Multiple Choice Answers ver 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Multiple Choice Answers
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A, A, D, A, D, B, A, D, D, B, D, B
Written Question Answers
21. (10 points) At noon, ship A is 50 km. west of ship B. Ship A is sailing east at 20 km./hr. and Ship B is
sailing south at 10 km./hr.
(i) (2 points) What is the distance in km. between the two ships t hours after noon?
(ii) (8 points) At what rate is the distance between the ships increasing at 4 pm. in km./hr.?
Solution: We set up the coordinate axes at the location of ship B at noon, with the y-axis pointing north.
The location of ship A is then (−50 + 20t, 0) and that of ship B (0, −10t) where t is the time after noon in
hours. The distance between the ships is
p
p
s = (50 − 20t)2 + (10t)2 = 10 25 − 20t + 5t2
in km. We get
1
1
ds
= 10(10t − 20) (25 − 20t + 5t2 )− 2
dt
2
To get the rate of change of separation at 4 pm. substitute t = 4
1
ds
= 5(40 − 20)(25 − 80 + 80)− 2 = 20.
dt t=4
The distance between the ships is increasing at a rate of 20 km/hr. at 4 pm.
22. (10 points) Let f (x) = x2 − 10x + 8 ln(x).
(i) (1 point) What is the domain of f ?
(ii) (4 points) Find f 0 (x) and f 00 (x).
(iii) (3 points) Find the intervals of increase and the intervals of decrease of f .
(iv) (2 points) Find the intervals where f is concave up and those where it is concave down.
Solution: The domain of f is x > 0, i.e. (0, ∞) in interval notation.
f 0 (x) = 2x − 10 + 8x−1 = 2x−1 (x2 − 5x + 4) = 2x−1 (x − 4)(x − 1) and f 00 (x) = 2 − 8x−2 =
− 4) = 2x−2 (x + 2)(x − 2).
The function is increasing on (0, 1] and on [4, ∞) and decreasing on [1, 4]. It is concave down on (0, 2]
and concave up on [2, ∞).
2x−2 (x2
23. (10 points)
Find the largest possible area of a trapezoid inscribed in a circle of radius 1 unit and with its base a
diameter of the circle.
Solution: We can compute the area of the trapezoid as the sum of the areas of two triangles. (There are many
other ways). The area is A = 12 (y)(2x) + 21 (y)(2) = (1 + x)y square units. The constraint is x2 + y 2 = 1
√
√
1
3
with y > 0. Therefore y = 1 − x2 and A = (1 + x) 1 − x2 = (1 + x) 2 (1 − x) 2 . Differentiating gives
1
1
1
1
3
1
dA
3
1
1
= (1 + x) 2 (1 − x) 2 − (1 + x) 2 (1 − x)− 2 =
3(1 − x) − (1 + x) (1 + x) 2 (1 − x)− 2
dt
2
2
2
1
1
= (1 − 2x)(1 + x) 2 (1 − x)− 2
From this, we see that A takes a maximum when x = 12 . Substituting back in A gives
3
2
q
3
4
=
√
3 3
4 .
24. (10 points) Let f (x) = 8|x| − (x − 1)2 .
Find the absolute maximum and absolute minimum values of f (x) for −5 ≤ x ≤ 9 and the locations
where they are attained.
Solution: Since the function is continuous on [−5, 9], the absolute maximum and absolute minimum values
are attained. We need to check the endpoints, x = −5 and x = 9, any point where f is not differentiable,
in this case x = 0 and any points where the derivative vanishes. On x > 0 we have f (x) = 8x −
(x − 1)2 = −x2 + 10x − 1 so that f 0 (x) = −2x + 10 which vanishes at x = 5. On x < 0 we have
f (x) = −8x − (x − 1)2 = −x2 − 6x − 1 and f 0 (x) = −2x − 6 which vanishes at x = −3.
Checking all the above values f (−5) = 4, f (9) = 8, f (0) = −1, f (5) = 24, f (−3) = 8. Therefore
the absolute maximum value taken is 24 taken at x = 5 and the absolute minimum is −1 taken at x = 0.
25. (10 points) Using L’Hospital’s Rule or otherwise determine
1
1
lim
− 2
.
x→1 ln(x)
x (x − 1)
Solution:
This is a limit of type ∞ − ∞, so the first task is to capture the cancellation. This is done by cross
multiplying:
1
1
x3 − x2 − ln(x)
− 2
= 3
.
ln(x) x (x − 1)
(x − x2 ) ln(x)
Now
x3 − x2 − ln(x)
x→1 (x3 − x2 ) ln(x)
lim
is of type 0/0 so we use L’Hospitals rule.
x3 − x2 − ln(x)
3x2 − 2x − x−1
3x2 − 2x − x−1
=
lim
=
lim
x→1 (x3 − x2 ) ln(x)
x→1 (3x2 − 2x) ln(x) + (x3 − x2 )x−1
x→1 (3x2 − 2x) ln(x) + (x2 − x)
lim
which is still of type 0/0 so we use L’Hospitals rule again.
6x − 2 + x−2
6−2+1
5
=
= .
x→1 (6x − 2) ln(x) + (3x − 2) + (2x − 1)
4 ln(1) + 1 + 1
2
lim

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