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(A)
(B)
(C)
200 + 100 = 300 Wrong
(D)
200 – 200 = 0 Correct!
(E)
2(100) – 100 = 100 Wrong
8. The correct answer is (C). [0] Since sin 
θ = 
1
2
:
And:
Since ABC is a right triangle:
AC
2
= AB
2
+ BC
2
Substituting 
AC
2
for BC:
AC
AB
AC
AC
AB
AC
AC
AC
AB
AC
AB
AC
AB
AB
AC
2
2
2
2
2
2
2
2
2
2
2
2
4
4
3
4
3
2
3
2
=
+




=
+
−
=
=
=
=
Since 
AB
AC
is sin 
φ, sin φ = 

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Mathematics Level IC/IIC Subject Tests
237
ARCO
■ SAT II Subject Tests
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9. The correct answer is (A). [–] Sometimes, one picture truly is worth a thousand words:
10. The correct answer is (A). [–] If x is an integer, then the expression |3x – 2| will have its minimum
value when x = 1:
|3(1) – 2| = |3 – 2| = |1| = 1
But 1 = 1. Therefore, there are no integral solutions to the inequality given.
11. The correct answer is (C). [0] One way of attacking this problem is to substitute the suggested
equivalences into the function. For choice (C), the result is:
x
2
– xy
2
+ y
2
= (–x)
2
– (–x)( –y) + (–y)
2
= x
2
– xy + y
2
You can reach the same conclusion by examining the structure of the function in light of the answer
choices. Since the first and third terms have variables of the second power, the sign before the substi-
tuted variable is irrelevant. The only question is whether the middle term—the xy term—will have the
same sign. Since (–x)( –y) = xy, (C) is necessarily true.
Finally, you can solve this problem very easily just by assuming some values for x and y—say 2 and
2. On that assumption, the function yields:
(2)
2
– (2)(2) + (2)
2
= 4
The correct answer choice, therefore, will also generate the value 4:
(A)
(2)
2
– (2)( –2) + (–2)
2
= 12 Wrong.
(B)
(–2)
2
– (–2)(2) + (2)
2
= 12 Wrong.
(C)
(–2)
2
– (–2)( –2) + (–2)
2
= 4 Correct!
(D)
(2)
2
– (2)(
1
2
) + (
1
2
)
2
= 
13
4
Wrong.
(E)
(2)
2
– (2)( –
1
2
) + (–
1
2
)
2
= 
21
4
Wrong.
12. The correct answer is (B). [0] When x = –2, (x + 2) = 0, and f(x) is undefined.
13. The correct answer is (C). [0] An angle has a measure of one radian if, when its vertex is placed at
the center of a circle, it intercepts an arc equal to the radius of a circle.

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Lesson 8
238
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ARCO
■ SAT II Subject Tests
The full circumference of a circle has a radian measure of 
2
πr
r
= 2
π. Now a simple proportion will
solve our problem: