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Lateral area of a right circular cone if the base has circumference C and slant height is /: 5
4
Volume of a sphere of radius r: ~ 37”
Surface area of a sphere of radius r: 5 =4n7
1
Volume of a pyramid of base area B and height h: ‘= 3
If point (a,b) lies on the graph of function f, which of the following points must lie on the graph of
the inverse f?
(A) (a,b)
(B) (-a,b)
(C) (a)
(D) (b,a)
(E) (-b-a)
NN
Harry had grades of 70, 80, 85, and 80 on his quizzes. If all quizzes have the same weight, what
grade must he get on his next quiz so that his average will be 80?
(A) 85
(B) 90
(©) 95
(D) 100
(E) more than 100
ail
pire
we
Which of the following is an asymptote 0
(A) x=-2
(B) x 7
(">
(D) x=1
(E) x=2
al
If log, x =p and log, y = q, then log, xy =
ge
(B) +
Rta
(a
(@) p-4q (E) p?
5. The sum of the roots of 3x? + 4x? — 4x = 0 is
4
(a) 3
3
(®) a
(o
4
@) 3
(E) 4
ee
6.18 then OG
(A) 0
B)
Z. If fx) = log(x + 1), what is f-1(3)?
(A) 0.60
(B) 4
(C) 999
(D) 1001
(E) 10,000
8. If f(x) 20 for all x, then f(2 — x) is
(A) 2-2
(B) 20
(©) 22
(D) <0
(E) <2
9, How many four-digit numbers can be formed from the numbers 0, 2, 4, 8 if no digit is repeated?
(A) 18
(B) 24
(©) 27 (D) 36
(E) 64
10. If x-1 is a factor of x? + ax — 4, then a has the value
(A) 4
(@) 3
(2
@)1
(E) none of the above
1. If 10 coins are to be flipped and the first 5 all come up heads, what is the probability that exactly 3
more heads will be flipped?
(A) 0.0439
(B) 0.1172
(C) 0.1250
(D) 0.3125
(E) 0.6000
12. Ifi=
Fi and nis a positive integer, which of the following statements is FALSE?
(a) if” =1
(B) i471 =
(
43
(> (E) Cannot be determined from the given information.
19. A cylindrical bar of metal has a base radius of 2 and a height of 9. It is melted down and reformed
into a cube. A side of the cube is
(A) 2.32
(B) 3.84
(C) 4.84
(D) 97.21
(E) 113.10
20. The graph of y = (x + 2)(2x—3) can be expressed as a set of parametric equations. If x = 2t— 2 and
y= f(t), then f() =
(A) 2(4t-5)
(B) (2t- 2)(4t-7)
(C) 2t(4t- 7)
(D) (2t-2)(4t-5)
(E) 20(4t + 1)
iS
21. If points (v2.3) and (~
.»:) lie on the graph of y = x3 + ax2 + bx +c, andy; —y = 3, then b=
(A) 1.473
(B) 1.061
(C) 0.354
(D) -0.939
(E) -2.167
22. Rent-a-Rek has 27 cars available for rental. Twenty of these are compact, and 7 are midsize. If two
cars are selected at random, what is the probability that both are compact?
(A) 0.0576
(B) 0.0598
(C) 0.481
(D) 0.521
(E) 0.541
23. If a and bare real numbers, with a > b and |a| 0
(B) a0
(D) b<0
(E) none of the above
by
24. If [x] is defined to represent the greatest integer less than or equal to x, and “> | the
maximum value of f(x) is g
Nero
(A) 0
(B) 1
(C) 2
() 3
(E)
26. A right circular cone whose base radius is 4 is inscribed in a sphere of radius 5. What is the ratio
of the volume of the cone to the volume of the sphere?
(A) 0.222:
(B) 0.256 :
(C) 0.288 :
(D) 0.333 :
(E) 0.864 :
Beaees
8
S
Tf x9 = Land %=1=
(A) 1.260
(B) 1.361
(C) 1.396
(D) 1.408
(E) 1.412
i
lee
28. The y-intercept of ©
(A) 0.22
(B) 0.67
(©) 1.41
(D) 1.49
(E) 4.58
i
eS
29. If the center of the circle x2 + y? + ax + by + 2 = 0 is point (4,-8), then a + b=
(A) 8
(B) 4
(Cc) 4
@) 8 (E) 24
30, If p(x) = 3x? + 9x + 7 and p(a) = 2, then a =
(A) only 0.736
(B) only -2.264
(C) 0.736 or 2.264
(D) 0.736 or -2.264
(E) ~0.736 or -2.264
Se
BL. If iis a root of x* + 2x3 — 3x? + 2x — 4 = 0, the product of the real roots is
(A) 4
(B) 2
(Cc) 0
(@) 2
«4
sina =3,90°< A<180°,c080= +.
32. If 5” 3 and 270° < B < 360°, sin(A + B) =
lo
IS
(A) 0.832
(B) 0.554
(C) -0.333
(D) 0.733
(E) 0.954
ee
ies
33. If a family has three children, what is the probability that at least one is a boy?
(A) 0.875
(B) 0.67
(©) 05
(D) 0.375
(E) 0.25
34, If sec 1.4 =x, find the value of csc(2 tan“ x).
(A) 0.33
(B) 0.87
(©) 1.00
(@) 1.06
(E) 3.03
eo
kn
35, The graph of ly — 1|= |x + 1| forms an X. The two branches of the X intersect at a point whose
coordinates are
(A) (11)
(B) (-1,1)
(C) (1-1) (D) (-L-1)
(E) (0,0)
36. For what value of x between 0° and 360° does cos 2x = 2 cos x?
(A) 68.5° or 291.5°
(B) only 68.5°
(C) 103.9° or 256.1°
(D) 90° or 270°
(E) 111.5° or 248.5°
32. For what value(s) of x will the graph of the function f= sin have a maximum?
38. For each positive integer n, let S, = the sum of all positive integers less than or equal ton, Then
Ss1 equals
(A) 50
(B) 51
(C) 1250
(D) 1275
(E) 1326
39. If the graphs of 3x? + dy? - 6x + 8y — 5 = 0 and (x — 2)? = 4(y + 2) are drawn on the same
coordinate system, at how many points do they intersect?
(A) 0
@®) 1
(C) 2
@) 3
() 4
40. If log, 2 = loggx is satisfied by two values of x, what is their sum?
(A) 0 (B) 1.73
(©) 2.35
(D) 2.81
(E) 3.14
41. Which of the following lines are asymptotes for the graph of "
Lx=-1
ILx=5
IL. y =3
(A) Tonly
(B) only
(©) Land Tl
(D) Land TIT
(E) 1, 1, and Ill
3sin20
1
A2. If 1cos2@ 2 and 0° < @< 180°, then @ =
(A) 0°
(B) 0° or 180°
(C) 80.5°
(D) 0° or 80.5°
(E) 99.5°
43. If f(x,y) = 2x2 - y* and g(x) = 2*, which one of the following is equal to 22*?
(A) fg)
(B) f(g»)
(C) f(g@%).9@))
(D) f(g@),0)
) g(fx))
44. Two positive numbers, a and b, are in the sequence 4, a, b, 12. The first three numbers form a
geometric sequence, and the last three numbers form an arithmetic sequence. The difference b —a
equals
1
(B) >
(Cc) 2
(D) *2
©) 3
45. A sector of a circle has an arc length of 2.4 feet and an area of 14.3 square feet. How many degrees
are in the central angle? (A) 63.4°
(B) 20.2°
(©) 143°
(D) 12.9°
(E) 115°
46, The y- coordinate of one focus of the ellipse 36x + 25y? + 144x—50y — 731 = O is
(A) -2
@)1
(©) 3.32
(D) 4.32
(E) 7.81
ew c
4 >
47. In the figure above, ABCD is a square. M is the point one-third of the way from B to C. N is the
point one-half of the way from D to C. Then @=
(A) 50.8°
(B) 45.0°
(C) 36.9°
(D) 36.1°
(E) 30.0°
48. If fis a linear function such that f(7) = 5, f(12) = -6, and f(x) = 23.7, what is the value of x?
(A) 3.2
(B) -1.5
©
(D) 2.4
(E) 3.1
xix=y)
49. Under which of the following conditions is » negative?
(A) xo
atest 4
50, The binary operation * is defined over the set of real numbers to be
oLit ach
t . What is the value of 2* (5 * 3)?
(A) 1.84
(B) 2.14
(©) 2.79
(D) 3.65
(E) 4.01
If there is still time remaining, you may review your answers, Bo Umea E Re TOR RRODD
m>mM>movawooOmUaOM
Answer Key
MODEL TEST 4
b
&
BrPRmOmMmaayoommns ANSWERS EXPLAINED
The following explanations are keyed to the review portions of this book. The number in brackets after
each explanation indicates the appropriate section in the Review of Major Topics (Part 2). If a
problem can be solved using algebraic techniques alone, [algebra] appears after the explanation, and
no reference is given for that problem in the Self-Evaluation Chart at the end of the test.
‘An asterisk appears next to those solutions in which a graphing calculator is necessary.
1. (D) Since inverse functions are symmetric about the line y = x, if point (a,b) lies on f, point (b,a)
must lie on f~1. [1.1]
70+ B04 _ g
2. * (A) Average ~ 3 Therefore, x = 85. [4.1]
3. * (C) Plot the graphy = (x? + 3x + 2)/(x + 2)tan(ex) in a [-2.5,2.5] by [-5,5] window and use
TRACE to approximate the location of asymptotes. The only answer choice that could be an
1
asymptote occurs at *~ 2.
An alternative solution is to factor the numerator of f(x) and observe that the factor x + 2
+204)
(x = E20)
divides out; / x42
tan nx=(& + 1) tan nx. The only asymptotes occur because of tan
2. (1.2, 1.3]
z x
nx, when nx is a multiple of 2. Setting ™*3 yields
4, (B) The bases are the same, so the log of a product equals the sum of the logs. [1.4]
5. * (A) Plot the graph of y = 3x3 + 4x? — 4x in the standard window and use CALC/zero to find the
three zeros of this function. Sum these three values to get the correct answer choice.
An alternative solution is first to observe that x factors out, so that x = 0 is one zero.
b_4
The other factor is a quadratic, so the sum of its zeros is “7
wef! Ve
a“ Therefore, “"/a)"°. [1.1]
1 1
6. (A) =" and a
1. (© By the definition of f, log(x + 1) = 3. Therefore, x + 1 = 108 = 1000 and x = 999. [1.4]
8. (B) The f(2 — x) just shifts and reflects the graph horizontally; it does not have any vertical effect on
the graph. Therefore, regardless of what is substituted for x, f(x) > 0. [2.1]
9. (A) Only 3 of the numbers can be used in the thousands place, 3 are left for the hundreds place, 2
for the tens place, and only one for the units place. 3 - 3 - 2 - 1 = 18. [3.1]
10. (B) Substituting 1 for x gives 1+ a—4 =0, and so a = 3. [1.2] 11. * (D) The first 5 flips have no effect on the next 5 flips, so the problem becomes “What is the
(3]= 10
probability of getting exactly 3 heads in 5 flips of a.coin?” I=" outcomes contain 3 heads
3
out of a total of 25 = 32 possible outcomes. P(3H) = 16” 0.3125. [4.2]
12, (BY £4" = 1; (4041 af GANZ Sg, A043 5 4914 = | AI 4) = (11), [3.2]
3 5 gua ot 4
13. (B) Since ¥* = ? tog, V3 =bog,3!? = log 3 =355 14 4)
14, (B) The 23° is superfluous because g(x) = f(sin x) + f(cos x) = 4 sin’x + 4 cos*x = 4(sin?x + cos?x)
=4(1) = 4. [1.3] —
15. * (D) This is a tricky problem. If you just plot the graph of y = (x - V2)(x2 - “5° +7), you will see
only one real zero, at approximately 1.414(~ V2). This is because the zeros of the quadratic
factor are imaginary. Since, however, they are imaginary conjugates, their sum is real—
namely twice the real part. Therefore, graphing the function, using CALC/zero to find the
zeros, and summing them will give you the wrong answer.
To get the, correct answer, you must use the fact that the sum of the zeros of the quadratic
factor is ~a~’* ~'7™?, Since the zero of the linear factor is ¥2 ~ 1.414, the sum of the zeros is
about 1.732 + 1.414 = 3.15. [1.2]
16. (D) Graph I consists of the lines x = 0 and y = 0, which are the coordinate axes and are therefore
perpendicular, Graph IT consists of y = |x| and y = —{x|, which are at +45° to the coordinate
axes and are therefore perpendicular. Graph III consists of the hyperbolas xy = 1 and xy = -1.
Therefore, the correct answer choice is D.
