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Material Type: Exam; Class: PRE-CALCULUS; Subject: Mathematics; University: University of California - Irvine; Term: Winter 2005;
Typology: Exams
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MATH 1B, Lecture C (44047) Midterm Exam 1 [100 points] - Key, Tuesday, February 8, 2005 Winter 2005, Dr. Masayoshi Kaneda, University of California, Irvine
Name (Printed):
Student ID:
1
(1) [2] (a) 5^50 · 550 ; (b) 5^100.
Solution: 550 ·^550 = 550+50^ = 5^100. (1) a. b. c. (2) [2] (a) 1 ; (b) 5
1 (^5).
Solution: The exponential function 5x^ is increasing since the base 5 is greater than 1. Thus 5
(^15)
50 = 1. (2) a. b. c. (3) [2] (a) 5−^1 ; (b) e−^1.
Solution: Since 5 > e, 5−^1 < e−^1. (3) a. b. c. (4) [2] (a) 5
(^25) ; (b) 5
2
Solution: 5
2 55 = 5
(4) a. b. c. (5) [2] (a) 1 ; (b) ( 15 )^0.
Solution: ( 15 )^0 = 1. (5) a. b. c. (6) [2] (a) ln 1 ; (b) 0.
Solution: ln 1 = 0. (6) a. b. c. (7) [2] (a) log 0.00001 ; (b) 15.
Solution: log 0.00001 = log 10−^5 = − 5 < 15. (7) a. b. c. (8) [2] (a) log 3 2 ; (b) log 2 3.
Solution: log 3 2 < log 3 3 = 1 = log 2 2 < log 2 3. (8) a. b. c. (9) [2] (a) ln 4 − ln 2 ; (b) ln 2. Solution: ln 4 − ln 2 = ln 42 = ln 2. (9) a. b. c. (10) [2] (a) (log 3 2)^3 ; (b) 3 log 3 2. Solution: Since 0 < log 3 2 < 1, (log 3 2)^3 < log 3 2. On the other hand, log 3 2 < 3 log 3 2. (10) a. b. c.
Solution: log 8 4 = log log^22 48 =
Solution: Take the logarithm with base 3. 3 x+2^ =9x ⇔ log 3 3 x+2^ = log 3 9 x ⇔ x + 2 = x log 3 9 ⇔ x + 2 = 2x ⇔ 2 = x.
Thus x = 2.
Solution: This problem is WeBWorK Assignment 2 #3 which has been done in the discussion class. e^2 x^ − 5 ex^ + 6 = ⇔ (ex^ − 2)(ex^ − 3) = 0 ⇔ ex^ = 2, 3 ⇔ x = ln 2, ln 3.
Thus x = ln 2, ln 3.
Solution: This problem is a homework problem (Exercise 2-6.37 on page 197 of the textbook), and done in the discussion class. Put y := 4 + x^2.
Then y = 4 + x^2 ⇔ y − 4 = x^2 ⇔
y − 4 = x (since x ≥ 0). Now switch x and y to get y =
x − 4.
Thus f −^1 (x) =
x − 4. Remark: Since it is easy to see that the range of f is [4, +∞), so is the domain of f −^1.
(1) [3] Find W (− 3 π/4). (2) [4] Find all solutions( (^) √ x with − 2 π ≤ x ≤ 2 π such that W (x) = 3 2 ,^ −^
1 2
Solution: (1)
W (− 3 π/4) =
(2) Since the wrapping function is a periodic function with period 2π, i.e., W (x + 2πk) = W (x), k ∈ Z, once you find one solution x = − π 6 which is easy to find, then the others are of the form π 6 + 2πk. But since there is a restriction − 2 π ≤ x ≤ 2 π, there are only 2 solutions:
x = −
π 6
11 π 6
(1) [2] Find the radian measure of the angle 18◦. (2) [2] Find the degree measure of the angle 1356 π radian. (3) [5] Convert the angle 354◦ 6 ′ 36 ′′^ to decimal degrees. (4) [5] Convert the angle 403. 225 ◦^ to degree-minute-second form.
Solution:
(1) 18 · 180 π = 10 π. Thus the answer is
π 10
radian.
(2) 1356 π · (^180) π = 8. Thus the answer is 8 ◦^. (3) This problem is similar to a homework problem (Exercise 5- 3.29 on page 364 of the textbook), but easier. 354 ◦ 6 ′ 36 ′′^ = (354 + 606 + 360036 )◦^ = (354 + 101 + 1001 )◦^ = (354 + 0.1 + 0 .01)◦^ = 354. 11 ◦^. (4) This problem is similar to a homework problem (Exercise 5- 3.33 on page 364 of the textbook), but easier.
