Aplß CALCULUS AS 2002 SCORING GUIDELINES
6~~~~,~",~
.~.,.~
"
Let f and g be the functions given by f(x) = eX and g(:r;) = In x.
1
(a.) Find the area of the region enclosed by the graphs of f and g between x = 2" and x = 1.
(b) Find the volume of the solid generated when the region enclosed by the graphs of f and 9 between
x = ! and a: = 1 is revolved about the line y = 4.
:2
(c) Let h be the function given by h(x) = f(x) - g(:i:). Find the absolute minimum value of h(:r.) on the
closod intcrvi.l ~ ~ 3; ~ 1, and find tho i.bsolutc rna,xímiim va,ll1o of h(:i:) on the closed inteI'va,l
~ ~ ;T, ~ 1. Show the f.iia.IYRís tha.t lr-ads to your a.m-iWeI'R.
(a) Area = r1/
(e'" -In:r;)d;i; = 1.222 or 1.223
J ,Y2
~ i:
')
t i:
(b) Volume = 7r .I~/ ( (4 - In;¡;)2 - (4 - e"; i ) d:r:
/2
integral
answer
i:
limits and constant
2:
integrand
7.515'7f or 23.609
.. - i )- each error
4
:lote: 0/2 if not of the fOlnl
k lb (R(:i;)' - ¡"(xl )dx
i:
(c)
IJ
answer
i: considers h'(;r.) = 0
I/(x) = j'(x) - g'(x) = e:" _.! = 0
x
1: identifies critical point
i: = 0.5G7143
3
and endpoints as candidates
i: answers
Absolute minimum value and absolute
inaximurii value occur at the critical point or
Kote: Errors in computation come off
at the endpoints.
the third point.
h(O.5G714:3) = 2.330
h(O.5) = 2.:H18
h(l) = 2.718
The absolute minimum is 2.330.
The absolute maximum is 2.718.
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iights resered.
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~
Aplß CALCULUS AB 2002 SCORING GUIDELINES
The rate at which people enter an amusement park on a given day is modeled by the function E defined by
E ()t(f'= -15600
.
24t + lßO)
The rate at which people leave the same amusement park on the same day is modeled by the function L defined by
T.
J (. t) '
= (t"9890
_ 38t + :370 )
Doth E(t) fwd L(t) are measured in people per hour and time t is measured in hours after midnight. These
functions are valid for 9 :: t :: 23, the hours during which the park is open. At time t = D, there are no people in
the parle
(a) How many people ha,ve en(;ered the park by 5:00 P.M. ( t = li )? Round answer (;0 (:h6 nearest whole number.
(b) The price of
admission to the park is $15 until
5:00 P.M. (t = 17). After 5:00 P.M., the price of admission to
i,he park is $ii. How Ilany dollars are collecLecl from admissions t,o t.he park on (,he given day? Round your
9 .
answer to the nearest whole number.
II.
(0) Let, H(t) = (E(x) - L(x))d:i: for g:: i,:: 23. The value of lIO?) t,o tbe nearest whole number is 3725.
Find Uie value of HI (17) and explain the meaning of H (17) and H' (i 7) in the cont8xt of the park.
(d) At what time t, for 9 ~ t ~ 23, docs the model predict that (:he number of people in the park is a rml.ximum?
(a.)117
E(t)dt
= 6004.270
!l
3 1: ill. tegrand
6004 people entered the park by 5 pm.
1: answer
i 1: limits
117
IJ i123
7
(b) 15 E(t)dt + 11 E(t)dt = 104048.1ß5
1 : setup
The amount collected was $104,048.
or
.2a
,J 17E(t)
df; = 1271.283
1271 people entered the park between 5 pm and
11 pm, so the amount collected was
$15 . (6004) + $11 . (1271) = $104,041.
1: value of H'(17)
(c) H'(17) = E(17) - L(l7) = -380.281
There were 3725 people in the pa,rk at t = 17.
2: meanings
The number of people in the park was decreasing
1: meaning of IJ(17)
3
at the rate of approximately 380 people/hr at
1: meaning of HI (17)
j;ime t = 17.
-( - 1 )- if no reference to t = 17
E(t) - L(t) = 0
(d) H'(t) = E(t) - L(t) = 0
.)
~1:
t = 15.794 or 15.795
t 1 :
answer
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Aplß CALCULUS AS 2002 SCORING GUIDELINES
Question 3
An object, moves along the :i'-axs with iiiitiaJ position :¡; (0) = 2. The velocity of the object at. time t, ~ 0 is given
by v (t) = sin ( it).
(a) What is the acceleration of the object at time t = L1?
(b) Consider the following two statements.
Statement I: For 3 .. l .. ~.5, the velocity of the object is decreasing.
Statement, II: For 3 .. l ,4,..5, t,he speed of the object is increasing.
Are either or both of these statements correct? For ec1ch statement provide a, reason why it is correct 01 not
cor.rect.
(c) What is the total distance traveled by the object over the time interval 0 S t ~ 4 ?
(d) What is the position of the object at time t = 4 '?
1: answer
(a) a(4) = v'(Ll) = iC08(~;)
7r
= -'6 or -0.523 or -0.524
(b) On 3 ~ t ~ "1.5 :
3 1: II eorrcct
3 3
aCt) = v'(t) = !!cos(!!t). ",0
Statement I is correct since aCt) .. O.
Stt1tement II is correct since 1)( t) ~ 0 and
for II
11'1:I reason
correct,
with reason
a(t),.. O.
'( c)
limits of 0 and 4 on an integral
Distance = 1"1111(t)ldt = 2.387
o
OR
of v(t) or I v(t)1
) 3:I('I)3 7r
3
or
1:
x( t = - - cos - t + - + 2
uses :1;(0) and :r(Ll) to compute
x(O) = 2
distanee
9
2'I
3
x(4) = 2 + - = 3."1239
1: handles change of direction a,t
v(l) = 0 when t= 3
student's turning point
6
:r;(;3) = - + 2 = a.90986
i: answer
7r
0/1 if incorrect turning point or
Ix(3) - x(O)1 + 1:r;(4) - :r;(3)1 = ~~ = 2.387
no turning point
integral
(d) :¡;(4) = 3;(0) + l'Í'v(t)dt
= 3.412
o
2 r 1 :
L 1: answer
OR
OR
3 (7r)
:c(t) = --cos
-;t +3
-+2
.r,(t) = -~cos (it) + C
7r .3 7r
21 ~
x(4) = 2 + ~ = 3.432
27r
answer
0/1 if no constant of integration
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Aplß CALCULUS AS 2002 SCORING GUIDELINES
f
Question 4
"
The graph of the function f shown above consists of two line segments. Let g be the
3
function given by g(x) = fox.f(t)dt.
(a) Find g(-l), g'(-i), and g"(-l),
(b) For what va.lues of x in the open interval (-2,2) is g increasing? Explain your
reasoning.
(c) For what values of :r: in the open interval (-2,2) is the graph of 9 concave
down? Explain your reasoning.
(-2, -3) (2, -3)
(d) On the axes provided, sketch the graph of g on the closed interval (-2,2J.
r-1 JO 3
(a) g(-l) = Jo f(t)dt = - -1 f(t)dt = -'2
Graph off
3 1: g'(-l)
g'(-1) = .f(-1) = 0
1: gll(-I)
g/I(-1) = f(-i) = 3
1 l' g( -1)
interval
(b) 9 is increasing on - 1 .. x .c 1 because
2
g'(x) = f("L) :)0 0 on this interval.
1:
t 1 :
reason
( c) The graph of 9 is concave clown on 0 .c ;1; .c 2
interval
because gll(X) = .f(x) , 0 on this interval.
2J1:
or
11:
reason
because gl(:i;) = f(x) is decreasing on this
interval.
(d)
)I
1: g(-2) = g(O) = g(2) = 0
1: appropriate increasing/decreasing
2
x
ancl concavity behavior
.c - 1 )- vertical asymptote
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Aplß CALCULUS AB 2002 SCORING GUIDELINES
Question 5
~
10 ern
.1
A container ha,s the shape of an open right circi.ùar cone, as shown in the figure
above. The height of the container is 10 cm and the diameter of the opening is 10
em. Water in the container is eva,pomting so that its depth h is changing at the
-3
í
constant rate of iO cm/hr.
.1 em
T
(The volmne of a cone of height h and radius r is given by V = ~7rr2h. )
(a) Find the volume V of water in the container when h = 5 em. Indicate units
of measure,
1
1
(b) Find the rate of change of the volume of water in the cont;ainer, with respeet to time, when II = 5 em. Tndicat.e
unitR of measure.
(c) Show that the rate of change of the volume of w~Lter in the container clue to evapora.tion is directly proportional
to the exposed sUl'fa,ce aJ:ea of the water. Vl'nat is the constant of proportíona.lity?
5 1 (5)2 125 '
i : 11 wlien h = 5
(a,). When h = 5, r =2-' V(5)
-7r em'!
' =3-7r2- 5 =12
2',
r 5 1
(b) - = - so l' =-h
1: 'I' = ~ h in (a.) or (b ì
h 10' 2
3 4 12 dt, 4 dt
Vi= (1
-7f I- 1.
~ h) =1-7rli:
'I dll
; -- =1 -;7rh"ry dh
V as a fund.ion of one variable
rlVI = .!1T(25)(-2.) = _15 7r em/(
cit h=.5 4 10 8 hr
in (a) or (b)
1:
OR
OR
5
d7'
dt
dV = ~7r(T"2 dh + 2T"h d'l); dT' = ~ dh
dt 3 dt dt dt 2 dt
dV
I 1 ((::,)1:
.) (+ '))2 (::\') 5( 3))'
= -1f
_..
--
2:
dt h=5.,.=~
3 ,1 10 2 20
, 2
dV
aT
.. -2 ? chain rule or product rule error
8 ll
_ 15 cm%'
1: cvalu8,tion at. h = 5
- --7r 1
dV
shows aT = k . area
dt 4 dt 40
(e) dV = ~1Th2 dh = _.21Tli2
3 'J 10
3 ',l10
3
40
21:
= --7r(2'l t = --7r1'" = -_. area
11 .
Th
, e const
,ant 0fproportiona
. r'
ity is -3
10 .
identifies constant of
proportionaliy
1: correct units in (a) and (b)
units of cm3 in (a) and cmlb in (b)
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Aplß CALCULUS AS 2002 SCORING GUIDELINES
Question 6
x
-1.5
-1.0
-0.5
0
0.5
1.0
1.
fix)
-1
-7
-4
-6
-7
-6
-4
-1
-5
-3
0
3
5
7
1'(x)
numbers. The ta,ble above gives the values of j' a,nd its
Let f be a function that is diferentiable for all real
derivative l' for selected points :1; in the closed interval -1.5 ~ x ~ 1.5. The second derivative of f has the
property that rex) ? 0 for -1.5 S x ~ 1.5.
(a) Evaluate 101. (3f' (x)+ 4)dx. Show the work that leads to your answer.
