935 \pm \sqrt{1080.2435}}{1.8} = \frac{-0.4935 \pm 32.867}{1.8}$
• Take the positive root: $y = \frac{32.3735}{1.8} \approx 17.985$
• Since $y = \sqrt{R}$, $R = y^2 = 17.985^2 \approx 323.46$
• Therefore, accept at most 323 reservations.

Vector Geometry Review

• Vector Geometry Topics

  • Notations and components of a vector
  • Norm of a vector
  • Vector operations (+, -, scalar multiplication)
  • Collinear vectors
  • Scalar product and vector product
  • Different forms of line equations in 2D/3D (vector, parametric, Cartesian/slope-intercept)
  • Different forms of plane equations (vector, parametric, Cartesian, normal, axes-intercepts)
  • Normal vector of a plane
  • Trace lines and axes intercepts of a plane
  • NHS form
  • Spheres and circles
  • Intersections of objects
  • Distances and angles between objects

• VG: Review exercises (from previous Maturas)

a) Equation of the sphere K:

• Given center M(5/2/7) and radius r = 9
• The formula for the sphere is: $(x - x<em>M)^2 + (y - y</em>M)^2 + (z - z_M)^2 = r^2$
• $(x - 5)^2 + (y - 2)^2 + (z - 7)^2 = 9^2$
• $(x - 5)^2 + (y - 2)^2 + (z - 7)^2 = 81$

b) Point P lies on the sphere K

• Given point P(2/8/1)
• Plug the point into the sphere equation:
• $(2 - 5)^2 + (8 - 2)^2 + (1 - 7)^2 = (-3)^2 + (6)^2 + (-6)^2 = 9 + 36 + 36 = 81$
• Since the result equals $r^2$, point P lies on the sphere K.

c) Intersection of the line g with the sphere K

• Given line g: $\vec{x} = \begin{pmatrix} 12 \ 8 \ 0 \end{pmatrix} + t \begin{pmatrix} 9 \ -1 \ 1 \end{pmatrix}$
• Sphere K: $(x - 5)^2 + (y - 2)^2 + (z - 7)^2 = 81$
• Substitute the line equation into the sphere equation:
• $(12 + 9t - 5)^2 + (8 - t - 2)^2 + (0 + t - 7)^2 = 81$
• $(7 + 9t)^2 + (6 - t)^2 + (t - 7)^2 = 81$
• $(49 + 126t + 81t^2) + (36 - 12t + t^2) + (t^2 - 14t + 49) = 81$
• $83t^2 + 100t + 134 = 81$
• $83t^2 + 100t + 53 = 0$
• Use the quadratic formula to solve for t:
• $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-100 \pm \sqrt{100^2 - 4 \cdot 83 \cdot 53}}{2 \cdot 83}$
• $t = \frac{-100 \pm \sqrt{10000 - 17516}}{166} = \frac{-100 \pm \sqrt{-7516}}{166}$
• Since the discriminant is negative, there are no real solutions for t. Therefore, the line g does not intersect the sphere K.

d) Center and radius of the circle of intersection of the sphere K with the plane ε1

• Given plane ε1: 12x + y + 2z - 8 = 0
• Sphere K: Center M(5/2/7), radius r = 9
• Find the distance d from the center M to the plane ε1:
• $d = \frac{|Ax<em>M + By</em>M + Cz_M - D|}{\sqrt{A^2 + B^2 + C^2}} = \frac{|12(5) + 1(2) + 2(7) - 8|}{\sqrt{12^2 + 1^2 + 2^2}} = \frac{|60 + 2 + 14 - 8|}{\sqrt{144 + 1 + 4}} = \frac{68}{\sqrt{149}}$
• $d \approx \frac{68}{12.207} \approx 5.57$
• The center of the circle is the projection of M onto the plane ε1. Let this point be C.
• The normal vector of the plane is $\vec{n} = \begin{pmatrix} 12 \ 1 \ 2 \end{pmatrix}$
• Parametric equation of the line through M in the direction of $\vec{n}$: $\vec{x} = \begin{pmatrix} 5 \ 2 \ 7 \end{pmatrix} + s \begin{pmatrix} 12 \ 1 \ 2 \end{pmatrix}$
• Substitute into the plane equation: $12(5 + 12s) + (2 + s) + 2(7 + 2s) - 8 = 0$
• $60 + 144s + 2 + s + 14 + 4s - 8 = 0$
• $149s + 68 = 0$
• $s = -\frac{68}{149} \approx -0.456$
• Coordinates of C: $\vec{C} = \begin{pmatrix} 5 + 12(-0.456) \ 2 + (-0.456) \ 7 + 2(-0.456) \end{pmatrix} = \begin{pmatrix} 5 - 5.472 \ 2 - 0.456 \ 7 - 0.912 \end{pmatrix} = \begin{pmatrix} -0.472 \ 1.544 \ 6.088 \end{pmatrix}$
• Radius of the circle: $r_{circle} = \sqrt{r^2 - d^2} = \sqrt{9^2 - (5.57)^2} = \sqrt{81 - 31.0249} = \sqrt{49.9751} \approx 7.07$
• Center of the circle: C(-0.472 / 1.544 / 6.088), radius of the circle: approximately 7.07.