There are two reasons why a graphing calculator solution is not recommended here. One is
that equations, not functions, are given, and solving these equations so that they can be
graphed involves two branches each. The other reason is that even with graphs, you would
have to make judgments about perpendicularity. At a minimum, this would require you to
graph the equations in a square window. [1.6]
12. (B) Points 6 in. from m form a cylinder, with mas axis, which is tangent to plane X. Points 1 in.
from X are two planes parallel to X, one above and one below X. The cylinder intersects only
one of the planes in two lines parallel to m. [2.2]
18. (E) Since the y-coordinate of the point V could be at any height, the slope of VO could be any
value. [2.1]
19. * (C) Volume of cylinder = mr2h = 367 = volume of cube = s3, Therefore, s = 2367 = 4.84, [2.2]
20. (C) Substitute 2t— 2 for x. [1.6]
21, * (D) y, = 23? + 2a + V2b + c and y) = (2)? + 2a — 2V2b +c. So, yy — yp = (232 + 232) + 2V2b =
3. Therefore, 5.65685 + 2.828b ~3 and” 20
22. * (E) The probability that the first car selected is compact is 77. There are 26 cars left, of which 19
(20), (18)
are compact. The probability that the second car is also compact is (3): (e)-om [4.2]
I
Ins
23, (D) Here, a could be either positive or negative. However, b must be negative. [algebra]
Na
iS
* (© Plot the graph of y = abs(x — int(x) ~ 1/2) in an xef~5,5] and ye{-2,2] window and observe
that the maximum value is 3.
An alternative solution is to sketch a portion of the graph by hand and observe the maximum
value. [1.6]
- * (D) Plot the graph y = (x? — 8)(x? — 4) in the standard window. Using CALC/value, observe that y
is not defined when x = 2. Therefore, enter 1.999 for x and observe that y is approximately
equal to 3.
no
a
An alternative solution is to factor the numerator and denominator, divide out x — 2, and
substitute the limiting value 2 into the resulting expression:
[1.5]
The volume of the cone is “ “°®, and the volume of the sphere is
i
”=7*') The desired ratio is V,:V,= 0.256 : 1. [2.2]
27. * (C) Enter 1 into your calculator. Then enter ¥2Ans three times, to accomplish three iterations that
result in x3 to get the correct answer choice E,
An alternative solution is to use the formula to evaluate x;, x2, and Xs, in turn. [3.4]
28. * (D) With your calculator in radian mode, plot the graph of y=abs(v2 (1/sin(x + 1/5))) in an xe[—
1,1] and ye[0,5] window. Use VALUE/X = 0 to determine that the y-intercept is
approximately 1.49. [1.3] 29. (D) The equation of the circle is (x -— 4)? + (y + 8)? = r?. Multiplying out indicates that a = -8 and b
= 16, and soa + b=8. [2.1]
30. (E) Since p(a) = 2, 3a” + 9a + 5 = 0. Solve by using the Quadratic Formula to get the correct
answer choice.
31. * (A) Since i is a root of the equation, so is i, so there are 2 real roots. Plot the graph of y = x4 +
2x3 — 3x? + 2x — 4 in the standard window. Use CALC/zero to find one of the roots, and store
the answer (X) in A. Then use CALC/zero again to find the other root, and multiply A and X
to get the correct answer choice. [3.2]
32. * (E) First find sin-!(3/5) = 36.87°, the reference angle for angle A. Since A is in the second
quadrant, A = 180° — 36.87° = 143.13°. Store 180-Ans in A. Similarly, find cos~1(1/3) =
70.53°, the reference angle for angle B. Since B is in the fourth quadrant, store 360-Ans in B.
Then evaluate sin(A + B) to get the correct answer choice. [1.3]
33, * (A) The probability that at least one is a boy is (1 — the probability that all 3 are girls) or 1 —
(0.5)? = 0.875. [4.2]
y “costa. Therefore, csc(2tan-! x) = 1/sin(2 tan“!(1/cos 1.4) =
34, * (E) Since secl.4 =
3.03. [1.3]
35. * (B) The equation ly — 1| = |x + 1| defines two functions: y=*+!!+1, Plot these graphs in the
standard window and observe that they intersect at (-1,1).
An alternative solution is to recall that the important point of an absolute value occurs when
the expression within the absolute value sign equals zero. The important point of this
absolute value problem occurs when y —1=0 and x + 1=0, ie., at (-1,1). [1.6]
36. * (E) With your calculator in degree mode, plot the graphs of y = cos2x andy = 2 cos x in an
xe[0,360] and ye[-2,2] window. Use CALC/intersect to find the correct answer choice E.
[1.3]
®
=(5) ana Therefore,
3 . . o
37. (E) Since sin@ has a maximum at
[1.3]
38, * (E) Enter LIST/MATH/sum(LIST/OPS/seq(X,X,1,51)) to compute the desired sum.