(b) Write an equa,tion of the lie tangent to the graph of .f a,t the point where x = 1. Use this line to
a,pprox:ímate the value of f(L.2). Is this approximation great;er than or less than the a.eual value of /(1.2)?
Give a, reason for your answer.
(c) Find a positive real number r having the property that there must exit a va,lue c with 0 .: c .: 0.5 and
f"(e) = r. Give a reason for your answer.
(d) Let g be the function given by g(x) = ,
2x2 + x - 7 for:c 2. O.
12X2 - x - 7 for x .: 0
The graph of 9 passes through each of the points (x,f(x)) given in the table above. Is iti possible that f and
9 are the S~1ine function? Give a reason for your answer.
(a)
aiitideriva.tive
r1.5( I . r1.. j 1.5
,)
J 0 3 J (x) + 4) dx = 3 J 0 f' (x) d,T, + 0 '1 dx
answer
- t 1:
1 :
= 31(x) + 4:i:li,G = 3( -1 - (-7)) + 4(1.5) = 24
o
(b)
y=5(a:-1)-4
¡1:
1(1.2) i":J 5 (0.2) - 4 = -3
3 1:
The approximation is less than 1(1.2) because l;he
graph of f is conca.ve up on the interval
i :
tangent. line
computes y on tangent line at :1; = 1.2
answer with reason
1 .: x .: 1.2.
reference to MVT for l (or differentiabilty
(c) By the Mei111 Value Theorem there is a c with
o .: c .. 0.5 such that
211 : of ¡')
1:
f"Ce) = t(0.5~ - ¡'(OJ = 3 - 0 = 6 = r
va.lue of 7' for interva.l 0 S x ~ 0.5
O.n - 0 0..5
(d)
lim g'(x) = lim (4x -1) =-1
$-.0- ;1;-70lim g/(x)
answers "no" \viLh reference Lo
= li (4:r+1)=+1
:r.-.o ,r.-.O I
2 ¡ 1 : g' or gil
1 : correct reason
Thus g' is not continuous at a; = 0, hut f' is
continuous a.t x = 0, so f +" 9 .
OR
gf/(:i;) = 4 for a.ll x+"O , but H; was shown in part.
(c) that r(c) = 6 for some c+"O 1 so 1 +" g.
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7
Entrce Sxaition Btlål'd.
I
AP(I CALCULUS AB
2001 SCORING GUIDELINES
QUestion 1
Let Rand S be the regions in the first quadrant shown in the y
figure above. The region R is bounded by the x-axis and the
graphs of y = 2 - x3 and y = tan x . The region S is bounded by
the y-axis and the graphs of y = 2 - x3 and y = tan x .
(a) Find the area of R.
x
(b) Find the area of S.
( c) Find the volume of the solid generated when S is revolved
about the x-axis.
Point of intersection
2 - x3 = tan x at (A, B) = (0.902155,1.265751)
rA J~
1 : limits
(a) Area R = J 0 tan x dx + A (2 - x3 ) dx = 0,729
3 ¡ 1 : integrand
or
1 : answer
Area R = loB ((2 - y?/3 - tan-1y)dy = 0.729
or
r~ rA
x )dx = 0.729
Area R = Jo (2 - x3 )dx - Jo (2 - x3 - tan
(b) Area S = foA (2 - x3 - tan x ) dx = 1. 1 60 or 1. 161
1 : limits
3' ¡ 1 : integrand
or
1 : answer
Area S = foB tan-1ydy + i: (2 - y?/3 dy = 1.160 or 1.161
or
Area S
r2 iE
= Jo (2 - y)1/3 dy - 0 ((2 - y)1/3 - tan-1y )dy
= 1.160 or 1.161
(c) Volume
1 : limits and constant
= 7rfaA((2_X3)2 -tan2x)dx
3 ¡ i : integrand
= 2.6527r or 8.331 or 8.332
1 : answèr
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ApCV CALCULUS AS
2001 SCORING GUIDELINES
//)
~n2
The temperature, in degrees Celsius (DC), of the water in a pond is a
t
W(t)
differentiable function Wof time t. The table above shows the water
(days)
a
(DC)
temperature as recorded every 3 days over a 15-day period.
(a) Use data from the table to find an approximation for W'(12). Show the
computations that lead to your answer. Indicate units of measure.
(b) Approximate the average temperature, in degrees Celsius, of the water
3
6
9
12
15
20
31
28
24
22
21
over the time interval 0 :: t :: 15 days by using a trapezoidal
approximation with subintervals or length b.t = 3 days.
(c) A student proposes the function P, given by P(t) = 20 + 10te(-/3), as a model for the
temperature of the water in the pond at time t, where t is measured in days and P(t) is
measured in degrees Celsius. Find P'(12). Using appropriate units, explain the meaning of
your answer in terms of water temperature.
(d) Use the function P defined in part (c) to find the average value, in degrees Celsius, of P(t)
over the time interval 0 :: t :: 15 days.
(a) Difference quotient; e.g.
1 : difference quotient
2 ¡
15 - 12 3 ay 0
W'(12) :: W(12) - W(9) = _~ DC/day or
12 - 9 3
W'(12) :: W(15) - W(12) = _! DC/d r
1 : answer (with units)
W/(12) :: W(lli = :(9) = -~ DC/day
1 : trapezoidal method
(b) ~ (20 + 2(31) + 2(28) + 2(24) + 2(22) + 21) = 376.5
2 : t
1 : answer
1
Average temperature :: 15 (376.5) = 25.1 DC
3 t=12
(c) P'(12) = 10e-t/3 _ 10 te-t/3!
2 ¡
= -30e-4 = -0.549 DC/day
1 : P'(12) (with or without units)
1 : interpretation
This means that the temperature is decreasing at the
rate of 0.549 DC/day when t = 12 days.
1 : integrand
DC
(d) 115Jo15(20+10te-t/3)dt = 25.757
1 : limits and
3:
average value constant
1 : answer
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AP~ CALCULUS AB
2001 SCORING GUIDELINES
Question 3
A car is traveling on a straight road with velocity
a(l)
(ftsec2)
55 ft/sec at time t = O. For 0 :: t :: 18 seconds, the
(18, 15)
15 (2,15)
car's acceleration a(t), in ft/sec2, is the piecewise
o
linear function defined by the graph above.
(a) Is the velocity of the car increasing at t = 2
I (seconds)
2
-15
seconds? Why or why not?
(10, -15) (14, -15)
\
(b) At what time in the interval 0 :: t :: 18, other than t = 0, is the velocity of the car
55 ft/sec? Why?
(c) On the time interval 0 :: t :: 18, what is the car's absolute maximum velocity, in ft/sec,
and at what time does it occur? Justify your answer.
(d) At what times in the interval 0 :: t :: 18, if any, is the car's velocity equal to zero? Justify
your answer.
(a) Since v'(2) = a(2) and a(2) = 15 ? 0, the velocity is
1 : answer and reason
increasing at t = 2.
(b)
At time t = 12 because
2:t
v(12) - v(O) = 0 a(t) dt = o.
1 : t = 12
1 : reason
112
1:t=6
(c) The absolute maximum velocity is 115 ft/sec at
t = 6,
1 : absolute maximum velocity
The absolute maximum must occur at t = 6 or
1 : identifies t = 6 and
t = 18 as candidates
at an endpoint.
4:
v(6) = 55 + J06 a(t) dt
or
indicates that v increases,
= 55 + 2(15) + ~(4)(15) 115? v(O)
decreases, then increases
1'18
1 : eliminates t == 18
J6 a(t)dt ~ 0 so v(18) ~ v(6)
(d) The car's velocity is never equal to O. The absolute
minimum occurs at t = 16 where
v(16) = 115 + 6 a(t)dt = 115 - 105 = 10)- O.
2:t
1 : answer
1 : reason
116
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the College Entrance Examination Board.
AP~ CALCULUS AS
2001 SCORING GUIDELINES
Question 4
Let h be a function defined for all x += 0 such that h( 4) = -3 and the derivative of h is given
2
by hi (x) =x2- -for
all x += 0 .
x
(a) Find all values of x for which the graph of h has a horizontal tangent, and determine
whether h has a local maximum, a local minimum, or neither at each of these values.
Justify your answers.
(b) On what intervals, if any, is the graph of h concave up? Justify your answer.
(c) Write an equation for the line tangent to the graph of h at x = 4.
(d) Does the line tangent to the graph of h at x = 4 lie above or below the graph of h for
x )0 4? Why?
( a)
l/(x) = 0 at x = :l.J
1 : x = :l.J
1 : analysis
h'(x)
0
+
I
x
-.J
und
I
0
0
4:
+
2 : conclusions
.: - 1 )0 not dealing with
I
.J
discontinuity at 0
Local minima at x = -.J and at x = .J
1 : hl/(x)
2
(b)
hl/(x) = 1 + 2 )0 0 for all x :; O. Therefore,
x
3 ¡
the graph of h is concave up for all x :; O.
1 : h'/(X) )0 0
1 : answer
4 2
(c) h'(4) = 16 - 2 = 7
7
Y + 3 = 2 (x 4)
i : tangent line equation
(d) The tangent line is below the graph because
1 : answer with reason
the graph of h is concave up for x )0 4.
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the College Entrance Examination Board.
ApCI CALCULUS AB
2001 SCORING GUIDELINES
Question 5
A cubic polynomial function f is defined by
f(x) = 4x3 + ax2 + bx + k
where a, b, and k are constants. The function f has a local minimum at x = - 1 , and the graph
of f has a point of inflection at x = - 2 .
(a) Find the values of a and b.
(b) If Ia1 f(x)dx = 32, what is the value of k?
(a) f'(x) = 12x2 + 2ax + b
1 : j'(x)
r(x) = 24x + 2a
1 : r(x)
5: l:f'(-l)=O
f' ( - 1) = 12 - 2a + b = 0
1 : r( -2) = 0
r(-2) = -48 +2a = 0
1 : a, b
a = 24
b = - 12 + 2a = 36
2 : antidifferentiation
(b) J01 (4x3 + 24x2 + 36x + k )dx
~ - 1 / each error
= x4 + 8x3 + 18x2 + kxl::~ = 27+ k
4:
1 : expression in k
1:k
27 + k = 32
k = 5
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AP(I CALCULUS AS
2001 SCORING GUIDELINES
Question 6
The function f is differentiable for all real numbers. The point (3, ¡) is on the graph of
~ 4
y = f(x) , and the slope at each point (x, y) on the graph is given by ~; = y2( 6 - 2x) .
evaluate
(a) Find d2; and
it
at
the
point (3,~).
(b) Find y = f(x) by solving the differential equation ~; = y2 (6 - 2x) with the initial
l' J . 1'/1:\ i
COnGltlOn 1\,j) = ¡ ,
d2y dy 2
d2y
2: -Z
dx
(a) dx2 = 2y dx (6 - 2x) - 2y
.: - 2 :? product rule or
= 2y3(6 - 2x? - 2y2
3:
d2y I (1)2
i
dx2 (3,¡) = 0 - 2 ¡
8
chain rule error
1 : value at (3, ¡)
1
(b) 2dy=(6-2x)dx
y
1 : separates variables
- -i
= 6x 2
-x+c
1 : antiderívative of dy term
1 : antiderivative of dx term
y
6:
- 4 = 18 - 9 + C = 9 + C
1 : uses initial condition f(3) = ¡
1 : solves for y
C = -13
Y=
1 : constant of integration
1
Note: max 3/6 ll-l-l-O-O-OJ if no
x2 - 6x + 13
constant of integration
Note: 0/6 if no separation of variables
CopyrìghtiQ200 i by c;ollege EI1tr~nce Exanunation Board, All rights reserved,
Advapced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
7
2000 GRADING STANDARllS
AP Calculus AB-l I BC-l
.,
Let R be tbe shaded region in the first quadrmit enclosed by the graphs of
x , mid the y-axis, as shown in the figue above.
u
2000
Dr (c).
integrand
, 10 20
= 6.666 or 6,667 meters/sec
(a) Runne" A: veloetty = 3".2 = 3"
i
Runner B: 11(2) = ~ = 6.857 meters/sec
(b) Runner .4: acceleration = 'Î = 3,333 metersfsec2
72 I
RmUler B: 11(2) = ,,'(2) = (2t + 3)2 1=2
72 / "
= ;¡ = 1.469 meters sec-
(c) Runner A: distance = i(j)(10) + 7(10) = 85 meters
II :!t +3
Runner B: distanceJ.1I
= '¿dl
= 83,336 meters
')41
(units) meters/sec in part (a), meters/sec2 in part (b), ,md
meters in part (c), Dr eqiu"iient.
"
"
"
l ~ I:; 1~.lln
; i II
,
II ~l~ - =.
,
n_
1_'-.
t 1: velocity for Runner .4
2000
t 1: accelemtioii for Runner A
2: distance for Runner A
1: iins'ver
1: method
1: Rnswer
1: inte&rral
2: distance for Runner B
1: unts
4
- 1: aeee1emtion for Runner B
.)
" 1: ve10city for Runner B
ry
(c) Find the total distance run by Runner A and the total distance ru by Runner B over the time
intel"al 0 :: i:: 10 seconds. Indicate units of mel\re.
Indicnte UIlt ts of rnetlsure.
(b) Find the Ilcceìemtioii of Runnel' it and the acceleration of Runner B nt time t = 2 seconds,
B n,t time t = 2 seonds. Indicatc units of mensure.
(a) Find the velocity of Runer .4 and the velocity of Runner
;.::111
2000 GRADING STANDARDS
AP Calculus AB-2 I BC-2
Two runners, A and B, fun on It straight racetrnck for
o :s 1 :s 10 seconds. The graph above, which consists of two line
segments, shows the velocity, in meters per second, of Runner
y = c-r, y = 1- cos
(a) Find the area of the region R.
.4, The velocity, in meters per seL'nd, of Runner B is given by
about the x-n.s.
.)
- t1:
answer
1:
1\ote: 0/2 if not of the form
~ - 1:0 each error
2: in tegrnnd
1: answer
12: integrand and const;it
3 ~ - 1:0 each error
3
. de med
fi by()
the function"
"I = 241
21 + 3 .
(b) Find the volume of the solid generited when the region R is revolved
d:
1: Co"reet limits in an integral in (a), (b),
(c) The region R is the bas of a solid. For this solid, each cross sedion
perpendicular to the J;iiis is a square. Find the volume of this solid.
Region R
x))
,,-,,:2 1 _ cos x at x = O.94l!44 = .4
"
(a) Area =1"(e-"" - (1- cos
= 0.590 or 0.591
(b) Volume = Jr L"((c-""t -(1-COSX)2)dx
= O,55596Jr = 1.7460" 1.747
x) dx
J."( " )2
(c) Volume = " e-"" - (1 - cos
= 0.461
k t (f(x) - g(X))2 d:
1: answer
Copyright () 200 by College Entmnce Exuminution Board and Educational Tei;ting Service. All rights resen-ed.
AP is a regi~uere trndeRwrk ofihe College Enmmce ExaoùnatÎon BoanL
Copyright () 2000 by College Entrnce Examinadon Boa and Educational. Testing Service. AU right'i reserved.
AP ili 11 register trndemiik of the College Entrnce Examnation Bown.
Repritedbyperm~:~~~n:O¡==I~:;CÀ=~~M:=~~(~~~rs Forlimile
use by the Nor Carlina Asiat of Advce Placent Matemaö T eacem (NCAAPMn.
Ae¡nle by pess 01 Edu Tes Sece an the Co En Exaton Boar, th copyht ownrs For limited
AP CalculusAB~3
2000
GRADING STANDARDS
The figure iilive shows the graph of ¡', the denviitive of
2000
AP Calculus AB-4
2000 GRADING STANDARDS
2000
Water is pumped into ii undergrund tank nt n constant rnte of 8 gal0us per minute. Water leaks out
of the tank at the rate of .J gallons per miute, for 0 :: t:: 120 minutes, At time t = 0, the tank
(a) How many gallons of water leak out of the tank from time t = 0 to t = 3 minutes'!
contAins 30 gallons of wnter.
""
horizontnl tangent lines at ;i = -3, ;i = 2, nnd x = 5, and
the function f, for -7:5 x:5 7. The graph of ¡' hIl
n vcrtiCRI tangent line at x = 3.
in) 1I1ethod 1: .f.Jt + 1 dt = if( t + 1)3tl
3 -3
(b) 30 + 8.3 _!: _ 148
A(t) = 30 + J.'¡s -.Jx + l)dx
(c) Method 1:
Method 1:
, 1: limits
anserte integral
i 2i :: defini
3 1 : integrand
Method 2:
-or-
r 1 : antiderivative with C
3 1, solves for C using L(O) = 0
1; answer
i : answer
Method 1:
1: - J.'.Jx + Idx
(1:30+8t
2t
Method 2:
- or-
"
= 30 + 8;' - J.'.Jx + 1 dx
- or-
ci = 8-.J
dt
A(t)=8t-J(t+1):i1 +c
.) 9"
30 = 8(0) - J(O + 1):l2 + c;
A(t) = 8t -~(t + 1)3/1 + f
(d) A'(t) = 8 -.J = 0 when t = 63
A'(t) is positive for 0 ~ t ~ 63 and negative tDr
63 ~ t ~ 120, Therefore the"e is a maxmum
1: justiliCltion
3 1: solves for t
¡ 1: sets A'(t) = 0
11 : antiderivative with C
.. 1: answer
'J
I\letliod 2:
-3
c _ 92
L(t) = ::(t + 1)3/2 _.:. L(3) = -
L(O) = 0: C = -.:
.) 3". 14
3
3' 3 ,
.)
~; = .J: L(t) = ~(t + 1):lt + C
I\lel.hod 2: L(t) = gnllous leaked in first t miu.tes
- or-
"
I," =~
, the total number of gal0ns of water in the tank at time t,
(c) \Vnte an expresion for A(t)
(b) How man)' gallons of water nre in the tank at time t = 3 minutes?
(d) At what time t, for 0 :: t:: 120, is the amount of water in the tank a maxmum'! Justify your
(~7,-3)
(2,3) U (3,5)
.. i)
(a) Find all values of x, for -7 ~ x ~ 7, at which I
attains fl relative minimum. Justi1).' your answer.
.)
f 1: answer
~ 1: justificntion
.)
ri :
~ f1:1:justification
answer
,)
- or -
as cnnàidiites
1: identifies;i = -5 and x = 7
1: answer
- 1 :
31
indicates that the graph of f
increRses~ detTeases, then increnes
1: justifies f(7) :; 1(-5)
n.nswer.
(h) Find all values of $, for -7 ~ x ~ 7, at which I attains a relative maxmum. Justify your nner,
(c) Find all values of $, for -7 'C ;i ~ 7, M which fl/(x) ~ 0,
x=-1
I
I
I
(d) At what. value oÍ Xl for -7 :: x S (1 doe J at,tain its HbsolUl,~~ !mu:Imum'! .histin" your answer.
(a)
fl(x) changes from negative to positive at x = -1
(b) x=-5
¡'(x) changes from positive to negative at x = -5
exists and ¡'is decreang on the intervals
(-7,-3), (2,3), aid(3,5)
(c) r(x)
(d) x= 7
The absolute mnxmum must occur at x = -5 or at an
endpoint.
1(-5) :; 1(-7) hecnuse f is increasing on (-7,-5)
negative change in ¡from x = -5 to x = -1 is smaller
The graph of ¡'shows that the magnitude of the
than the positive change in f from x = -1 to x = ï.
;i = 7, nndf(7):; f(-5). So f(7) is the iibsolute
Therefore the net change in f is positive from x = -5 to
mflum.
AP is a registere n-ademnk of the College Entrce Exnaiion Board.
Copyright (Q 200 by College Entrnce Examination Boar and Educuiional Testing Service. AU rights reserved.
at t=63.
AP is ¡i register trdemark of the College Entrance Exmninaiion Board.
Reprinted byperm=~i~~C":a~~~ cl~~V~~ii~a~M=:~-:~~(~~n~eff. Forlimít
Copyñght 19 2000 by College Entrnce Examination Board mid Educational Testing Servce. All rights reserved.
Aepinledbyperm~:~~~~.i¡~= i;~~~'i=';M==::~'1~~frJO'er, Foliriled
2000 GRADING STANDARDS
AP Calculus AB-5 / BC-5
Consider the eurve given by :i2 - x3y = 6.
dll 3:r1l _ 112
(ii) Show thiit dx =--,
2.xy - x"
2000
(ii) Find all points on the cure whose x,coordinate is 1, iid write an L"Iuation for the tiigent line at
eaeh of these points.
.)
.
- . ,àu
formy2-y=k
1, sets denominator of : equal to 0
1: substitutes y = .! x2 or x = ::..
in to the eaufttion for the- cure
AP Calculus AB-6
2000 GRADING STANDARDS
, , , . dy 3x2
dx e-Y
Consider the differentin1 equation -' = -".
(a) Find a solution 11 = J(x) to the differential equation satislYing J(O) = i,
1: sepiiates viiiables
1: Itntiderivative of dx ter
1: antidervntive of dy ten
1: constant of integrntion
1: uses initial condition J(O) = ~
1: solves for y
2000
Kate: 0/1 if 1/ is not a logarithmc function of x
1: 2x"+e::O
Kote: 0/1 if 0 is not iu the domain
1: domain
1: range
Note: 0/3 if 11 is not a logarithmic function of X
:i
J\"ote: 0/6 if no separation of variables
integration
Note: max 3/6 (1..1-1-0-0-0) if no constant of
I)
(b) Find the domain ,md range of the fuction Jfound in piit (a).
(a) c2. dy = 3,2 d:
~e"!l: = ;r" + c;
e2P=2x"+C
1 (" )
1/ ='21n 2x" +C
22'
.! = .!In(O + C), C = e
1 ., )
Y = '21n(2x" + c
(b) Domain: 2x" + e :; 0
1
x":; --e
2
x:;(-~er =-(~er
Range: -'X": Y , -'X
Copyright ig 2000 by College Entnince Exnminotion Boar imd Educational Testing Servce. All ñghi.. reserved.
