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MATH 1009-A TEST 1: Solutions January 2011

PART A: MULTIPLE CHOICE QUESTIONS.

A1. The domain of the function () = 9 is

(a) [3, 3], or, equivalently, {3 3}.(b) (9, ), or, equivalently, { > 9}.(c) [9, ), or, equivalently, { 9}.(d) (, 9], or, equivalently, { 9}.(e) none of the above.

Answer: (c)

A2. If () = 2

+ 1 and () = 3 + 2, then (()) is

(a) 2

3 + 2 + 1. (b) (2

+ 1)3 + 2. (c) 3 + 2

+ 3.

(d) 23

+ 2. (e) none of the above.

Answer: (a)

A3. The expression (52/3

52)3/4 evaluates to

(a) 5. (b) 5. (c) 15

. (d) 15

. (e) None of the above.

Answer: (b)

A4. The domain of the function () = ln( + 1) is

(a) [1, ), or, equivalently, { 1}.(b) (1, ), or, equivalently, { > 1}.(c) , or, equivalently, (, ).(d) (1, ), or, equivalently, { > 1}.(e) [1, ), or, equivalently, { 1}.

Answer: (d)

A5. The expression1.2 (0.1)2

5simplifies to

(a) 4.3. (b)

5.7. (c) 4. (d) 5. (e) 6.

Answer: (e)

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A6. The expression (836)1/3 simplifies to

(a) 22. (b) 22. (c)836

3. (d) 82 . (e) None of the above.

Answer: (b)

A7. The expression log3

1

3evaluates to

(a) 1. (b) 1. (c) 2. (d) 3. (e) None of the above.

Answer: (a)

A8. Write the expression (ln 2 +1

3ln 2 ln ) as the logarithm of a single quantity.

(a) ln(

3

). (b) ln

(21/3

2

). (c) ln

(2 +

3 2

). (d) log2

(21/3

2

).

(e) None of the above.

Answer: (b)

PART B: LONG STYLE QUESTIONS.

[6 marks] B1. The weekly demand and supply equations for a company are given by

= 68 2 and = 14 + 12

2, respectively, where is the price measured in dollars and is

measured in units of a thousand.

[2] (a) For the demand equation = 68 2, determine the quantity demanded, when theprice is set at 4 dollars.

4 = 68 2, 2 = 64, = 8.We reject the negative root = 8, since only positive values of the quantity demanded aremeaningful.

[1] (b) For the supply equation = 14 +1

22, determine the price at which the supplier will

make 2 thousand units available in the market.

= 14 +1

2 22 = 14 + 2 = 16.

[3] (c) Find the equilibrium quantity and price.

At the equilibrium point the supply is equal to the demand, and therefore

68 2 = 14 + 12

2.

Solving this equation for yields3

22 = 68 14 = 54, 2 = 36, = 6. (We reject

the negative root = 6, since positive values of demanded are meaningful.) Thus, the

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equilibrium quantity is 6 thousand units, and the corresponding price is = 68 62 = 32dollars per unit.

[11 marks] B2. Solve each of the following equations for .

[4] (a) 1 = 5 [3] (b) log2(2 5) = 0 [4] (c) ln + ln 6 ln 2 = 2Solution:

NOTE: There is more than one way of solving each of the equations.

(a) Take the natural logarithm of both sides of the equation and use the laws of logarithms:

ln(1) = ln 5 1 = ln 5 = ln 5 1 = 1 ln 5.

(b) Since the logarithm equals zero, the argument must be equal to 1:

2 5 = 1 2 = 6 = 3.

(c) Simplify the LHS of the equation and then exponentiate both sides of the relation:

ln(

62

)= 2 ln(3) = 2 ln(3) = 2 3 = 2 = 2/3.

[7 marks] B3. The amount of $20, 000 is deposited in a bank that pays interest at the rateof 6% per year compounded semiannually. Using the compound interest formula

() = (1 + )

,

answer the following questions. (Round each answer to one decimal.)

[2] (a) What is the accumulated amount on deposit in 5 years?

(5) = 20, 000(

1 +0.06

2

)25= 20, 000(1.03)10 = 26, 878.3

[1] (b) What is the interest earned in 5 years?

Interest (5) = (5)

= 26, 878.3

20, 000 = 6, 878.3

[4] (c) How long will it take to double the investment?

40, 000 = 20, 000(

1 +0.06

2

)2 2 = (1.03)2 ln 2 = 2 ln(1.03) = ln 2

2 l n 1.03= 11.7