Download Final Exam Review Part 1

Honors Math 3
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Final Exam Review Problems Part 1
Chapter 1: Functions and polynomials
1. a. Find a polynomial that agrees with each table below. Use the two methods
discussed in class at least once: Lagrange interpolation and “agree-to-disagree”
(Sasha’s method).
x
f(x)
x
g(x)
x
h(x)
-3
-27
-2
2970
-1
2
4
64
0
-378
2
23
6
108
3
1440
4
47
9
729
7
-504
6
527
10
1000
9
5940
8
2711
b. Suppose one student uses Lagrange to find the polynomial g(x) while another
student uses Sasha’s method. They each get polynomials of degree 4. Assuming
neither student made errors in applying their method, should the two students’
polynomials be equivalent? Explain why or why not.
2. Given a polynomial of degree 5, list all the possible numbers of distinct real zeros it
could have. For each possible number of zeros, sketch a graph showing what the
polynomial could look like.
3. a. Divide x 5 -19x 3 - 26x 2 - 36x +80 by x 2 + 2x + 4 .
b. Factor x 5 -19x 3 - 26x 2 - 36x +80 over R.
c. Factor x 5 -19x 3 - 26x 2 - 36x +80 over C.
4. Factor each of the following expressions (over Z unless otherwise specified).
a. x 3 - 3x 2 - 7x + 21 (over R)
b. (2y -1)3 -8
c.
x 3 + 3x 2 - 25x + 21
d. x 4 -14x 2 + 45 (over R)
e.
t 4 -11t 2 + 25
f.
y9 -8y6 - 9y3 + 72
g. 27x 3 + 27x 2 + 9x +1
5. a. Let g(x) = x 3 + 2x 2 + x - 5 . Divide g(x) by (x + 2). Use your result to find g(– 2)
without substituting –2 into the formula for g.
b. Let h(x) = x107 + 6x 52 - 2x 33 -19x + 2. Find the remainder when h(x) is divided by
(x – 1).
6. a. Prove that if (x – a) is a factor of polynomial f(x) then x = a is a zero of f(x).
b. Prove that for any polynomial f(x) and constant k, (x – k) is a factor of f(x) – f(k).
c. Find a constant k such that (x + 2) is a factor of x 4 - 2x 3 + kx 2 + 3x - 6 .
7. Sketch and write an equation for a function that has the following attributes:
o degree of 5
o two x-intercepts, at x = –2 and at x = 3
o y-intercept at y = 4
o as x approaches infinity, y approaches negative infinity
o as x approaches negative infinity, y approaches infinity
8. Write a polynomial function formula that could fit each graph below. You will need
to approximate a point to find a.
a.
b.
b.
Chapter 2: Sequences and series
9. Evaluate each sum below using Gauss’ method or Euclid’s method. Show all steps of
each method you use!
a. 4 + –2 + –8 + … + –62
b. 4 + –12 + 36 + … + –972
10. Write each sum below in sigma notation, then evaluate using sum identities.
a. the sum of the odd integers from 3 to 97, inclusive
b. the sum of the first 8 terms of the geometric sequence with initial term 9 and
common ratio -2
c. the sum of the first n terms of the arithmetic sequence with third term 10 and fifth
term 2
d. the sum of the first n terms of the geometric sequence 18, 6, 2, …
11. Find a closed form formula for the sum
1
1
3
9
27
81
+ 5+ +1+ - 3+ - 7 + -11+ +... + 5- 4n . Hint: Write this series as the
3
2
4
8
16
32
sum of two other series.
16
8
4
i, - , - i, ... to answer the following questions.
3
9
27
a. Imagine you plotted each term of the sequence on the complex plane. Identify the
geometric relationship between subsequent terms.
12. Use the sequence 32,
b. Suppose we wanted to know the sum of the entire infinite sequence. Based on the
conditions we discussed for when an infinite real number series has a limit, under
what conditions do you think an infinite complex series has a limit?
c. After checking your answer to part c, determine whether the sum has a limit. If so,
find it.
13. A variation of the Koch snowflake is formed as follows. Start with the unit square.
On the middle third of each line segment attach a square pointing outward, then
remove the original middle third. Repeat this process forever. Stages 0 through 2 are
shown below.
a. Write a series that represents the total area after infinitely many iterations. Find
the value of this series if it has a limit.
b. Write a series that represents the total perimeter after infinitely many iterations.
Find the value of this series if it has a limit.
