MATH 1300-MIDTERM # 1(a)-2007
NAME and I.D.#
Instructions: This midterm exam consists of 4 multiple choice questions and 3 long answer questions. The multiple choice questions are worth 5 points each, and the long answer questions are as indicated. The total value of the exam is 60 points. Place your answers to the multiple choice questions in the boxes below. All your work on the long answer questions must be clearly marked. You may use the backs of pages.
For long answer questions, YOU MUST SHOW YOUR WORK
NO CALCULATORS. NO BOOKS. NO NOTES.
If you need additional scrap paper, it will be provided by the proctors.
Answers: C #1 B #2 E #3 A #4
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Multiple Choice Section Questions (1-4) Question 1 For what value of a is the following function continuous? f (x) = x2 + 4x + 1 if x > 2 ax + 2 otherwise
A)
12 5
B)
13 5
C)
11 2
D)
1 7
E)
4 3
Solution We must have limx→2− f (x) = limx→2+ f (x). The function f (x) is continuous for x < 2 and x > 2. Therefore
x→2−
lim f (x) = 2a + 2,
x→2+ 11 2
lim f (x) = 4 + 8 + 1 = 13
Therefore, a satisfies 2a + 2 = 13. So, a =
Question 2 Find the equation of the tangent line of the function f (x) =
x x2 +2
at x = 1.
A) y =
3x 4
−
3 4
B) y =
x 9
+
2 9
C) y =
2x 5
−
2 5
D) y = x − 1
E) y =
x 2
−
1 2
Solution 1 x = 1; then y = f (1) = 3 . Equation of tangent line is y = mx + b. (x2 + 2) − x · 2x 1 m = f (1) = = 2 + 2)2 (x 9 So y = 1 x + b 9 Replacing x = 1, y =
1 3
gives 1 1 = · 1 + b, 3 9 b= 2 9
1 2 So, y = 9 x + 9 .
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Question 3 Solve the following equation for t. 5t = 3t+1 A) t =
ln(3) ln(5)
B) t =
ln(3) ln(2)
C) t =
ln(3) 2
D) t =
ln(5) ln(3)
E) t =
ln(3) ln(5)−ln(3)
Solution Take ln of both sides: ln(5t ) = ln(3t+1 ) t ln(5) = (t + 1) ln(3) t(ln(5) − ln(3)) = ln(3) ln(3) t= ln(5) − ln(3)
Question 4 Suppose the profit for a given product is given by P (x) = the marginal profit when x = 4. A)
9 10
√
5 + x + x2 . Find
B)
4 7
C)
5 8
D)
7 11
E)
3 10
Solution marginal profit at x = 4 is given by 1 + 2x 9 P (4) = √ |x=4 = 10 2 5 + x + x2
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Long Answer Section Questions (5-7) Question 5 (14 points) Using only the definition of derivative as a limit, calculate f (x) where f (x) = 3x2 − x − 2 Solution We have (setting h = ∆x) f (x) = lim = = = = f (x + h) − f (x) h→0 h (3(x + h)2 − (x + h) − 2) − (3x2 − x − 2) lim h→0 h 3x2 + 6xh + 3h2 − x − h − 2 − 3x2 + x + 2 lim h→0 h 2 6xh + 3h − h lim h→0 h lim (6x + 3h − 1)
h→0
= 6x − 1.
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Question 6 (14 points) Suppose that a deposit of 2,000 $ is made into a bank that gives 3% interest. Suppose that interest is compounded 6 times per year. (a) (2 points) Write a formula for A(t), the value of the investment, after t years in this case. (b) (4 points) How much will the investment be worth after 4 years? (c) (8 points) How long will it take for the investment to double? Solution We have p0 = 2000, r = 0.03, n = 6. (a) A(t) = 2000 1 +
0.03 6 6t
= 2000 · 1.0056t .
(b) A(4) = 2000 · 1.00524 . (c) Find t such that A(t) = 2000 · 1.0056t = 2 · 2000. So 1.0056t = 2 6t ln(1.005) = ln(2) ln(2) t = 6 ln(1.005)
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Question 7 (12 points) Use implicit differentiation to find equation: x2 y − 3xy 3 = −2xy + 2
dy dx
at the point (2,1) for the
Solution Differentiate the equation w.r.t. x; solve y ; replace x = 2, y = 1. 2xy + x2 y − 3y 3 − 9xy 2 y (9xy 2 − x2 − 2x)y (18 − 4 − 4)y 10y −2y − 2xy 2xy − 3y 3 + 2y 4−3+2 3, 3 y (2) = 10 = = = =
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Space for additional work
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