Gre Math Subject Worksheet Page 2
ADVERTISEMENT
1
2
3
4
5
614. C (Logic) Careful with the qualifiers!
15. B (Algebra) Since g(x) ∈ (a, x) and x > a, we see that g can not be constant.
(Otherwise, use counterexamples to eliminate false answers.)
16. D (Linear Algebra) We solve A(0, 1, 1, 1) + B(0, 0, 0, 1) + C(1, 1, 2, 0) = (1, 2, m, 5)
for A, B, C, D, m. Equating like entries, we find that A = C = 1, and thus m = 3.
17. E (Discrete Math) Write out the difference table, working downward and to the
right, as needed. First, deduce that f (2) = 3 and f (3) = 1 in succession from the ∆f
2
column. Next, we find that ∆
f (1) =
6 and ∆f (3) = 4. Finally, f (4) = 5.
18. E (Calculus) Draw the radii and label their lengths. We find that
2
2
A(r)
π(1
r
)
. Letting r → 1 yields ∞ as the limit.
1+r
=
=
2
a(r)
π(1 r)
1 r
19. B (Abstract Algebra) Note that each row and column should have a permutation of
all elements of the group; this eliminates III. As for why II is invalid, note that bc = d
2
2
2
2
2
yields (bc)
= d
, but (bc)
= b(cb)c = bdc = (bd)c = c
= a, but d
= b. (Observe
that I is a multiplication table for the cyclic group of order 4.)
20. D (Calculus) I is true by the definition of the derivative, II is not true (try
f (x)
· x (and apply the product rule for limits).
f (x) = x), and III is true since f (x) =
x
21. D (Calculus) First of all, we compute the equation of the tangent line at x = 0 to
x
be y =
+ 1. Sketching the region now gives a right triangle with base and height
2
· 2 · 1 = 1.
1
having lengths 2 and 1; so the area equals
2
22. B (Abstract Algebra) Note that the subset in (B) is not closed under inverses.
23. C (Geometry) Draw the figure, and note that ∆OBA and ∆COA are equilateral
◦
triangles. Hence, ∠BAC = 120
.
24. C (Linear Algebra) Let (a, b, c, d) be orthogonal to both vectors. Then, the
corresponding dot products equal 0: b + c + d = 0 and a + b + c = 0. Solving this
system yields the desired answer.
25. D (Linear Programming) The critical points occur at the vertices of the region;
1
1
these are (
,
), (1, 0), (2, 0), and (2, 2). Substituting these into f , we find that the
2
2
1
maximum is 10 (and the minimum is
).
2
26. B (Calculus) This is immediately seen from sketching the graph.
′
′
27. E (Calculus) Differentiating the relation yields f
(x) =
f
(1
x). Now, let x = 0.
∩ V
28. C (Linear Algebra) Use dim(V
+ V
) = dim V
+ dim V
dim(V
). This yields
1
2
1
2
1
2
∩ V
) ≤ 10, and so dim(V
) ≥ 2.
6 + 6
dim(V
+ V
1
2
1
2
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education