AP is;i registered trdemark of ¡he College Emroce Examination Board.
1: solves for x-coordinate
Reprntedbypermus~~;=~%.T~~=:~~erla~Mi:~=l=~Ji~efS. Forlimrted
3
2. '
Kote: 0/4 if not solving ii ec¡uation of the
2: tangen t lines
11:
1/-y=6
4 1: solves for 1/
1. : venfire expression for ;t
11 : implicit differentiation
(c) Find the :icoo"diate of ellh point on the curve where the taugent line is verticaL.
.,' .. d: d:
(a) ,," +2xu'!_3:r2y_x"dY = 0
dii
dx (2:i - x") = 3x2y - y2
dy 3x21/ _ 1/2
-=-d: '1:i-x
(b) \Vhen x= 1, ¡f- 1/ = 6
y2 y-6=0
(1/ - 3)(1/ + 2) = 0
y= 3, y=-2
dx 6-1
At (1,3). '! = 9 - 9 = 0
T Algent lie equation is 11 = 3
, . dx -4-1 -5
At (L-'1). dy = -6 - 4 = -10 = 2
Tangent lie equation is y + 2 = '1(x -1)
(c) Tangent line is vertical when 2x¡1 - x3 = 0
( ?) 1 "
X 21'- X" = 0 gives X = 0 or y = '2x-
".coortiinate 0,
There is no point on the curve with
-.jx" = 6
When Yi = ~x2. .!a;,' - ~xii = 6
2 . 4 2
Copyñght f9 2000 by College Entrnce E"amination Board and Educutional Te.'Iting Servce. All ñghts reserved.
AP is a registered trademark of (he College Enuimce Examination Board.
x = ii
Aeprbype~~%:=011~ ~~~anc:i;=e: M~:= =ê::N~rJn~el Forlimited
Answers to the 1998 AP Calculus AB and
Calculus BC Examinations
Section I: Multiple Choice
~1 Section I: Multiple Choice
II Blank Answer Sheet
Listed below are the correct
ii Section II: Free Response
answers to the multiple-
choice questions and the percentage of AP candidates who
II Student Preparation for the Exams
II Free-Response Questions, Scoring Guidelines,
and Sample Student Responses with Commentar
II Section II, Calculus AB
II Section II, Calculus BC
answered each question correctly. A copy of the blank
answer sheet appears on the following pages for reference.
Section 1 Answer ((ey and Percent Answering Correctly
Calculus AB
95
Section 1 Answer Key and Percent Answering Correctly
Calculus BC
ie¡ q y
Calculus AS Part
~ A,
CALCULUS AB
E
SECTION I, Part A
Time-55 minutes
E
Number of questions - 28
S
¡f
A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION.
Directions: Solve each of the following problems, using the available space for scratchwork. After examining the form of the choices, decide which is the best of the choices given and fil in the corresponding
avalon the answer sheet. No credi1t will be given for anything written in the test book. Do not spend too
much time on anyone problem.
d
In this test: Unless otherwise specified, the domain of a function f is assumed to be the set of all real
numbers x for which f(x) is a real number.
I. What is the x-coordinate of the poii¡t of inflection on the graph of y = i x3 + 5x2 + 24 ?
it
:1"
t.
(A) 5
(B) 0
(C) -L.
3
(D) -5
(E) -10
11
Y
1
e
.,
,.
s
GO ON TO
13
'THE NEXT PAGE
y
2
3 4
-1 0
-1
x
-2
2. The graph of a piècewise:.linèar function f, for -1 .:: x :: 4, is shown above. What is the value of
4
I-If(x) dx ?
(B) 2.5
(A) 1
(C) 4
(D) 5.5
(E) 8
, 2
3.
x2 dx =
fi .l
i
(A) -2
7
(B) 24
(C) 1.
2
(D) 1
(E) 2 In 2
GO ON TO THE NEXT PAGE
14
L
n
Calculus AB ~ J
¡~
~
~
t;,
¡(
l\
Part A
4. If f is continuous for Cl :s x ~ b and differentiable for a .: x .: b, which of the following could be
false?
r~
s
Ji
f:
l:
i
i
r
(A) f'(c) = f(b)b --f(a)
for some c such that Cl .: c .: b.
a
(B) f'(c) = 0 for some c such that a .: c .: b,
l:
r
l
l¡,
i.
i.
1
~
;
(C) f has a minimum value on a ~ x :s b.
(D) f has a maximum value on a:S x :s b.
I
'~'
(E) Jb
f(x) dx exists.
a
8
5. t( sin t dt =
o
(A) sin x
'(B) -cos x
(C) cos x
(D) cos x-i
(E) 1 - cos x
2
,
.,.
,.
GO ON TO THE NEXT PAGE
15
tp
dy
6. If x2 + xy = 10, then when x = 2, dx =
7
(B) -2
(A) -2:
(C) ~7
(D) ;?2
(E) 2.2
7. ¡e(X2;1)dX=
i
1
(A) e - e
(B) e2 - e
e21
2 2
(C) - - e +-
(D) e2 - 2
(E) e2
2 _2~
~
i:
'.
f;
¡
¡;
GO ON TO THE NEXT PAGE.
i
l
16
Calculus AS ~ I
A
__Part
.
8. Let I and g be differentiable functions with the following properties:
(i) g(x):; 0 for all x
(ii) 1(0)= 1
If hex) = I(x)g(x) and h'(x) = I(x)g'(x), then i(x) =
-------1------I
200
3
2
i-
§
:iiQ)
ci
(D) 0
(C) eX
(B) g(x)
(A) j'(x)
1
1
I
I
1
1
I
1
(E) 1
I 1I 1
1 1
1
I1II I1I1
1I
-------~----- -~-
_______1 ______J_
100 ------
Vi
1o:
o
6
12
18
24
Hours
9. The flow of oil, in barrels per hour, through a pipeline on July 9 is given by the graph shown above.
Of the following, which best approximates the total number of barrels of oil that passed through the
pipeline that day?
(A) 500
(B) 600
(C) 2,400
.
(D) 3,000
(E) 4,800
GO ON TO THE NEXT PAGE
17
,." "\/.\
x2 - 2
10. What is the instantaneous rate of change at x = 2 of the function f given by f(x)
x - = I?
(A) -2
(B) l
6
(C) l
2
(D) 2
12. If f(
(E) 6
, .,
b
1 1. If f is a iinear function and a .( a .( b, then f f" (x) dx =
a
(A) 0
(B) 1
(C) ab
2
(D) b - a
(E)b2-2 a2
13. 1
t
r:
~,'
\:
M
~"
fí:
~,
~
~
~;
If
'I
WA
GO ON TO THE NEXT PAGE
~
I
~~
~"
~~
tr:,
iì
~
ii
-~~
18
í~f.
--b.
I-~lJ
Calculus AS Part
~ AI
12. If ¡(x) = tIn
x2 Inx 2for
for02 .(
.( xx S;
4, then lim/ex) is
~ 2
x--2
J,
(A) In 2
(B) In 8
(C) In 16
(D) 4
(E) nonexIstent
y
2
a2
x
-2
13. The graph of the function / shown in the figure above has a vertical tangent at the point (2, 0) and
horizontal tangents at the poInts (1, -l) and (3, 1). For what values of x, -2 .( x .( 4, is / not
differentiable?
(A) 0 only
(B) 0 and 2 only
(C) 1 and 3 only
,
19
(D) 0, 1, and 3 only
(E) 0, 1, 2, and 3
14. A particle moves along the x-axis so that its position at time t is given by xU) = t2 - 6t + 5. For
what value of t is the velocity of the particle zero?
(B) 2
(A) 1
(C) 3
(E) 5
(D) 4
15. If F(x) = r 1i dt, then F '(2) =
o
(A) -3
(B) -2
(C) 2
(D) 3
I (E) 18
i
, .
GO ON TO THE NEXT PAGE
20
-
I:
f
ì
Calculus AS ~part A I
16. If /(x) = sin (e-X), then 1'(x) =
(A) -cos(e-X)
5
(B) cos(e-X) + e-x
(C) cos(e-X) - e-x
(D) e-x cos(e-X)
(E) -e-x cos(e-X)
18
y
x
17. The graph of a twice-differentiable function / is shov:n in the figure above. Which of the following is
true?
(A) f(l) 0:1'(1) 0:/"(1)
(B) f(1) 0: f"(1) 0: 1'(1)
(C) 1'0) 0: /(1) 0: /"(1)
(D) f"(1) o:f(1) 0:1'0)
(E) f"(1) o:f'(1) 0:/(1)
-
. GO ON TO THE NEXT PAGE
21
18. An equation of the line tangent to the graph of y = x + cos x at the pOInt (0, 1) is
(A) y = 2x + 1
(B) y = x + 1
(C) y = x
(D) y = x-I
(E) y = 0
19. If f"(x) = x(x + 1)(x - 2)2, then the graph of f has inflection points when x =
(A) -1 only (B) 2 only (C) -1 and 0 only CD) -1 and 2 only (E) -1, 0, and 2 only
22
Calculus AB ~. I
Part
k
20. What are all values of k for which J x2 dx = 0 ?
-3
(A) -3
(B) 0
(D) -3 and 3
(C) 3
(E) -3, 0, and 3
Ay
21. If dt = ky and k is a nonzero constant, then y could be
(A) 2ekty
(B) 2ekt
(C) ekt + 3
t
(D) kty + 5
1 1
2 2
(E) _ky2 + -
GO ON TO THE NEXT PAGE.
23
A
22. The function f is given by f(x) = x4 + x2 - 2. On which of the following intervals is f increasing?
(A) (- Ji, 00)
(B) (- Ji, Ji)
23.1
(C) (0, 00)
d
(D) (-00, 0)
c
(E) (_00, - Ji)
GO ON TO THE NEXT PAGE
24
Calculus AS ~part A J
y
,g?
x
23. The graph of f is shown in the figure above. Which of the following could be the graph of the
derivative of f?
(A)
y
(B)
x
(D)
y
x
,
i
\
25
L
o
b
x
y
x
x
a
(E)
a
y
(C)
t
24. The maximum acceleration attained on the interval 0 ~ t ~ 3 by the particle whose velocity is given
by vet) = tJ - 3t2 + 12t + 4 is
(A) 9
(B) 12
(C) 14
(D) 21
(E) 40
25. What is the area ûf the region between the graphs of y = x2 and y = -x from x = 0 to x = 2 ?
(A) ~
3
(B) §.
3
(C) 4
(D) ~
3
(E) 1.
3
/
GO ON TO THE NEXT PAGE
r~
t,.
tl~
l."
l
26
E'
~
I!
f4
w,
ii
Calculus AS Part
~ AI
x
f(x)
! ~ EÆ
26. The function f is continuous on the closed interval (0, 2) and has values that are given in the table
above. The equation f(x) = ~ must have at least two solutions in the interval (0, 21 if k ==
(A) a
(B) l
2
(C) 1
(D) 2
(E) 3
(D) 52
3
(E) 24
27. What is the average value of y = X2-.X3 + 1 on the interval (0, 2) ?
(A) 26
9
(B) 52
9
(C) 26
3
~
27
28. If ¡(x) = tan(2x), then f'(~) =
(A) 13
(B) 213
(C) 4
(D) 4..