14. a. Write out rows 0 through 8 of Pascal’s triangle.
b. Circle two different sets of numbers that sum to 70, using the properties of
Pascal’s triangle.
c. Let f(n) be defined as in the table below. First, fill in the first cumulative sum
column. Then, describe how you could use Pascal’s triangle to find the
cumulative sums ΣΣf(n)and ΣΣΣf(n).
d. Expand (2x - y)4 . Simplify your answer.
e. Find the coefficient of the a 5b3 term of the expansion of (a + 4b)8 without writing
out the entire expansion.
n f(n)
0
1
1
3
2
6
3
10
4
15
5
21
6
28
Σ
Chapter 3: Statistics
15. Suppose we gave a course wide survey about how many weeks of summer students
will spend away from Lexington. The data collected are as follows:
Weeks Out of Town
Number of Students
0
44
1
33
2
36
3
8
4
5
5
12
6
27
7
3
8
2
a. Is this data approximately normally distributed? Justify your answer.
b. Find the mean, variance, and standard deviation for the number of weeks a
student spends out of town.
16. A standard number cube is rolled, and a spinner with equal fifths labeled 2, 5, 6, 8,
and 9 is spun. Find the mean, variance, and standard deviation for the sum of the
values obtained from the roll and spin.
17. A sunscreen manufacturer hires a researcher to develop and implement a welldesigned experiment to gather information about whether their new formula of
sunscreen is more effective than their old formula in preventing sunburns.
Researchers collect data about whether study participants get a sunburn after being
outside for an extended time and wearing one of the two formulas of sunscreen. The
results are shown in the table below.
Sunburn
No Sunburn
Total
Old Formula
30
220
250
New Formula
20
230
250
Total
50
450
500
Do you believe the new formula is more effective than the old formula? Show
calculations to support your answer.
18. Let {x1, x2, … xn} be a data set.
a. How does adding a constant c to each element of the data set change the mean and
variance? Justify your answer.
b. How does multiplying each element of the data set by a constant k change the
mean and variance? Justify your answer.
19. Suppose the Honors Math 3 teachers have a serious lapse in judgment and decide to
give a multiple choice final exam. (Note: The probability of this actually happening is
zero.) The final consists of 100 multiple choice questions, each with five possible
answers, one of which is correct. Now suppose that you also have a serious lapse in
judgment and neglect to study, and end up randomly guessing on all questions.
a. Find the mean, variance, and standard deviation for the number of questions you
get correct.
b. Find the exact probability that you get between 20 and 30 questions correct,
inclusive. Leave your answer expressed as a sum.
c. Explain why and how you can approximate the number of questions you get
correct using a Normal distribution. Then use this Normal distribution to find an
approximate probability that you get between 20 and 30 questions correct.
d. Find the approximate probability that you get more than 50% of the questions
correct.
20. In 2013, the average SAT math score was 514 with a standard deviation of 118, while
the average ACT math score was 21.1 with a standard deviation of 5.2. Assume
scores on these two assessments are normally distributed.
a. If a student earned a 650 on the SAT and a 28 on the ACT, on which test did the
student perform better?
b. Approximately what percentage of students nationwide scored higher than a
student who earned a 632 on the SAT?
21. An ice cream shop advertises that its small frappes are 16 ounces, but the amount
customers actually get is normally distributed with a mean of 14.5 ounces and a
standard deviation of 0.4 ounce.
a. Suppose you treat yourself to a small frappe after your math final. Find the
approximate probability that your frappe is between 13.7 and 15.3 ounces.
b. Find the approximate probability that your frappe is less than 14 ounces.
c. Imagine the Honors Math 3 teachers decide to treat all their students and
themselves to a frappe after their math final. (Unfortunately, like the hypothetical
multiple choice final, this is not going to actually happen.) Find the approximate
probability that out of all 175 students and 3 teachers, the average we each get is
between 15 and 16 ounces. Explain why your answer makes sense.
Answers
1
1. a. Forms may vary: f (x) = x 3 - (x + 3)(x - 4)(x - 9)(x -10)
2
g(x) = 3x(x - 3)(x - 7)(x - 9) +1(x + 2)(x - 3)(x - 7)(x - 9) + 4(x + 2)(x)(x - 7)(x - 9)
+1(x + 2)(x)(x - 3)(x - 9) + 5(x + 2)(x)(x - 3)(x - 7)
h(x) = 2 + 7(x +1)+1(x +1)(x - 2)+8(x +1)(x - 2)(x - 4)+ 2(x +1)(x - 2)(x - 4)(x - 6)
b. The two polynomials may look different, but are equivalent. The students have
come up with two degree four polynomials that agree at five points (the points in
the table). By Corollary 1.2.2, the two polynomials are equivalent.