An alternative solution is to observe that the sequence is arithmetic with ¢; = 1 andd
Using the formula for the sum of the firstn terms of an arithmetic sequence,
. st
+50:1) =51-26 = 1326
2 [3.4]
5S
5 39. * (C) Complete the square in the first equation to get 3(x — 1)? + 4(y + 1)? =
_[B=30-pF 2!
equation for y yields” 4 . Solving for y in the second equation, ”* . Plot
the graphs of these three equations in the standard window to see that the graphs intersect in
two places.
2. Solving this
An alternative solution can be found by completing the square in the first equation and
(i=? (v4
dividing by 12 to get the standard form equation of an ellipse, 4 > 3
The second equation is the standard form of a parabola. Sketch these two equations and
observe the number of points of intersection. [2.1]
40, * (D) Let y = log, 2 = log, x. Converting to exponential form gives x” = 2 and 3” = x. Substitute to
182. peane
get 3” = 2, which can be converted into” ~ oz3” ”. Thus, y = +0.7943. Therefore, 3°-7943
=x = 2.393 or 3-0-7949 = x 0.4178. Therefore, the sum of two x’s is 2.81. [1.4]
(2x+2)(x-5)
41, (D) Factor the numerator and denominator: (*+(«-5), Since the x— 5 divide out, the only vertical
asymptote is atx = —1. Since the degree of the numerator and denominator are equal, y
approaches 3 as x approaches +, so y = 3 is a horizontal asymptote. [1.5]
\
42, * (C) With your calculator in degree mode, plot the graphs of y = 1/2 andy = (3 sin(2x))/(1 -
cos(2x)) in the window [0,180] by [-2,2]. Use CALC/intersect to find the one point of
intersection in the specified interval, at 80.5°.
An alternative solution is to cross-multiply the original equation and use the double angle
formulas for sine and cosine, to get
6sin20 = 1 — cos26
12 singcos@ = 1 — (1 — 2sin*9) = 2sin26
6sin@cosé — sin? @
sin9(6cos@ — sing) = 0
Therefore, sin = 0 or tan@ = 6. It follows that @ = 0°,180°, or 80.5°. The first two solutions
make the denominator of the original equation equal to zero, so 80.5° is the only solution.
[1.3]
ts
43. (C) Backsolve until you get f(g(x),9(x)) = 2(2*)? — (2*)? = (2%)? = 2?*. [1.1]
44, (E) From the geometric sequence ”~“\+) From the arithmetic sequence, b — a = 12 - b, or 2b —a =
°)
12, Substituting gives "| +)
Solving gives a = 6 or —4. Eliminate -4 since a is given to be positive. Substituting the 6
gives 2b — 6 = 12, giving b = 9. Therefore, b - a = 3. [3.4] 24 gala Las
45. * (E) Since s = r6, then 2.4 =r, which implies that "~ ¢° “~>’* and so '*3~ >" which implies
2 286 (24) _ 286
that" ~"e Therefore,(@] ~ “@ which implies that 2.42 = 28.60. Therefore,
7867 020148 = IL * [1.3]
46, * (D) Complete the square and put the equation of the ellipse in standard form:
36x74 4 144%—S0y-731
The center of the ellipse is at (2,1), with a? = 36 and b? = 25, and the major axis is parallel
to the y-axis. Each focus is c= va? -6* units above and below the center.
Therefore, the y-coordinates of the foci are i= il ~4.32 and -2.32. [2.1]
47. * (B) Because you are bisecting one side and trisecting another side, it is convenient to let the
length of the sides be a number divisible by both 2 and 3. Let AB = AD = 6. Thus BM = 2,
MC = 4, and GN = ND = 3. Let ZNAD = x, so that, using right triangle NAD, tan*~ 6~ °°,
which implies that x
an“! 0,5 = 26.6°, Let ZMAB = y, so that, using right triangle MAB, tan
8 ‘Therefore, @ ~ 90°— 26.6° — 18.4° ~ 45°. [1.3]
6, which implies that ”
48. * (B) The slope of the line is 12-7 ~ s. An equation of the line is therefore’ ~*
Substitute 23.7 for y and solve for x to get x =-1.5. [1.2]
49, (A) You must check each answer choice, one at a time. In A, x < 0,y < 0,x —y <0, so the
expression is negative. In B, x <0, y <0, x—y > 0, so the expression is positive. At this point
you know that the correct answer choice must be A. [algebra]
3
. * (B) Put your calculator in radian mode: 5 * 3 =°""5~
g
B23 cos 2.8%
2*(5*3)=2* 2,823 = 2 [1A]