(E) 8
F
t
rj
r
¡c'
f,
~
r~
:'~!
~
t1
END OF PART A OF SECTION I
i
1-;':
IF YOU FINISH BEFORE TIME is CALLED, YOU MAY CHECK YOUR WORK ON THIS PART ONLY. i
DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO. ~!l
ß',
~l
i
,~~
28
'\1
Calculus AB ~ Part B
..~
1
CALCULUS AB
~
SECTION I, Part B
:rl
-~
,~
Time - 50 minutes
;'~
..~
.,~
Number of questions - 17
:~
A GRAPHING CALCULATOR is REQUIRED FOR SOME QUESTIONS ON
THIS PART OF THE EXAMINATION.
~
-l!
Directions: Solve each of the following problems, using the available space for scratchwork. After exam-
ining the form of the choices, decide which is the best of the choices given and fill in the corresponding
oval on the answer sheet. No cædit wil be given for anything written in the test book. Do not spend too
much time on anyone problem.
BE SURE YOU ARE USING PAGE 3 OF THE ANSWER SHEET TO RECORD YOUR ANSWERS
TO QUESTIONS NUMBERED 76-92.
YOU MAY NOT RETURN TO PAGE 2 OF THE ANSWER SHEET.
In this test:
(1) The exact numerical value of the correct answer does not always appear among the choices given.
When this happens, select from among the choices the number that best approximates the exact
numerical value.
(2) Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x
for which f(x) is a real number.
-fLY.
"
GO ON TO THE NEXT PAGE .
Copyright (Ç 1998 College Entrance Examination Board and Educational Testing Service. All Rights Reserved.
Certain test materials are copyiighted solely in the name of ETS.
29
y
.
x
76. The graph of a function f is shown above. Which of the following statements about f is false?
(A) f is continuous at x = a.
(B) f has a relative maximum at x = a.
(C) x = a is in the domain of f.
(D) lim f(x) is equal to lim f(x).
x~a+ x~a
(E) lim f(x) exists.
x~a
t
t~
r:
f
¡t¡
;,-
l
~
GO ON , TO THE NEXT PAGE
I
tti
j'"'
~.
30
f~
i
1.
~,
1~
æ
Calculus AB ~part B I
1Y
:t
;t
;~
,;
:~
&
77. Let f be the function given by f(x) = 3e2X and let g be the function given by g(x) = 6x3. At what
value of x do the graphs of f and g have parallel tangent lines?
(A)
(B)
(C)
(D)
(E)
-0.701
-0.567
-0.391
-0.302
-0.258
78. The radius of a circle is decreasing at a constant rate of 0.1 centimeter per second. In terms of the
circumference C, what is the rate of change of the area of the circle, in square centimeters per
second?
(A) -(0.2)n:C
(B) -(O.l)C
(C) _ (O.l)C
2n:
(D) (0. i )2C
(E) (0.1 in:c
,
31
y
y
y
x
x
x
a
b
79. The graphs of the derivatives of the functions f, g, and h are shown above. Which of the functions
f, g, or ii have a relative maximum on the open interval a 0(. x 0( b ?
(A) f only
(B) g only
(C) Ii only
(D) f and g only
(E) f, g, and 1i
i
\
80. The first derivative of the function f is given by r(x) = CO;2X - l. How many critical values does f
\
i
.1
:1
II
have on the open interval (0, 10)?
Iii
(A) One
II
i1
l
I,
(B)
(C)
(D)
(E)
Three
Four
Five
Seven
r
IIII
ill!
¡ii
ii'i
h
J
'.','1
II
i:1
ii1
¡I
II
"
I'
.t.
GO ON TO THE NEXT PAGE '
!
iil
.i.lli....
I,:
III
¡'ì
Ii;
32
Calculus AS Part
~ BI
81. Let f be the function given by f(x) = 'x I. Which of the following statements about f are true?
i. f is continuous at x = O.
II. f is differentiable at x = O.
III. f has an absolute minimum at x = O.
(A) I
only
(B) II only
(C) III
only
(D) I and III only
(E) II and III only
.ns
, 1
3
82. If f is a continuous function and if F'(x) = f(x) for all real numbers x, then J f(2x )dx =
(A) 2F(3) - 2F(1)
2 2
(B) IF(3) - IF(l)
(C) 2F(6) - 2F(2)
(D) F(6) - F(2)
es f
1 1
2
2
(E) -F(6) - -F(2)
~
33
83. If a :t 0, then lim
x2 - a2
x--G x4 - a4
is
1
(A) ~2
(B) 2a2
1
(C) 6a2
CD) 0
(E) nonexistent
84. Population y grows according to the equation ~~ = le, where k is a constant and t is measured in
years. If the population doubles every 10 years, then the value of k is
(A) 0.069
(B) 0.200
(C) 0.301
34
(D) 3.322
(E) 5.000
Calculus AS ~part S i
I f ~ ) ~ i~ I :0 I :0 I 2~ I
ent
85. The function f is continuous on the closed interval (2, 8) and has values that are given in the table
above. Using the subintervals (2, 5), (5, 7), and (7, 8), what is the trapezoidal approximation of
8
12 f(x)dx ?
(A) 110
(B) 130
(C) 160
(D) 190
(E) 210
d in
)00
y
x
8
86. The base of a solid is a region in the first quadrant bounded by the x-axis, the y-axis, and the line
x + 2y = 8, as shown in the figure above. If cross sections of the solid perpendicular to the x-axis are
semicircles, what is the volume of the solid?
(A) 12.566
(B) 14.661
(C) 16.755
(D) 67.021
(E) 134.041
GO ON TO THE NEXT PAGE
35
'. ...
8S
87. Which of the following is an equation of the line tangent to the graph of ¡(x) = x4 + 2x2
at the point where r(x) = i ?
(A)
(B)
(C)
(D)
Y
y
y
y
=
=
=
=
8x - 5
x + 7
x + 0.763
x - 0.122
(E) Y = x - 2. 146
90
88. Let F(x) be an antiderivative of (In
x X)3 . If F(l) = 0, then F(9) =
(A) 0.048
(B) 0.144
(C) 5.827
(D) 23.308
(E) 1,640.250
h
~
lj
l'.:
¡¡:
if
i
i
t
N
rt
l''~
~i;
¡r
~
~.
I
~
t~
GO ON TO THE NEXT PAGE
. ,,'
~I':
~;
36
~
Calculus AS ~part B I
89. If g is a differentiable function such that g(x) -( 0 for all real numbers x and if
f'(x) = (x2 - 4)g(x), which of the following is true?
(A) f has a relative maximum at x = -2 and a relative minimum at x = 2.
(B) f has a relative minimum at x = -2 and a relative maximum at x = 2.
(C) f has relative minima at x = -2 and at x = 2.
(D) f has relative maxima at x = -2 and at x = 2.
(E) It cannot be determined if f has any relative extrema.
90. If the base b of a triangle is increasing at a rate of 3 inches per minute while its height h is
decreasing at a rate of 3 inches per minute, which of the following must be true about the area A
of the triangle?
50
(A) A is always increasing.
(B) A is always decreasing.
(C) A. is decreasing only when b -( h.
(D) A is decreasing only when b :? h.
(E) A remains constant.
.
37
91. Let f be a function that is differentiable on the open interval (1, 10). If f(2) = -5, f(5) = 5, and 9.
f(9) = -5, which of the following must be true?
1. f has at least 2 zeros.
II. The graph of f has at least one horizontal tangent.
III. For some c, 2 .: c .: 5, f(c) = 3.
(A) None
(B) I only
(C) I and II only
9'
(D) I and II only
(E) I, II and III
n . n
9:
92. If 0 ~ k .: '2 and the area under the curve y = cos x from x = k to x = '2 is 0.1, then k =
(A) 1.471
(B) 1.414
(D) 1.120
(C) 1.277
(E) 0.436
9(
END OF SECTION I
IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY
CHECK YOUR WORK ON PART B ONLY.
DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.
MAKE SURE YOU HAVE PLACED YOUR AP NUMBER LABEL ON YOUR
ANSWER SHEET AND HAVE WRTTEN AND GRIDDED YOUR AP NUMBER
IN THE APPROPRIATE SECTION OF YOUR ANSWER SHEET.
AFTER TIME HAS BEEN CALLED, ANSWER/ QtJESTIONS 93-96.
//
38
In the fist thee par of tl corron problem, the student is asked to
ABIBC Question 2
analyze the behavior of a given fuction over its entie domai. In par (b),
the stùdent must give an explicit arguent for why the critical point is an
absolute mium; relative arguents are not sufcient. The absolute
mium found in par (b), together with the lits in the fist par, lead
diectly to fidig the range in par (c). In the last par, the fuction is generaled to a famy of fuctions and .stdents are requied to deal with the
parameter b correctly to receive credit; studyig a specifc member of the
famy can help develop intution about the problem but does not show the
result for the entie famy.
The mean scores were 3.76 for Calculus AB and 5.41 for Calculus BC.
Students had diculty with justication of the absolute mium of 2xe2x
and handlg bxebx for general nonzero b.
X---OO X-'(,
bxebx is the same for al nonzero values of b.
a nonzero constat. Show that tlie absolute mium value of
(d) Consider the famy of fuctions defied by y = bxebx, where b is
(c) Whatistherangeoff?
(b) Fid the absolute mium value of f. Justi that your answer
is an absolute mium.
(a) Fid li f(x) and lif(x).
2. Let f be the fuction given by f(x) = 2xe2x.
46
Solittion
(a) x~-oo
lim 2xé~ = 0
li 2xe2x = 00 or DNE
"-+00
(b) ¡'(x) = 2e2x + 2x. 2 .e2x = 2e2"(1 + 2x) = 0
ifx= -1/2
f(-1/2) = -l/e or -0.368 or -0.367
-1/ e is an absolute mium vaue becuse:
(i) f'(x).c 0 for al x .c -1/2 and
--ri
-1/2
+
¡'x):; 0 for al x:; -1/2
(Ii) f'(x)
and x = -1/2 is the only critical
number
(c) Rage of f = (-l/e,(0)
or (-0.367,00)
or (-0.368, (0)
(d) y' = bebx + b2xeb" = beb"(l + bx) = 0
ifx = -lIb
At x = -lIb, Y = -lIe
y has an absolute minium vaue of - 1 / e for
al nonzero b
Board Note
Part (d)
Points
student's derivative
Scoring Scale
0/1 if not local minium from
I: evaluates f at student's critical point
1: solves f'(x) = 0
r 1: 00
0 as
-t -00
2 l1:
or xDNE
as x -t 00
3
0/1 for a local arguent
1: justifes absolute minium vaue
0/1 without exlicit sybolic
derivative
Note: 0/3 if no absolute mium based on
student's derivative
I: answer
exclude the right-hand "endpoint"
Note: must include the left-hand endpoint;
¡i: sets y' = beb~(l + bx) = 0
3 I: solves student's ii = 0
and gets a vaue independent of b
I: evauates y at a critical number
Note: 0/3 if only considering specic
vaues of b
1. The graph of y = bxeb" is a horiontal compression or expansion (with a reflection acoss
3/3 Argument with the followig three ingredients:
the y-axs if b .c 0) of the graph of y = xi".
