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MATHEMATICS 1220 MIDTERM 1 SOLUTIONS
Overview The exam went well. The average was 23 but that is due to there being an outlier. The median is a better measure of central tendency in this case. The median is 26.5. Here is the score distribution. Score Number of Students 28 4 25 1 24 1 23 1 3 1 I do not curve exams. I do sometimes curve classes if I think it is appropriate. It would not be appropriate to curve the class if the class scores were along these lines. The preset scale outlined on the syllabus tells us that the grade scale when scores are given out of 30 is as follows.
Score Grade 27-32 A, A- 23-26 B+, B, B- 20-22 C+, C 18, 19 C- 14-17 D+, D, D- 0-13 E
Thus, 7 out of 8 passed with 4 people earning an ‘A’. Well done! Here are the solutions.
- Problem 1 Let f (x) = xx^ for x > 0. Compute f ′(e).
Solution 1.1. Write f (x) = ex^ ln^ x. By the chain and product rules,
(1.1) f ′(x) = ex^ ln^ x
(ln x +
x x
= xx(ln x + 1).
Hence,
(1.2) f ′(e) = ee(ln e + 1) = 2ee. 1
- Problem 2 Compute
ln xdx.
Solution 2.1. We integrate by parts. Let u = ln x and dv = 1. Then du = (^) x^1 and v = x. Hence,
ln xdx = x ln x −
xdx x
= x ln x −
1 dx = x ln x − x + C
where C is an arbitrary constant.
- Problem 3 Show that for every x, y ∈ R
(3.1) cosh(x + y) = cosh x cosh y + sinh x sinh y.
Solution 3.1. From the definition of cosh and sinh we have:
cosh x cosh y + sinh x sinh y
=
ex^ + e−x 2
ey^ + e−y 2
ex^ − e−x 2
ey^ − e−y 2
=
exey^ + exe−y^ + e−xey^ + e−xe−y^ + exey^ − exe−y^ − e−xey^ + e−xe−y 4
=
2 ex+y^ + 2e−x−y 4
=
ex+y^ + e−x−y 2 = cosh(x + y).
- Problem 4 Compute
0
(2x^ − x^3 )dx.
Solution 4.1. We compute:
0
(2x^ − x^3 )dx =
2 x ln 2
x^4 4
4 0
ln 2
ln 2
ln 2
for some constant c. At x = 1,
(7.4) ln(y) = ln(1 · y) = g(1) = ln 1 + c = c.
Hence,
(7.5) g(x) = ln x + ln y
as required. Since this equation holds for any y > 0, our claim has been shown.
- Problem B It is well known that
2 is an irrational number. What sort of psychoanalysis should be given to straighten the poor number out?
Solution 8.1. Answers will vary.