2. A fifth degree polynomial could have 1, 2, 3, 4, or 5 distinct real roots. See below for
possible graphs.
3. a. x 3 - 2x 2 -19x + 20
b. (x 3 - 2x 2 -19x + 20)(x 2 + 2x + 4) ® (x -1)(x + 4)(x - 5)(x 2 + 2x + 4)
c. (x -1)(x + 4)(x - 5)(x - (-1+i 3))(x - (-1-i 3))
4. a. Factor by grouping. Group the first two terms and last two terms together, then
use GCF factoring. (x - 3)(x + 7)(x - 7)
2
b. Use difference of cubes identity then simplify. (2y - 3)(4y + 3)
c. Use find a root, find a factor. Possible factors of 21 are 1, 3, 7, 21. One root is 1,
because
13 + 3(12 ) - 25(1)+ 21=1+ 3- 25+ 21= 0 . So
x 3 + 3x 2 - 25x + 21= (x -1)(some quadratic) . Use synthetic or long division to
find the quadratic, then factor completely.
x3 + 3x 2 - 25x + 21= (x -1)(x 2 + 4x - 21) = (x -1)(x + 7)(x - 3)
d. Make a substitution (ex: u = x2), rewrite as a quadratic then factor. Always
remember to substitute back at the end, and continue factoring further if possible.
(x + 5)(x - 5)(x + 3)(x - 3)
e. Difference of squares in disguise: t 4 -11t 2 + 25 ® u2 -11u + 25 = u2 -10u + 25- u
= (u - 5)2 - u ® (t 2 - 5)2 - t 2 = (t 2 - 5+ t)(t 2 - 5- t) = (t 2 + t - 5)(t 2 - t - 5)
f. Let u = y3. Rewrite as a cubic, then factor by grouping.
y9 -8y6 - 9y3 + 72 ® u3 -8u2 - 9u + 72 = u2 (u -8) - 9(u -8) = (u2 - 9)(u -8)
= (u + 3)(u - 3)(u -8) ® (y3 + 3)(y3 - 3)(y3 -8) = (y3 + 3)(y3 - 3)(y - 2)(y2 + 2y + 4)
g. Factor by grouping, but group first and last term together, and second and third
term together. 27x 3 + 27x 2 + 9x +1= (27x 3 +1)+ (27x 2 + 9x) . Factor the first piece
using sum of cubes identity, and the second using GCF factoring, then simplify.
(27x 3 +1) + (27x 2 + 9x) = (3x +1)((3x)2 - 3x +1) + 9x(3x +1)
= (3x +1)(9x 2 - 3x +1) + 9x(3x +1) = (3x +1)(9x 2 + 6x +1) = (3x +1)(3x +1)2 = (3x +1)3
5. a. Quotient: x 2 +1, Remainder: -7. By remainder theorem, g(–2) = – 7.
b. h(1) = –12, so by remainder theorem, the remainder is –12.
6. a. We’re given that x – a is a factor of f(x). It follows that f(x) can be written as the
product of (x – a) and some other polynomial, q(x). So f(x) = (x – a)(q(x)). But
when x = a, we have f(a) = (a – a)(q(x)) = 0. Therefore, by definition of zero, x =
a is a zero of f(x).
b. By the Remainder Theorem, the remainder when f(x) is divided by (x – k) is f(k).
The Euclidean property states that f(x) = (x – k) q(x) + f(k). By subtracting f(k)
from both sides to get f(x) – f(k) = (x – k) q(x), you can see that f(x) – f(k) can be
written as a product of (x – k) and some other function with no remainder. It
follows that (x – k) is a factor of f(x) – f(k).
c. In order for (x + 2) to be a factor of x 4 - 2x 3 + kx 2 + 3x - 6 , the remainder when
we divide x 4 - 2x 3 + kx 2 + 3x - 6 by (x + 2) should be 0. Use synthetic division to
divide and get a remainder in terms of k, then set equal to 0. Remainder should be
20 + 4k, so that k = -5 would make (x + 2) a factor.
1
7. y = - (x + 2)2 (x - 3)3 (graph to the right)
27
1
3
2
8. a. f (x) = - ( x +1) ( x -1) ( x - 3)
2
b. g(x) = 2(x + 4)(x +1)3 (x - 2)
9. a. –348, see steps for Gauss’ method on p. 91
b. –728, see steps for Euclid’s method on p. 92
10. a.