e) Angle of intersection between the line g and plane ε2

• Given line g: $\vec{x} = \begin{pmatrix} 12 \ 8 \ 0 \end{pmatrix} + t \begin{pmatrix} 9 \ -1 \ 1 \end{pmatrix}$
• Given plane ε2: 2x + y - 2z + 2 = 0
• Direction vector of line g: $\vec{v} = \begin{pmatrix} 9 \ -1 \ 1 \end{pmatrix}$
• Normal vector of plane ε2: $\vec{n} = \begin{pmatrix} 2 \ 1 \ -2 \end{pmatrix}$
• Angle between line and plane: $\sin(\alpha) = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| \cdot |\vec{n}|}$
• $\vec{v} \cdot \vec{n} = (9)(2) + (-1)(1) + (1)(-2) = 18 - 1 - 2 = 15$
• $|\vec{v}| = \sqrt{9^2 + (-1)^2 + 1^2} = \sqrt{81 + 1 + 1} = \sqrt{83}$
• $|\vec{n}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$
• $\sin(\alpha) = \frac{|15|}{\sqrt{83} \cdot 3} = \frac{15}{3\sqrt{83}} = \frac{5}{\sqrt{83}} \approx \frac{5}{9.11} \approx 0.549$
• $\alpha = \arcsin(0.549) \approx 33.33°$
• The angle of intersection between line g and plane ε2 is approximately 33.33 degrees.

f) Shortest distance between the sphere K and the plane ε3

• Given plane ε3: x - 2y + 2z + 27 = 0
• Sphere K: Center M(5/2/7), radius r = 9

(i) Shortest Distance

• Distance from M to plane ε3: $d = \frac{|Ax<em>M + By</em>M + Cz_M + D|}{\sqrt{A^2 + B^2 + C^2}}$
• $d = \frac{|1(5) - 2(2) + 2(7) + 27|}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{|5 - 4 + 14 + 27|}{\sqrt{1 + 4 + 4}} = \frac{|42|}{\sqrt{9}} = \frac{42}{3} = 14$
• Shortest distance between the sphere and the plane: $d_{shortest} = d - r = 14 - 9 = 5$

(ii) Coordinates of the point on the sphere K with the shortest distance to plane ε3

• The point lies on the line through M normal to plane ε3.
• Normal vector of plane ε3: $\vec{n} = \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix}$
• Parametric equation of the line through M in the direction of $\vec{n}$: $\vec{x} = \begin{pmatrix} 5 \ 2 \ 7 \end{pmatrix} + t \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix}$
• We want a point on the sphere, so the distance from M is 9. The point will be in the opposite direction of the plane (since d > r).
• Since d = 14 and r = 9. We want t such the distance is 9 in the -normal direction. Set the vector length 9, pointing away from Plane 3
• $\text{Point} = \begin{pmatrix} 5 \ 2 \ 7 \end{pmatrix} - \frac{9}{14} * \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix}$
• Coordinate on Sphere closest to Plane 3: $\text{Point} = \begin{pmatrix} 5-9/3 \ 2+18/3 \ 7-18/3 \end{pmatrix} = \begin{pmatrix} 2 \ 8 \ 1 \end{pmatrix}$

