=========================preview======================
(MATH005)2003fall.pdf
Back to MATH005 Login to download
======================================================
Math005 Algebra and Calculus I, Fall 2003 Mid-term Exam Solution
Part I: Multiple Choice Questions
Version A
Question 1 2 3 4 5 6 7 8
Answer c e a b c c d R = f(E) = 8.8930E.20060 8.950
Version B
Question 1234567 8
12.7590P .18060
Answer b a c d abbR=f(P)=
12.8480
Solution for Version A MC
12%
1. The monthly interest rate is = 1%. We compute the present values of the down payment and the 4
12
payments at 3, 6, 9 and 12 months. We set the sum of these present values equal to US$1,400.00. We get 1400 = D + P (1.01).3 + P (1.01).6 + P (1.01).9 + P (1.01).12 . The answer is (c).
30000 . 300000
2. The constant rate of decrease is given by = .27000. After 6 years, the car would depreciated
10
by 6 27000 = 162000 dollars. Hence the value of the car after 6 years would be 300000 . 162000 = 138000
dollars.
Or, in the form of a linear function, the value (y) of the car after x years is given by y = .27000x + 300000.
Putting in x = 6 results in y = 138000. The answer is (e)
3.
We have logb x + logb(x . 4) = logb 21, logb[x(x . 4)] = logb 21, therefore x satis.es x(x . 4) = 21. Solving this equation, we obtain x =7 or x = .3. We reject the root x = .3 because logb(.3) doesnt exist. The answer is (a)
4.
When moving towards x = 2 from the right hand side of x = 2, i.e. x . 2+ , f(x) . 1 (found by moving
..2 along the graph on the right of x = 2). lim ([f(x)]2 + 1) = lim f(x) +1=12 + 1 = 2. The answer is
x.2+ x.2+
(b).
5.
The solution is Pe3r + Se.2s (continuous future value + continuous present value). The answer is (c).
6.
At t = 10th day, R.(10) = . 10 (10)2 + 40(10) + 50 = 116.67. Thus we have
3
R(10 + h) . R(10) h=1
R.(10) = lim R(11) . R(10),
h.0 h
or equivalently, by using the equation of the tangent line to the graph of y = R(x) at t = 10, i.e., y = R(10) + R.(10)(x . 10),
R(11) R(10) + R.(10) = 43, 400 + 116.67 = 43, 516.67.
The answer is (c).
7. From the de.nition g(x +1) = x f(x + 1) . x, we see that
g(x)= g((x . 1)+1) = (x . 1) f((x . 1) + 1) . (x . 1) = xf(x) . f(x) . x +1
Take the derivative: g.(x)= f(x)+ xf.(x) . f.(x) . 1. Therefore g.(1) = f(1) + f .(1) . f.(1) . 1=
from product rule
f(1) . 1. The answer is (d).
8.
8.8930E . 20060
f(E)= 8.9500
Part II: Long Questions
9. (a)
Months 1 2 121314 2223 24
Figure showing the payments over the 24 months
(i)
The PV for the .rst 12 months: PV = P (1 + 0.005).1 + + P (1 + 0.005).12 .
(ii)
The PV for the last 12 months:
PV =2P (1 + 0.005).1 + + (1 + 0.005).24. . 2P (1 + 0.005).1 + + P (1 + 0.005).12. .
(iii) The total amount from (i) and (ii) above is
2P . 1 . P . 1 .
0.005 1 . (1 + 0.005)24 . 0.005 1 . (1 + 0.005)12 .
(iv) We substitute P = $2, 500 into the expression obtained in (iii) to obtain $83, 767 to the nearest
dollar.
(b)
Months 1234567
3435 36 Figure indicati