48
48
48
48
48
k=1
k=1
k=1
k=1
k=1
æ 48(49) ö
÷ + 48 = 2400
2 ø
å2k +1 = å2k + å1 = 2åk + å1 = 2 çè
b.
7
æ (-2)8 -1 ö
k
k
9(-2)
=
9
(-2)
=
9
ç
÷ = -765
å
å
-2
-1
è
ø
k=0
k=0
c.
æ n-1 ö æ n-1 ö
æ (n -1)(n) ö
2
2
18
4k
=18
çå1÷ - 4 çå k ÷ =18n - 4 ç
÷ =18n - 2(n - n) = -2n + 20n
å
è
ø
2
è k=0 ø è k=0 ø
k=0
7
n-1
æ æ 1 ön ö
æ æ 1 ön ö
ç
÷
ç ç ÷ -1 ÷
-1
ç ÷
n-1
n-1
æ n ö
æ1ö
æ1ö
è 3ø
è 3ø
ç
÷
÷ = -27 çæç 1 ö÷ -1÷
d. å18 ç ÷ = 18åç ÷ = 18
= 18 ç
çè 3 ø
÷
ç 1
÷
ç -2 ÷
è 3ø
è ø
è
ø
k=0
k=0 3
-1
ç 3
÷
ç 3 ÷
è
ø
è
ø
k
k
11. Alternating terms starting with the first term belong to a geometric series with a = 1/3
and
r = 3/2. Alternating terms starting with the second term belong to an arithmetic series
with
a = 5 and d = –4. Sum these two series separately, then add together.(continued on
next page)
æ æ 3 ö n +1 ö
ç ç ÷ -1÷
k
k
n
n
n
n
æ n(n +1) ö
1æ 3ö
1 n æ 3ö
1 çè 2ø
å 3çè 2 ÷ø + å5 - 4k = 3 åçè 2 ÷ø + å5 - 4å k = 3 ç 3 ÷÷ + 5(n +1) - 4çè 2 ÷ø
k =0
k =0
k =0
k =0
k =0
ç 2 -1 ÷
è
ø
2 ææ 3ö
çç ÷
3 èè 2 ø
n +1
n +1
n +1
ö
ö
ö
2 ææ 3 ö
2 ææ 3ö
-1÷ + 5n + 5 - 2n(n +1) = çç ÷ -1÷ + 5n + 5 - 2n 2 - 2n = çç ÷ -1÷ - 2n 2 + 3n + 5
3 èè 2 ø
3 èè 2 ø
ø
ø
ø
12. a. From one term to the next we multiply by (1/6)i. Multiplying by i results in a
rotation by 90 degrees on the complex plane, while multiplying by the constant
1/6 results in one term being 1/6 as far from the origin as the previous term.
b. Still need r <1, but recall that a + bi = a 2 + b2 .
æ1ö 1
a
32
1
=
or
i = 0 2 + ç ÷ = <1 , so it does have a limit of
1- r 1- 1 i
è6ø 6
6
6
1152 192
+
i.
37
37
2
c.
r=
æ1ö
æ1ö
æ1ö
13. a. area =1+ 4 ç ÷ + 20 ç ÷ +100 ç ÷ +...
With the exception of the
è 3ø
è9ø
è 27 ø
first term, this is geometric with a = 4/9, r = 5/9. Since |r| < 1, the geometric part
4
a
= 9 = 1 . Therefore, the entire series (with the
of the series has a limit of
1- r 1- 5
9
first term) has a limit of 2.
2
2
2
æ1ö
æ1ö
æ1ö
b. perimeter = 4 + 4(2)ç ÷ + 20(2)ç ÷ +100(2)ç ÷ +...
è 3ø
è9ø
è 27 ø
With the exception of the first term, this is geometric with a = 8/3, r = 5/3. Since
|r| > 1, the geometric part of the series increases to infinity. So the shape has a
finite area but infinite perimeter!
14. a.
b. Could circle the two 35 entries right above to the
left and right of 70, and a “hockey stick” with the tip of
the stick on one of the outer diagonals of the triangle
(entries 1, 4, 10, 20, 35).
c. See table for first cumulative sum column.
Notice that the f(n) column is the third diagonal of
Pascal’s triangle, and the first cumulative sum
column is the fourth diagonal of Pascal’s triangle.
