Solutions to MI 4 Problem set 2 F11
Introduction: This problem set continues some ideas from the MI 3 problem sets (see comments
on MI 3 problem set 6 from fall 10) as well as begins some new themes. At this point in the
course students are well into their unit on sequences and series, so several problems extend that
work using technology. At IMSA solutions to problem sets are posted so students can study
them. We encourage this for problems they may miss but also for learning new techniques or
gaining insight as is shown in problem #1 below.
1) Comment: This problem begins a series of problems on partial fractions. This first attempt is
given with no instruction. Most students manage to solve this with a system of equations or trial
an error. When the set is returned would be a good time to demonstrate the general idea of partial
fractions. At IMSA we do not ‘formally’ teach partial fractions, but find this series of problems
extended over several sets is enough work on the topic.
Soln:
2 x  11
A( x  2)  B( x  1)

( x  1)( x  2)
( x  1)( x  2)
2 x  11  A( x  2)  B( x  1)
Let x  2 then 15  3B  B  5
Let x  1 then 9  3A  A  3
2) Comment: This problem continues a theme begun in MI 3 of using interval graphs to solve a
problem.
Soln: log 2  x 2  4 x   5
x 2  4 x  25
x 2  4 x  32  0
( x  8)( x  4)  0 and x( x  4)  0 (from original domain constraints)
8  x  4 or 0  x  4
3) Comment: This is a good example of taking advantage of CAS. IMSA students have TI-89
access. Wolframalpha will also work with the factor command.
Solutions to MI 4 Problem set 2 F11
Soln: ( x  2)( x  3)(2 x  5)(3x  7)  0
7
5
 x  2 or x 
3
2
x  3 or
4) Comment: This problems has students think about sequences that get generated by an
iterative geometric process.
Soln: a) 2
 1 
b) 4, 2 2, 2, 2,... An Bn  4 

 2
n 1
 2
or 4 

 2 
n1
2
n 1
  1 n1 
 1  
c)  4 
   16 
 2 
 2  
  2  
1
16  
2
n1
or
32
2n
5) Comment: This problem asked students to think about generating a formula for a sequence of
partial sums. In their unit in class IMSA students will be asked to consider the lim S n .
n0
Soln: a) 1, 2, 4, 8, 16, 32, 64
b) 1, 3, 7, 15, 31, 63, 127
n 1
1
c) S n  2n  1 or Sn  
n 1
Sn1  2 n  1
6) Comment This is a review from MI 3 log properties.
Soln:
1
1
1
1



4
log 3 x log 6 x log8 x log 9 x
Solutions to MI 4 Problem set 2 F11
log x 3  log x 6  log x 8  log x 9  4
log x 1296  4
x 4  1296
x6
7) Comment: This problem is more practice at sequence generation with the calculator. The are
not many opportunities to use the sequence mode on the calculator in the typical class. This skill
on the calculator will come to good use as the theme begun in problem 16 carries through the
next few problem sets.
Soln: a) 1, 0.5403, 0.8576, 0.6543
b) 0.7391
8) Comment: A unit in MI 2 used 2X2 matrices to do some basic transformations on figures in
the plane. This problem extends using matrices for other transformations found in linear algebra,
although the problems sets do not go any further with this topic, it does leave that possibility.
 7 3 2 
Soln: a) V1   2 4 5 


 1 1 1 
 2 4 5 
V2   7 3 2 
 1 1 1 
 7 3 2
V3   2 4 5
 1 1 1 
b) This transformation rotates points 90° clockwise or 270° counterclockwise.
9) Comment: More interval practice, but this time considering issues arising from the
composition of functions.
Soln:
( x  5)( x  7)
0
( x  4)( x  2)
 ,  7 5, 2  4, 
10) Comment; IMSA students learned half of their trigonometry in MI 3, including graphing
sinusoidal functions. This problem is review in the context of an application.
Solutions to MI 4 Problem set 2 F11
Soln:
Amp: 220 Period: 6 phase shift: 4.75 vertical shift: 220


h(t )  220 cos   t  4.75    220
6

11) Comment: Students are practicing formulas developed in their first unit of MI 4 in this
problem.
10
Soln: a)
3
n
is geometric with 11 terms
n 0
Sn 
11  3n 
1 3
 88573
i

2
 2
b)  6    is an infinite geometric series with r  
3
3
i 1 
S 
a
4
12


1 r 1 2 3
5
n
n
n
i 1
i 1
i 1
c) 2 i 2  2 i  31 
2n(n  1)(2n  1) 2n(n  1)

 3n 
6
2
Solutions to MI 4 Problem set 2 F11
2n3  6n 2  5n
3
12) Comment: This problem gets students to use CAS to do basic summations. Hopefully they
will think to use it in future problem sets and calculus. Students should be surprised by their
exact answer to part b. Part c opens the opportunity to discus using the finance application found
on TI calculators
Soln: a)
9304682830147
  4  (5n), n,1,30   2911361953500
b)
  3  i 2 , i,1,   
c)
  55000 1.08
n 1
or 3.196
2
2
, n,1, 20   2,516,908.04
13) Comment: This problem continues what was begun in part c of problem 12. Hopefully there
is some motivation given by the application as to why one might study sequences and series.
u  1.082  u1 (n  1)  5000
Soln: a)  1
ui1  5000
5000, 10410, 16264, 22597, 29450
b) $ 1,477,408
c) n  37 we have $ 1,065,092
During his 36th year of saving.
14) Comment: This problem introduces the concept of the limit in a recursive sequence. It can
lead to the idea of a fixed point which is what this limit is. This opens the possibility of further
study in an application of managing a forest.
u1  0.92  u1(n  1)  700
Soln: a) 
ui 1  10000
10000, 9900, 9808, 9723, 9645
b) On home screen type: u1(20)  9006
c) u1(100)  8750
It appears to be approaching a steady population of 8750 trees.
Solutions to MI 4 Problem set 2 F11
15) Comment: Most students have encountered the Fibonacci sequence in middle school. This
problem gives students more practice at using the recursive features of the clalculator and sets
them up for a theme begun in the next problem.
u1  u1(n  1)  u1(n  2)
Soln: 
ui 1  {1,1}
u1(37)  24,157,817
16) Comment: This problem begins a series of problems that explore ratios in Fibonacci type
sequences and more generally Lucus sequences. In a future problem set it will be tied to
extended fractions and eventually the explicit formula for the Fibonacci sequence and other type
sequences. This theme shows some very rich mathematics can be developed through student
exploration using technology.
Soln: a) 1, 2, 1.5, 1.667, 1.6
b)
u1(21)
 1.618
u1(20)
c) Gn is approaching the Golden Ratio   1.618...
17) Comment: This problem is a review problem from the first half of trigonometry students did
in MI 3. We would expect students to take a geometrical approach to the problem. These
problems will be explored further as students do their analytical study of trig later in the term.
Soln:
a)  1  x
1


b) tan      cot 
2

X
- 1-x

 1 x
x