2. The range of y = bxeb" is therefore the same as the range of y = xe".
of b.
3. Therefore the absolute mium vaue of y = bxeb" is the same for all (non-zero) values
0/3 Analyzing the horizontal compression/expansion of graphs of y = bxeb:i for specifc values of b.
47
ff Question 3
'h problem ilustrates one of the new topics in the course description,
iamely multiple representations of fuctions. Here, the velocity of a car is
jven both graplucaly and numericaly, but never algebraicaly. Some of
he par requie the graph, some requie the table, and some can be suc-
essfuy anwered using either: the graph is needed for par (a), the table
Dr par (b), either for (c), and the table for (d). The anwer to par (b) is
I
I
//
i
\/
t
vet)
75
60
72
81
20
30
55
70
78
0
12
(feet per second)
30
35
40
45
50
25
20
0
5
10
15
(seconds)
n exact aner, whie the anwers to (c) and (d) are approximations.
The mean score was 3.82 with a varety of nustakes spread over al
vet)
90
80
70
60
50
40
30
/ /'
1/
iar of the question.
~L
3 ui
~ ..
li&
'0)
~
20
10
0
Time (seconds)
values for v(t), at 5
3. The graph of the velocity vet), in ftsec, of a car traveling on a straight
road, for 0:5 t:5 50, is shown above. A table of
second interval of tie t, is shown to the right of the graph.
interval 0 :5 t :5 50.
(a) Durg what interval oftie is the acceleration of
the car
positive? Give a reason for your answer.
(b) Fid the average acceleration of the car, in ftsec2, over the
50
(c) Fid one approximation for the acceleration of the car, in ftsec2,
at t = 40. Show the computations you used to arve at your
anwer.
length. Using correct unts, explai
(d) Approxiate f v(t)dt with a Rieman sum, using the midpoints
o
of five subintervals of equal
the meang of th integral.
Solution
the velocity vet) is increasing on (0,35) and (45,50)
(a) Acceleration is positive on (0,35) and (45,50) because
(b) Avg. Acc. = v(50)
- v(O)
72 - 050
72
50-0
50=
or 1.44 ft/sec2
(c) Difference quotientj e.g.
v(45) - v(40) 60 -75 = -3 ft/sec2 or
5 5
5 - 5 5
v(40) - v(35) _ 75 - 81 = _.! ft/sec2 Or
10 - 10 10
v(45) - v(35) _ 60 - 81 = _ 21 ft/sec2
-or-
90-75
through (35,90) and (40,75): 35 _ 40 = -3 ft/sec2
Slope of tangent line, e.g.
¡50
(d) 10 vet) dt
"' 1O(v(5) + v(15) + v(25) + v(35) + v( 45))
= 10(12 + 30 + 70 + 81 + 60)
= 2530 feet
Thi integral is the total distance traveled in feet over
the time 0 to 50 seconds.
Points
3 1: (45,50)
1: (0,35)
reason
t 1:
Scoring Scale
Note: ignore inclusion of endpoints
1: answer
2 r 1: method
ll: answer
Note: 0/2 if fist point not eared
~ 1: midpoint Rieman sum
3 1: anwer
1: meang of integral
49
il Question 4
n tls problem, students are given inormation about the slope of the point
x, y) on a graph, which is equivalent to being given a dierential equation.
:'e exact value of f(1.2) is found from the solution of the separable dier~ntial equation, which compares closely with the approxiation to th
'alue found by using the tagent lie to f at x = 1. Unle recent year, tls
The mean score was 4.61. Some students were able to solve the dier-
lierential equation is not presented in a physical context.
~ntial equation in par (c) but did not anwer the presumably easier quesions in pars (a) and (b) about slope and tangent lie at a parcular point.
(c) Fidf(x) by solvig the separable dierential equation
(d) Use your solution from par (c) to fid f(1.2).
dy = 3x2 + 1 with the iitial condition fer) = 4.
dx 2y -
and use it to approxiate f(1.2).
(b) Write an equation for the lie tangent to the graph of f at x = 1
( a) Fid the slope of the graph of f at the point where x = 1.
thope
1 .
b 3x2
+ 1
graph 0ff
e s
IS .glVen
y 2y
4. Let f be a fuction with f(1) = 4 such that for al points (x, y) on the
50
Solution
dx 2y
(a) dy = 3x2 +. 1
:Ix=i
y=4 =3/;=~=~
1
1
(b) y - 4 = -(x - 1)
2
f(1.2) -,4 ~ '2(1.2 - 1)
f(1.2) ~ 0.1 +4 =4.1
(c) 2y dy = (3x2 +. 1) dx
l 2ydy = l(3x2 + l)dx
y2=X3+X+C
42 = 1+1 +C
14=C
y2 =x3+x+ 14
f(x) = Jx3 +.x+ 14
y = J x3 +. x +. 14 is branch with point (1,4)
(d) f(1.) = v'1.3 +1.2 + 14 ~ 4.114
Points
1: answer
Scoring Scale
r 1:
equation
of tangent
lie
2 1.
1: uses
equation
to approxiate
f(1.2)
1: separates variables
1: antiderivative of dy term
1: antiderivative of dx term
fuction out of a famy of fuctions
5 ~ 1: uses y = 4 when x = 1 to pick one
1: solves for y
0/1 if solving a liear equation in y
0/1 if no constant of integration
antidierentiation
Note: max 0/5 if no separation of variables
Note: max 1/5 (1-0-0-0-0J if substitutes
vaue(s) for x, y, or dy/dx before
the given dierential equation in (c)
1: answer, from student's solution to
51
In the second common problem, a fuction representig the outside tem-
ABIBC Question 5
peratue for a 24-hour period is given. Students are asked to sketch the
graph of the fuction as an aid to helping them get a feel for the question.
Defute integrals are used to compute both the average temperatue over
an 8-hour tie interval and the cost of ai conditionig the house durg
integral as an accumulator. It is expected that students compute the inte-
the 24 hour. The las par of the problem is an example of using a defite
gral using a calculator.
The mean scores were 4.09 for Calculus AB and 5.67 for Calculus BC.
Many students did not give the anwers, as requested, to the nearest
degree and nearest cent. !
hours.
(a) Sketch the graph of F on the grd below.
6
12
18
~ 80 ----------¡---------~----~.----t---------
60
o
TIme in
24
values of t was the ai conditioner coolig the house?
(d) The cost of coolig the house accumulates at the rate of $0.05
per hour for each degree the outside temperatue exceeds 78
degrees Fahenheit. What was the total cost, to the nearest cent,
to cool the house for th 24-hour period?
temperatue was at or above 78 degrees Fahenheit. For what
(c) An ai conditioner cooled the house whenever the outside
(b) Fid the average temperatue, to the nearest degree Falrreneit,
between t = 6 and t = 14.
Hours
II
1 I 1
Q
70 ----------r---------~----~.----t--------i
,
I
I
I
,
,
I
I
I
~ 90 ----------¡---------¡---------t---------
100
__ I I I,
C
l
t
~
:
i
:
=
i
I
I
4l
i
e
i
I
ti
t
I I
I
I
I
I
where F(t) is measured in degrees Fahenheit and t is measured in
F(t) = 80 - 10 cos (~~), 0 s t s 24,
5. The temperatue outside a house durg a 24-hour period is given by
52
100
Solution
(a)
l
~
~
:¡
i:
60
o
6
12
Tune in Hour
18
24
(b) Avg. = 14 ~ 614 (80 - lOcos GÐ) dt
= ã(697.2957795)
= 87.162 or 87.161
~ 87" F
(c) (80 - lOcos (~) J -782: 0
2 -lOcos (~) 2: 0
or -:t-: or
5.230J r18.769
5.231 - - 18.770
18.770
or
(d) C = 0.05 h~;:69 ((80 - lOcos (~) J -78) dt
0"
5.230
= 0.05(101.92741) = 5.096 ~ $5.10
Points
Scoring Scale
1: bell-shaped graph
mimum 70 at t = 0, t = 24 only
maxum 90 at t = 12 only
2: integral
1: limits and 1/(14 - 6)
1: integrand
0/1 if integral not of the form
3 ~ 1: anwer
b~a t F(t)dt
ll: solutions with interva
2 r 1: inequality or equation
2: integral
1: lits and 0.05
1: integrand
3 ~ 1: anwer
0/1 if integral not of the form
k t (F(t) - 78) dt
53
AB Question 6
An implicitly defied fuction is given, and in par (a) the student is asked
to veri its derivative. The derivative was given so that each student
(c). In the second par, there is both an x-value and a y-value to be ana-
would be dealg with the correct derivative throughout the other, more
complicated par of the problem and be able to fid the point P in par
a point on the cure. In the last par, students can use either the cure or
lyzed, and the y-value where the derivative is zero does not correspond to
the derivative to fid the point P.
The mean score on ths question was 2.86 with 80 percent of the scores
scores discriated very well at al grade levels.
4 or below and over 50 percent of the scores 2, 3, or 4. In spite of th, the
J
dy _ 4x - 2xy
(a) Show that
(b) Write an equation of each horizontal tangent lie to the cure.
(c) The line though the origin with slope - i is tangent to tlie cure
at point P. Fid the x- and y-coordiates of point P.
dx - 2 2 .
X+Y+i
6. Consider the cure defied by 2y3 + 6x2y - 12x2 + 6y = 1.
54
Solution
,dy 2dy dy
(a) 6y- dx +6x dx +12xy-24x+6dx =0
~~ (6y2 + 6x2 + 6) = 24x -12xy
dy 24x - I2xy 4x - 2xy
dx = 6x2+6y2+6 = x2+y2+1
(b) dy = 0
dx
4x - 2xy = 2x(2 - y) = 0
x=O or y=2
Wben x = 0, 2y3 + 6y = 1 i Y = 0,165
Points
Scoring Scale
1: uses solutions for x to fid equations
of horizontal tangent lines
1: solves dx = 0
dy
1: sets dx = 0
dy
1: verifes expression for dx
r 1: implicit dierentiation
2 dy
4
Note: max 1/4 (l-O-o-OJ if dy/dx = 0 is not
1: verifes which solutions for y yield
equations of horizontal tangent lines
of the form g(x,y)/h(x, y) = 0 with solutions
y coordinate of 2.
There is no point on the cure with
y = 0.165 is the equation of the only
-or-
incorrect derivative from part (a)
Note: max 2/3 (I-I-OJ if importing
11: solves
dyfor x and y
3 1: substitutes y = -x into ~~
1: sets dx =-1
1: solves
11:
y =-xfor x and y
of cure
3 1: substitutes y = -x into equation
for both x and y
horizontal tangent lie.