The entries in the second cumulative sum column
can be found in the fifth diagonal of Pascal’s
triangle, and the entries in the third cumulative sum
column can be found in the sixth diagonal.
d. (2x - y)4 =16x 4 - 32x 3 y + 24x 2 y2 -8xy3 + y 4
æ
ö
e. a 5b3 term will be of the form ç 8 ÷a 5 (4b)3 so coefficient is
è 3 ø
æ 8 ö 3
ç
÷× 4 = 56(64) = 3584 .
è 3 ø
15. a. No. The data are not symmetrical, and there are two peaks in the data (one 0-2
weeks and one 5-6 weeks.
b. mean: 2.4 weeks, variance: 5.275, standard deviation: 2.297
16. Since the outcome from the spin and the die roll are independent, we know the mean
and variance of the sum will be equal to the sum of the individual means and
variances respectively. It will be more efficient to calculate these individual means
and variances then add them than to use other strategies. Also, note that the standard
deviation is not additive, so you need to calculate the variance using additivity, then
just take the square root to get the standard deviation. mean: 9.5, variance: 8.917,
standard deviation: 2.986
17. The control group is the participants who used the old formula. Therefore,
30
p = 250
= 0.12 , which is the probability of getting a sunburn assuming the new
formula is not any more effective than the old. We need to determine the probability
that the treatment group would get the results it did (or better) assuming this p value.
normalcdf(0, 20, 250 × 0.12 , 250(0.12)(1- 0.12) ) = 0.0258 or 2.58% Since this is
lower than our 5% cut-off, we say there is enough evidence to show the new formula
is more effective than the old formula at preventing sunburns.
18. a. new mean = old mean + c; new variance = old variance
b. new mean = old mean * k; new variance = old variance * k2
19. a. The number of questions you get correct is the number of successes in 100
independent Bernouilli trials, each with probability 1/5 of success. Using the
formulas we derived for statistics about successes in Bernouilli’s trials:
mean: 100(0.2) = 20, variance: 100(0.2)(0.8) = 16, standard deviation = 4
æ 100 ö
å ç k ÷(0.2)k (0.8)100-k
ø
k=20 è
30
b.
c. The Central Limit Theorem states that if X is a random variable with mean m and
standard deviation s , the distribution for the sum of all the outputs of X over n
experiments as n gets large is approximately Normal with mean m n and standard
deviation s n . If you consider a single experiment to be guessing on a single
question and the output to be the number correct you get on that guess (0 or 1),
then the sum of these outputs after 100 experiments gives you the number of
questions you get right in total. The mean m for a single trial is p = 0.2 and the
standard deviation s is p(1- p) = 0.2 × 0.8 = 0.4 The CLT says the number of
questions you get right should be normally distributed with mean 100(0.2) = 20
and standard deviation 0.4 100 = 4 .
normcdf(20, 30, 20, 4) = 0.49
d. normcdf(50, 100, 20, 4) = about 0
æ (650 - 514)
ö
20. a. By calculating z-scores for the student’s scores on the SAT ç
» 1.153÷
è
ø
118
æ 28 - 21.1
ö
and on the ACT ç
» 1.327÷ you can determine that the student scored
è 5.2
ø
1.153 standard deviations above the mean on their SAT as compared to 1.327
standard deviations above the mean on their ACT. Therefore, they performed
better on their ACT than SAT.
b. A 632 is exactly 1 standard deviation above the mean. We know that about 68%
of scores will fall within 1 standard deviation of the mean, such that about 32% of
scores will fall outside 1 standard deviation of the mean. Half of these scores will
be lower than 1 standard deviation below the mean, and half will be higher than 1
standard deviation above the mean. Therefore, about 16% of students nationwide
scored higher than 632.
21. a. 13.7 ounces is exactly two standard deviations below the mean and 15.3 is exactly
two standard deviations above the mean. We know about 95% of the time
normally distributed data should fall within two standard deviations of the mean.
So the probability is 0.95.
b. normcdf(0, 14, 14.5, 0.4) = 0.11
c. By second interpretation of CLT, the mean of the outputs over 178 “experiments”
should be approximately normally distributed with mean 14.5 and standard
0.4
» 0.02998 . P(avg. between 15 and 16) = normcdf(15, 16, 14.5,
deviation
178
0.02998) » 0 . This makes sense because with a larger sample size, the standard
deviation decreases and the spread of possible results narrows. Therefore, it is
extremely unlikely to have our result fall within 15 and 16 ounces.