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MATH 1B, Lecture B (44034) Midterm Exam 1 [100 points] - Key, Tuesday, February 8, 2005 Winter 2005, Dr. Masayoshi Kaneda, University of California, Irvine
Name (Printed):
Student ID:
- You have 50 MINUTES to answer the following problems.
- SHOW ALL YOUR WORK. NO PARTIAL CREDIT will be given for an answer without work except for #1.
- SIMPLIFY YOUR ANSWERS AS MUCH AS POSSIBLE.
- NO CALCULATOR of any type is allowed.
- [20] In each problem, which is bigger? Circle “a” if (a) is bigger. Circle “b” if (b) is bigger. Circle “c” if both are equal.
(1) [2] (a) 3^150 ; (b) 3^3 · 350.
Solution: 33 · 350 = 33+50^ = 3^53 < 3150. (1) a. b. c. (2) [2] (a) 3
(^13) ; (b) (^13).
Solution: 3
1 (^3) = 3
(2) a. b. c. (3) [2] (a) 3−^1 ; (b) e−^1.
Solution: Since 3 > e, 3−^1 < e−^1. (3) a. b. c. (4) [2] (a) 3
(^53) ; (b) 3
5
Solution: 3
5 33 = 3
(4) a. b. c. (5) [2] (a) e^0 ; (b) 2^0.
Solution: e^0 = 1 = 2^0. (5) a. b. c. (6) [2] (a) ln 1 ; (b) log 3 1.
Solution: ln 1 = 0 = log 3 1. (6) a. b. c. 1
(7) [2] (a) log 0.01 ; (b) 12.
Solution: log 0.01 = log 10−^2 = − 2 < 12. (7) a. b. c. (8) [2] (a) ln 2 + ln 3 ; (b) ln 5.
Solution: ln 2 + ln 3 = ln(2 · 3) = ln 6 > ln 5. (8) a. b. c. (9) [2] (a) log 3 6 − log 3 2 ; (b) log 3 4.
Solution: log 3 6 − log 3 2 = log 3 62 = log 3 3 < log 3 4. (9) a. b. c. (10) [2] (a) log 3 (3^2 ) ; (b) 3 log 3 2.
Solution: log 3 (3^2 ) = log 3 9 > log 3 8 = log 3 (2^3 ) = 3 log 3 2. (10) a. b. c.
- [10] An 16-gram sample of radioactive material has a half-life of 400 years. After 800 years, how much are remaining?
Solution: In general, if a radioactive sample containing A 0 units has a half-life of k years, then the amount A(t) remaining after t years is given by
A(t) = A 0
) (^) kt .
In this problem,
A(800) = 16
Thus the answer is 4 gram. Note: You also need to know that if the bacteria culture starts with P 0 and doubles every k minutes, then after t minutes the population P (t) is given by
P (t) = P 0 · 2
t k (^).
- [10] Solve 2 log 3 x = log 3 2 + log 3 (4 − x).
Solution: This problem is a homework problem (Exercise 4-3.83 on page 313 of the textbook), and done in the discussion class.
2 log 3 x = log 3 2 + log 3 (4 − x) log 3 x^2 = log 3 2(4 − x) x^2 = 2(4 − x) x^2 = 8 − 2 x x^2 + 2x − 8 = 0 (x + 4)(x − 2) = 0 x = − 4 , 2.
Check whether these satisfy the original equation since not every step above is reversible. log 3 (−4) is not defined (in real numbers). So x = −4 is not a solution. x = 2 certainly satisfies the original equation. Thus x = 2 is the only solution.
- [5] Find f −^1 for f (x) = 2 + 5
3 x − 7. You do not have to show that f is one-to-one. Also you do not have to specify the domain of f −^1.
Solution: This problem is a homework problem (Exercise 2-6.55 on page 197 of the textbook), and done in the discussion class. Put y := 2 + 5
3 x − 7.
Then y = 2 + 5
3 x − 7 ⇔ y − 2 = 5
3 x − 7 ⇔ (y − 2)^5 = 3x − 7 ⇔ (y − 2)^5 + 7 = 3x ⇔ (y−2)
5 3 +^
7 3 =^ x. Now switch x and y to get
y =
(x − 2)^5 3
Thus
f −^1 (x) =
(x − 2)^5 3
Remark: Since it is easy to see that both the domain and the range of f are the set of all real numbers, so are the domain and the range of f −^1.
8. [7]
(1) [3] Find W (− 5 π/6). (2) [4] Find all solutions( x with − 2 π ≤ x ≤ 2 π such that W (x) = √^1 2 ,^
√^1 2
Solution: (1) This problem is a homework problem (Exercise 5-1.17 on page 347 of the textbook).
W (− 5 π/6) =
(2) This problem is a homework problem (Exercise 5-1.47 on page 347 of the textbook). 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 4 which is easy to find, then the others are of the form π 4 + 2πk. But since there is a restriction − 2 π ≤ x ≤ 2 π, there are only 2 solutions:
x = −
7 π 4
π 4
- [10] Given cos θ = 1/2 and tan θ < 0. Find the value of the other five trigonometric functions.
Solution: This problem is a homework problem (Exercise 5.2.75 on page 356 of the textbook and done in the discussion class. Since cos θ > 0 and tan θ < 0, we know that θ is in Quadrant IV. Thus sin θ < 0. By using Pythagorean identity cos^2 θ + sin^2 θ = 1,
sin θ = −
1 − cos^2 θ = −
tan θ = (^) cossin^ θθ = −
√ 3 / 2 1 / 2 =^ −
csc θ = (^) sin^1 θ = (^) −√^13 / 2 = −
sec θ = (^) cos^1 θ = (^11) / 2 = 2.
cot θ = (^) tan^1 θ = (^) −^1 √ 3 = −
Note: You can also solve graphically. Then you will need to use Pythagorean theorem in which you have to do essentially same calculation as when finding sin θ above. Note that Pythagorean identity cos^2 θ + sin^2 θ = 1 is essentially Pythagorean theorem.
