Put down your arguments next to the answers of the multiple choice problems for partial credits.
(1) Students in a calculus class were given an exam and then retested monthly with equivalent exams. The average score S 8%
(on a 100-point scale) for the class can be modeled by
S= 68 − 12 · ln(t + 1), 0 ≤ t ≤ 11
where t+ 1 is the time in months (so t = 0 is the first time of test). After how many months was the average score below 45 for the first time?
Solution. Note that S is decreasing. Rewrite 45= 68 − 12 · ln(t + 1) as ln(t + 1) = 45−68−12 =2312. Solve the last equation to get t= e23/12− 1 = 5.798259795. So, when t = 6, i.e. after 7 months, the average score fall below 45 for the first time.
(2) After t years, the value V of a car purchased for $80, 000 is 8%
V= 80, 000 · (0.75)kt.
It is known that immediately after 5 years, the car worths $50, 000. If immediately after s years, the car worths between
$25, 000 and $26, 000, then s is
Solution. First, solve k by the equation 50, 000 = 80, 000 · (0.75)5k: Rewriting, we have 58= (0.75)5k, and so k= ln(5/8)
5·ln0.75 = 0.3267521158. Now, we note that V is a decreasing function since 0.75 < 1. Rewrite 26, 000 = 80, 000 · (0.75)0.3267521158·t as (0.75)0.3267521158·t =2680 and solve for t to get t= 11.95661083.
Therefore, immediately after 12 years, the value of the car will drop below 26,000 for the first time.
(3) You want to deposit in a saving account that is compounded monthly at an annual interest rate 2.76%. If you wish to have 8%
$100, 000 in your account after 12 years, how much do you need to put into the account now?
Solution. Let P be the amount to be put in the account initially. Then we have to solve for t in the following equation: 100, 000 = P · (1 + 0.0276/12)12·12= P · (1.0023)144= 1.392108777 · P. Thus, P = 100, 000/1.392108777 = 71833.47.
(4) For the following questions, use the data given in the graph (where 0 corresponds to they year 1990). (Note. A million = 9%
106, and a billion = 109.)
(a) The percent increase in revenue (or sales) between 1995 and 1998 was about: (1600 − 1020)/1020 ≈ 0.5686 (b) Which of the following statements about the sales of jogging and running shoes can be concluded from the graph?
(I) In 1993, the sales were greater than $1.2 billion. CORRECT
(II) Between 1995 and 1997, the average increase in sales per year was about $55 million. WRONG ((1460 − 1020)/2 = 220 (million))
(III) The greatest one-year increase in sales occurred between 1996 and 1997. CORRECT
(c) Let f(x) represent the revenue (sales) function. Then most likely f′(5) = 1
(5) The demand and cost functions for a product are 8%
p= 6000 − 45x and C = 2x2+ 28800 where p is the price per unit, x is the number of units, and C is the total cost.
What will the revenue be when the level of production minimizes the average cost per unit?
Solution. The average cost per unit is give by A= C/x = 2x +28800x . Solving 0= A′= 2 −28800x2 for x, we get x= 120. Since A′′< 0, A(120) is a local minimum. For x = 120, the revenue is 120 · p(120) = 120 · (6000 − 45 · 120) = 72000.
(6) The graph of f(x) = x2−2 xis 8%
(7) Find the indefinite integrals. Write details for full mark.
14%
(a) Z (√
x+ 1)2
√x d x
= Z √
x+ 2 + 1
√x
d x
=23x32+ 2x + 2√ x
(b) Z
(x4− 2x)3(2x3− 1)d x
=12 Z
u3d u=1 2·1
4u4
=18(x4− 2x)
(8) Find the derivative of the function. Write details for full mark.
21%
(a) y=ex x2
y′=ex·x2x−2x·e4 x =ex(x−2)x3
(b) y= lnx+ 1 x − 1
2
y= 2 · (ln(x + 1) − ln(x − 1)) y′= 2 ·
1 x+1−x−11
(c) y= log3(e2xp e2x − 1) y= log3e2x+12log3(e2x − 1)
=ln13· 2x + 1
2·ln3ln(e2x − 1) y′=ln23+ e2
2·ln3·e2x−11
(9) Let g(x) =1−2ee−xx. Find the local maxima and local minima of the graph of g(x) if they exist. You need to explain why 8%
they are local maxima or minima.
Solution. g′(x) = −(−1+2·ee−x−4x)2; g′(x) = 0 when x = −ln(4). Since g′(x) > 0 if x > −ln(4) and g′(x) < 0 if x< −ln(4), g(−ln(4)) = 8 is a local minimum.
(10) The marginal cost for producing x units of a product is modeled by dC
dx = 128 − 0.03x. Suppose that it costs $12500 to 8%
produce 100 unit.
(a) Find the cost function.
Solution. C= Z
(128−0.03x)d x = 128x−0.015x2+K; 12500 = C(100) = 128·100−0.015·10000+K = 12650+ K; K = 150.
(b) Suppose that the demand functions is given by p= 1280 − 5x. Find the profit when the production level is 200.
Solution. profit= 200 · p(200)−C(200) = 200(1280−5·200)−(128·200−0.015·40000+150) = 30850