(c) y = -x is equation of the lie.
2(-x)3 +6x2(-x) -12x2+6(-x) = 1
_8x3 - I2x2 - 6x - 1 = 0
x=-1/2, y=I/2
-or-
-=-1
dy
dx
4x - 2xy = _x2 - y2 - 1
4x2+4x+I=0
4x + 2x2 = _x2 - x2 - 1
x=-I/2, y=I/2
55
ic¡c1r
CALCULUS AB
SECTION I, Part A
Time - 50 minutes
Number of questions - 25
A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION.
Directions: Solve each of the following problems, using the available space for scratchwork. After exam-
ining the form of the choices, decide which is the best of the choices given and fill in the corresponding
oval on the answer sheet. No credit wil be given for anything written in the test book. Do not spend too
much time on anyone problem.
In this test: Unless otherwise specified, the domain of a function f is assumed to be the set of all real
numbers x for which f(x) is a real number.
2
1. f (4x3 - 6x) dx =
1
(A) 2
(B) 4
(C) 6
(D) 36
(E) 42
~
f,
1,
I
t
¡
!
E
~
~
~
I
i
.
i
¡¡
~
'I
GO ON TO THE NEXT PAGE
\
'i
II
I:
i
I
~
~
:fi
0\
~:
\1
~
~.
ï
*~
l;
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lï
Il
12
ii
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~
Calculus AS ~part AI
2. If f(x) = x..2x - 3, then f'(x) =
(A) 3x - 3
..2x - 3
(B) x
-fx - 3
(C) I
..2x - 3
(D) -x + 3
..2x - 3
(E) 5x - 6
, , 2..2x - 3
b
b
a a
3. If J f(x) dx = a + 2b, then J (f(x) + 5) dx =
(A) a + 2b + 5
(B) 5b - Sa
(C) 7b - 4a
(D) 7b - Sa
(E) 7b - 6a
om ON TO THE NEXT PAGE
13
1
4. If f(x) = -x3 + x + -, then 1'(-1) =
x
(B) 1
(A) 3
(C) -l
(D) -3
(E) -5
5. The graph of y = 3x4 - i 6x3 + 24x2 + 48 is concave down for
(A) x ~ 0
(B) x;: 0
(
2
(C) x ~ -2 or x ;: -"3
2
(D) x c: "3 or x ;: 2
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~ AI
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7.
(B) e-2 + C
t
(C) e2 + C
t
(D) 2e2 + C
(E) et + C
d
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(A) 6x2 sin(x3)cos(x3)
(B) 6x2 cos(x3)
(C) sin2(x3)
(D) -6x2 sin(x3)cos(x3)
(E) -2 sin(x3)cos(x3)
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15
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o ~ t ~ 8, is given by the function whose graph is shown above.
8. At what value of t does the bug change direction?
(A) 2
(B) 4
(C) 6
(D) 7
(E) 8
9. What is the total distance the bug traveled from t = 0 to t = 8 ?
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is
(A) y - i = -(x - ¡)
(B) y - i = -2(X - ¡)
(C) y = 2( x - ¡)
(D) y = -(x - ¡)
(E) y = -i(x - ¡)
GO ON TO THE NEXT PAGE
17
l.
y
x
i 1. The graph of the derivative of f is shown in the figure above. Which of the following could be the
graph of f?
(A)
y
(B)
x
x
(C)
y
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(D)
x
(E)
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of f at x = 3 is used to find an approximation to a zero of f, that approximation is
(B) 0.5
(A) 0.4
(C) 2.6
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(E) 5.5
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(A) X-lQ
lim f(x) x-'b
= lim f(x)
(B) lim f(x) = 2
X-'Cl
(C) x-'b
lim ¡(x) = 2
(D)x-'b
lim f(x) = i
(E) lim f(x) does not exist.
X-'Cl
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21
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16. The area of the region enclosed by the graph of y = x2 + 1 and the line y = 5 is
(A) ~
3
(E) 8n 8.
(D) 32
3
(C) 28
3
(B) 1&
3
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17. If x2 + y2 = 25, what is the value of ~;~ at the point (4, 3)?
\19. If f(x)
7
(A) _ 25
27
(B) -27
7
(C) 27
(E) ~~ ¡ i
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(B) 1
(A) 0
(C) e - 1
(D) e
(E) e + 1
19. If f(x) = In\x2 - 1 \, then f'(x) =
(E) 25
27
(A) x2 - 1
\ 2x \
2x
(B) L x2 - I I
(C) 2,1
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(D) x,2x
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23
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20. The average value of cos x on the interval (-3, 5) is
I
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2
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(C) I
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24
Calculus AS ~part A ,
22. What are all values of x for which the function .f define~ by ¡(x) = (x2 - 3)e-X is increasing?
(A) There are no such values of x.
(B) .; ;( -1 and x )0 3
(C) --3 ;( x ;( 1
(D) -1 -: x -: 3
(E) All values of x
23. If the region enclosed by the y-axis, the line y = 2, and the curve y = -I is revolved about the
y-axis, the volume of the solid generated is
(A) 32n
5
(B) i 6n
3
(C) i 6n
5
(D) 81t
3
(E) n
:nt
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2
24. The expression 510 ( ~ + ~ + & + ... + ~) is a Riemann sum approximation for
i
(A) J fF dx
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(B) J -l dx
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(C) 50 "V5õdx
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~ Ai
25. J x sin(2x) dx =
x 1
x 1
(A) -"2 cos(2x) + 4 sin(2x) + C
(B) -"2 cos(2x) - 4 sin(2x) + C
(C) ~ cos(2x) - ~ sin(2x) + C
(D) -2x cos(2x) + sin(2x) ;t C
(E) - 2x cos(2x) - 4 sin(2x) + C
END OF PART A OF SECTION I
IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS PART ONLY.
. DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO.
~
27
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CALCULUS AB
SECTION I, Part B
Time-40 minutes
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Number of questions-IS
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A GRAPHING CALCULATOR IS REQUIRD FOR SOME QUESTIONS ON
THIS PART OF THE EXAMINATION.
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Directions: Solve each of the following problems, using the available space for scratchwork. After examining the fûim of the choices, decide which is the best of the choices given and fil in the corresponding
avalon the answer sheet. No credit wil be given for anything written in the test book. Do not spend too
1III
much time on anyone problem.
III
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BE SURE YOU ARE USING PAGE 3 OF THE ANSWER SHEET TO RECORD YOUR ANSWERS
TO QUESTIONS NUMBERED 76-90.
I
YOU MAY NOT RETURN TO PAGE 2 OF THE ANSWER SHEET.
In this test:
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(l) The exact numerical value of the correct answer does not always appear among the choices given.
When this happens, select from among the choices the number that best approximates the exact
numerical value.
(2) Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x
for which f(x) is a real number.
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Certain test materials are copyrighted solely in the name of ETS.
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28
Calculus AB ~part 81
e2X
76. If ¡(x) = 2x' then F(x) =
(A) 1
e2X(l - 2x)
(B) 2x2
(C) e2X
(D) e 2X(2;2 + 1)
e2X(2x - 1)
imng
00
(E) 2x2
:RS
'S x
t
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29
77. The graph of the function y = x3 + 6x2 + 7 x - 2 cos x changes concavity at x =
7
(A) -1.58
(B) -1.63
(D) -1.89
(C) -1.67
(E) -2.33
y
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81
x
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78. The graph ~f f is shown in the figure above. If J f(x) dx = 2.3 and F'(x) = f(x), then
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(A) 0.3
(B) 1.3
(C) 3.3
(D) 4.3
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30
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79. Let I be a function such thafi lim 1(2 + Iii - 1(2) = 5. Which of the following must be true?
h..o
1. f is continuous at x = 2.
.33
II. f is differentiable at x = 2.
III. The derivative of I is continuous at x = 2.
(A) I
only
(B) II only
(C) I and II only
(D) I and III only
(E) II and III only
80. Let I be the function given by i(x) = 2e4X2. For what value of x is the slope of the line tangent to
the graph of I at (x, ¡(x)) equal to 3 ?
(A) 0.168
(B) 0.276
(C) 0.318
(D) 0.342
(E) 0.551
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31
8l. A railroad track and a road cross at right angles. An observer stands on the road 70 meters south of the
crossing and watches an eastbound train traveling at 60 meters per second. At how many meters per
second is the train moving away from the observer 4 seconds after it passes through the intersection?
(A) 57.60
(B) 57.88
(C) 59.20
(D) 60.00
8
(E) 67.40
8
82. If y = 2x - 8, what is the minimum value of the product xy ?
(A) -16
(B) -8
(C) -4
(D) 0
(E) 2
t
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Calculus AB Part
~ BI
, the
83. What is the area of the region in the first quadrant enc1os~d by the graphs of y = cos x, y = x, and
the y-axis?
?
(A) 0.127
LQ
(B) 0.385
(C) 0.400
(D) 0.600
(E) 0.947
84. The base of a solid S is the region enclosed by the graph of y = -f, the line x = e, and the
x-axis. If the cross sections of S perpendicular to the x-axis are squares, then the volume of S is
2
(A) 1.2
1
(B) ~
3
(C) 1
(D) 2
(E) -(e3 - 1)
3
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I 33
85. If the derivative of ¡ is given by f'(x) = eX - 3x2, at which of the following values of x does f
have a relative maximum value?
(A) -0.46
(B) 0.20
(C) O.9l
(D) 0.95
87.
(E) 3.73
86. Let ¡(x) = -I. If the rate of change of ¡ at x = c is twice its rate of change at x = 1, then c =
(A) 1.
4
(B) 1
(C) 4
(D) -L
-.
(E) 2~
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34
I
Calculus AB ~part B I
87. At time t;: 0, the acceleration of a particle moving on the x-axis is a(t) = t + sin t. At t = 0, the
velocity of the particle is -2. For what value of t wil the'velocity of the particle be zero?
73
(A) 1.02
(B) 1.48
(C) 1.85
(D) 2.81
(E) 3.14
2
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35
I
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x
8Ç
x
88. Let ¡(x) = J h (t) dt, where h has the graph shown above. Which of the following could be the
a
graph of f?
x
a 0
b
a
c
y
(C)
a
y
(B)
y
(A)
(D)
x
o
o
x
b
c
y
x
c
(E) Y
a
x
b
c
GO ON TO THE NEXT PAGE '
36
Calculus AB ~p
x
0
0.5
1.0
1.5-
2.0
f(x)
3
3
5
8
13
89. A table of values for a continuous function f is shown above. If four equal subintervals of (0, 2) are
2
used, which of the following is the trapezoidal approximation of f f(x) dx ?
o
e
(A) 8
(B) l2
(C) .16
(D) 24
(E) 32
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37
art B
90. Which of the following are anti
derivatives of f(x) = sin x cos x?
. 2
I
,(
L F(x) = Sl~ X
r'
.t
2
II. F(x) = co~ x
III. ' F(x) = -CO~(2x)
(A) I
only
(B) II only
(C) II only
(D) I and II only
(E) II and II only
2
2
END OF SECTION I
IF YOU FINISH BEFORE TIME is CALLED, YOU MAY
CHECK YOUR WORK ON PART B ONLY.
DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.
38
2'
Answers to the 1997 AP Calculus AB and
Calculus BC Examinations
.
.
ii
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Section I: Multiple Choice
Section I: Multiple Choice
II Blank Answer Sheet
.,0)
Listed below are the correct answers to the multiple-
Section II: Free Response
II Student Preparation for the Exams
II Free-Response Questions, Scoring Guidelines,
and Sample Student Responses with Commentary
II Section II, Calculus AB
II Section II, Calculus BC
choice questions and the percentage of AP candidates
who answered each question correctly. A copy of the
blank answer sheet appears on the following pages
~.
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for reference.
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Section I Answer Key and Percent Answering Correctly
Calculus AB
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98
1997
AB-1
1. A particle moves along the x-axis so that its velocity at any time t '2 0 is given by
vet) = 3t2 _ 2t - 1. The position x(t) is 5 for t = 2.
(a) Write a polynomi expression for the position of the particle at any time t :: O.
(h) For what values of t, 0 $ t $ 3, is the particle's instantaneous velocity the sareas its
average velocity on the closed interval (0,3F
(c) Find the total distance traveled hy the particle from tbne t = 0 until time t = 3.
x (t) = t3 - t2 - t + C
(a) x(t) = J vet) dt = J (3t2 - 2t - '1) dt
er1'or(s) in t3 - t2 - t
3L -(-( ~1?
_ I? no constant of integration
\ 2:
= t3 _ t2 - t + C
evaluates constant of integration
x(2) = 8 - 4 - 2 + C = 5; C = 3
x(t)=t3_t2_t+3
1: average velocity = 5
ad of
Erage
(b)
avg.
. - 3 -x(O)
0
L _ ve
x(3)
1: sets vet) equal to student's
not
= =5
18 - 3
1: answer
3
~ Il.
uted
rded.
average velocity
3
0/1 if not solving
3t2 - 2t - 1 = an avg. velocity
3t2 - 2t -1 = 5
1 +.J
t = or
1.786
3
1: limits of 0 and 3 on an
integral of vet) or \v(t)\
(c) distance == fo3 \v(t)\ dt
or
substitutes t == 0 and t = 3 in x(t)
3
== fo3 l3t2 - 2t - 1\ dt = 17
or
\ 1:
vet) = 3t2 - 2t - 1 = 0
ll:
1
t = -3 l t = 1
x(O) = 3
x(l) = 1 - 1 - 1 + 3 == 2
x(3) = 27 - 9 - 3 + 3 == 18
distance == (3 - 2) + (18 - 2) = 17
( 105 J
handles change in direction
at student's turning point
answer
AB-2
1997
2. Let f be the function given by f(x) = 3cosx. As
shown ..
y
above, the graph of f crosses the y-axis at point P and the
P (0, 3)
x-axis at point Q.
(a) Write an equatipn for the line passing through points P
and Q.
(b) Write an equation for the line tangent to the graph of f at
point Q. Show the
analysis that leads to your equation.
(c) Find the x-coordinate of the point on the graph of f,
between points P and Q, at which the line tangent to the
graph of f is parallel to line PQ.
x
(d) Let R be the region in the first quadrant bOu....lded by the
graph of f and line segment PQ. Write an integral
expression for the volume of the solid generated by
revolving the region R about the x-axis. Do not evaluate.
3 - 0
(a) slope = 0 _ 1f /2 =
6
7r
2 J 1: slope
L 1: equation
6
y ~ 3 = --(x - 0)
1f
(b) j'(x) = -.3sinx
1: equation
j'(1f/2) =using
-3 j'(1f/2) and j(1f/2)
2 LJ 1:
f'(7r/2) = -3sin(1f/2) =-3
y - 0 = -3(x -1f/2)
( c)
j'(x) = -3sinx =-~1f
1/2 if equation only
2 J 1: equates derivative to slope
. 2
sinx = 1f
L 1: solution in (O,1f/2J
1/2 if solution only
x = 0.690
2: integrand
(d) V = 1f17l/2 l(3COSX)2 - (-~x + 3)2J dx
0/2 if not difference of 2 squares
3
1/2 if incorrect but of form
(acosx)2 - (bx + c)2 ; O¡, b =1 0
1/2 if reversal
1: constant and limits
113 I
;t-
îi
1997
AB-3
3. Let f be the function given by f(x) = JX - 3.
lineofx j
= 6.
(a) On the axeS pr6vided below, s!rtch the graph
and shade the region R enclosed by
the graph of f, the x-axis, and the vertical
(b) Find the area of the region R described in part (a).
(c) Rather than usin tliè line x = 6 as in pax (.), consider the line" = w, where w can be
integral
A(w).
any number greater than 3. Let A(w)
be tlie ere' ofanthe
regionexpression
enclosed for
by the
graph
lie x = w. Write
of j, the x-axis, aid tlie vertica
(d) Let A(w) be ,as described in part (c). Find the rate of che of A with respect to w
when w = 6.
(a)
graph of f, (domain is x 2: 3,
goes through (3,0), is increasing,
positive, and concave down)
y
2 \1:
1:
correct region relative to graph of f
1 2
13 3 3
1: limits
(b) area = (6 .¡ dx = t(x - 3)3/216
= 2-/ = 3.464
(c)
A(w) = hW .¡ dx
dA = -fw-- 3
(d)
dw
dA\
.
W w=6 ..
-d = J3 = 1.732
3
1: integrand
1: answer
0/1 if second point is not earned
2 r 1: limits
ì. 1: integrand
~1: dA
2 dw
1: evaluation at 6
0/2 if ~~ is constant
C 121 J
1997
AB-4
4. Let f be the function given by f(x) = x3 - 6x2 + p, where p is an arbitrary constant.
maximum and minimum
(a) Write an expression for 1'(x) ald use it to find the relative
values of f in terms of p. Show the analysis that leads to your conclusion.
(b) For what values of the constant p does f have 3 distinct roots?
(c) Find the value of p such that the average value ofjover the closed interval (-1,2) is 1.
1: finds fiex)
(a) f(x) = x3 - 6x2 + p
1: solves l' (x) = 0
f'(x) = 3x2 - 12x = 0
3x(x - 4) = Q
1: indicates the location of a maximum
and a minimum, with analysis
0/1 if explicitly chooses to work
only with a specific value for p
4
x=O , x=4
l' (x) changes sign from positive to negative
1: finds the maximum and minimum
at x = 0
values in terms of p
f'(x) changes sign from negative to positive
at x =4
or
f"(x) = 6x - 12, f"(0) = -12, t'(4) = 12
relative maximum at x = 0, f(O) = p
relative minimum at x = 4, f (4) = p - 32
(b) f (x) has three distinct real roots
when p )- 0 and p - 32 .: 0,
so 0.: p .. 32
2 r 1: upper bound
1, 1: lower bound
.. -1)- including 0 or 32, or both
1 12
( c)
2 - (-1) -1
average value
(x3 - 6x2 + p) dx = 1
r
~ ri~x4 - '),~3 -I nrr12 = 1
3 4 --, .r- j -1
3
1: integrand and limits
1: appropriate constant for
definite integral
1: sets average value equal to 1
and solves for p
~ (( 1: _ 16 + 2P) - (l + 2 - p) 1
equation must be of the form
k r2 (x3 _ 6x2 + p) dx = 1
1-1
= ~ (3P - 5:) = 1
1/3 solution only
23
P = -4 = 5.75
129
1997
AB-5 , BC-5
y
5. The graph of a function f consists of a semicircle and two line
segments as shown above. Let 9 be the function given by
3
g(x) = 10: f(t) dt. .
x
(a) Find g(3).
-5 -4 -3 -2 -1 0
-1
(b) Find all values of x on tne open interval (-2) 5) at which
9 has a relative maximum. Justify your answer.
(c) Write an equation for the line tangent to the graph of 9
-2
-3
at x = 3. .
(d) Find the x-coordinate of each point of inflection of the
graph of 9 on the open interval (- 2) 5). Justify your
answer.
answer
(a)
2f'
g(3) = 13 Jet) dt
-( -1:; each incorrect area
-( - 1:; error in summing
4 2 2
== ~ . 1f . 22 - ~ = 1f - ~
1: relative maximum at x = 2 only
(b) g(x) has relative maximum at ,x = 2
1: g'(x) = ¡(x) or interprets
because g'(x) = ¡(x) clîånges from positive
3 g (x) as area accumulator
to negative at x = 2
1: justification
'e
(ignore discussion at x = 5)
1
ts
( c)
f 1: g'(3) = -1
g(3) = 1f - "2
2 ll: equation using g(3) and g'(3)
g'(3) = ¡(3) = -1
y - (7l - ~) = -l(x - 3)
points of infection with
(d) graph of 9 has points of infection with
2ri'
x-coordinates x = 0 and x = 3
because g"(x) = f'(x) changes from positive
x-coordinates o
,and 3 only
justification
L 1: (ignore discussion at x = 2)
to negative at x = 0 and from negative to
i
positive at x = 3
1/2 if x = 0) 3 selected as candidates
or
and x = 3 discarded because
because g'(x) = ¡(x) changes from increasing
g" (3) does not exist
to decreasing at x = 0 and from decreasing
1/2 if x = 0,2,3 selected as candidates
to increasing at x = 3
and x = 2 and x = 3 discarded because
g" (2) and g" (3) do not exist
137 I
1997
AB-6, BC-6
her
6. Let vet) be the velocity, in feet per second, of a skydiver at time t seconds, t :2 O. After
parachute opens, her velocity satisfies the differential equation ~~.. = -2v - 32, with initial
condition v(O) = -50.
(a) Use separation of variables to find an expression for v in terms of t, where t is measured
in seconds.
(b) Terminal velocity is defined as t~ vet). Find thevterminal velocity of the skydiver to
the nearest foot per
second.
(c) It is safe to land when her speed is 20 feet per second. At what time t does she reach
this speed?
dv
(a)
1: separates variables
dt = -2v - 32 = -2(v + 16)
1: antiderivative of dv side
dv == -2dt
Oil if not fav+
~b ' a # 0
v + 16
In Iv + 161 = -2t + A
al
1: antiderivative of dt side
Iv + 161 = e-2t+A = eAe-2t
1: constant of integration
6
1: uses initial condition v(O) = -50
v + 16 = Ce-2t
1: solves for vet)
-50 + 16 = Ceo j C = -34
0/1 if not solvingav+
dv b = k dt
v = _34e-2t - 16
where a, b, k nonzero
Oil if no constant of integration
0/6 if variables not separated
(b) lim
vet) t-+DO
= lim (_34e-Zt - 16) = -16
t-+oo
(c) vet) = _34e-Zt - 16 = -20
1: limit value
must be exponential vet) with finite limit
sets vet) = -20
r 1:
2 t1:
2
e-2t = 17
_ .'
solution
mûst be exponential v (t)
t = -~ In (2.) = 1.070
2 17
145 I