Area and Volume

A2 Area and Volume

• Given function: $f(x) = -2x^2 + 8x - 6$
• Task: Calculate the area A enclosed by the function graph and the x-axis.
• First, find the roots of $f(x) = 0$: $-2x^2 + 8x - 6 = 0 \implies x^2 - 4x + 3 = 0$
• $(x-1)(x-3) = 0 \implies x = 1, 3$
• The area can be found by integrating the function between the roots:
• $A = \int<em>{1}^{3} (-2x^2 + 8x - 6) dx = [-\frac{2}{3}x^3 + 4x^2 - 6x]</em>{1}^{3}$
• $A = [-\frac{2}{3}(3)^3 + 4(3)^2 - 6(3)] - [-\frac{2}{3}(1)^3 + 4(1)^2 - 6(1)] = [-18 + 36 - 18] - [-\frac{2}{3} + 4 - 6] = 0 - [-\frac{2}{3} - 2] = 0 - [-\frac{8}{3}] = \frac{8}{3}$
• Area A: $\frac{8}{3}$ square units
• Task: Calculate the volume of the solid formed when area A rotates about the x-axis by 360°.
• Volume of solid of revolution: $V = \pi \int<em>{a}^{b} [f(x)]^2 dx = \pi \int</em>{1}^{3} (-2x^2 + 8x - 6)^2 dx$
• $V = \pi \int_{1}^{3} (4x^4 - 32x^3 + 88x^2 - 96x + 36) dx$
• $V = \pi [\frac{4}{5}x^5 - 8x^4 + \frac{88}{3}x^3 - 48x^2 + 36x]_{1}^{3}$
• $V = \pi [(\frac{4}{5}(3)^5 - 8(3)^4 + \frac{88}{3}(3)^3 - 48(3)^2 + 36(3)) - (\frac{4}{5}(1)^5 - 8(1)^4 + \frac{88}{3}(1)^3 - 48(1)^2 + 36(1))]$
• After calculating, $V = \frac{64\pi}{15}$ cubic units

A3 Family of Curves

• Given: Family of curves $f_c(x) = x^3 - \frac{3}{c}x^2 + 2$, with $c \neq 0$.

• Determine the x-coordinate of the inflection point $I_c$ as a function of c.

• First, find the first derivative: $f_c'(x) = 3x^2 - \frac{6}{c}x$

• Second derivative: $f_c''(x) = 6x - \frac{6}{c}$

• Set the second derivative to zero to find the inflection point: $6x - \frac{6}{c} = 0 \implies 6x = \frac{6}{c} \implies x = \frac{1}{c}$

• x-coordinate of inflection point: $x = \frac{1}{c}$

• Now, let $c = 2$. Then, $f_2(x) = x^3 - \frac{3}{2}x^2 + 2$

• Determine the roots, extrema, and inflection point of $f_2(x) = x^3 - \frac{3}{2}x^2 + 2$
  First, find the roots.
  The roots for this situation is difficult to estimate as only one root is apparent, the other solutions are a non-terminating decimal, approximate numbers can be found using the calculator.
  Roots ≈ (-0.839, 1.339)
  After calculating, $x = 1/2$
  *Determining the local maximas, we can do this by substituting possible values into the first equation.
  We get that there should be a turning point at 0 and at 1.

• Area A rotates about the x-axis by $2\pi$. Calculate the volume of the solid of revolution.

A5 Finding the Function Equation

• Let f be a cubic function. Parabola $p(x) = \frac{1}{2}x^2$ touches the graph of f at $x = 0$. Graph of f has a maximum at point P(3 | 4.5).

Find the function equation of f.
Since p(x) touches at graph f at x=0. p(0)=f(0). Therefore we substitute zero into p(x) which will give = $\frac{1}{2} * 0^2 =0$.
this means that the y intercept for f(x) is also zero. This means that $f(x) = ax^3 + bx^2 + cx + 0$
We also know that p'(0)=f'(0) for the graph, and where there is a maximum on this graph f'(x) = 0
So finding the first derivative to make it easy
$f'(x) = 3ax^2 + 2bx + c$
$p'(x) = 1x$

Therefore $p'(0) = 1*0 = 0$
Thus, $f'(0) = 0$
If that’s the case, then C= 0. This leaves us with “$f(x) = ax^3 + bx^2$"
Current information: Graph has maxima at the point p(3 | 4.5), where Maxima should be f’(x)=0
therefore
$f'(3) = 3a(3)^2 + 2b(3) = 27a + 6b = 0$
f(3)= 4.5, therefore
$f(3) = a(3)^3 + b(3)^2 = 27a + 9b = 4.5$

We now have a system of equation
27a+6b=0
27a+9b=4.5
We subtract the system of equation from one another to find b.
3b=4.b=1.5
Subsisting b=1.5 into 27a+6b=0.
27a + 9 = 0, a = - ⅓
Therefore the equation = $- \frac{1}{3} + 1.5x^2$

Functions: Revision

• Definition of a function
  • Relation between two quantities (variables)
  • For every value of the first variable, there is only one value of the second variable