Gre Math Subject Worksheet Page 4
ADVERTISEMENT
1
2
3
4
5
6t
t
t
43. B (Linear Algebra) Since (6, 7, 8)
=
(0, 1, 2)
+ 2(3, 4, 5)
, we see that
t
t
t
t
t
t
A(6, 7, 8)
=
A(0, 1, 2)
+ 2A(3, 4, 5)
=
(1, 0, 0)
+ 2(0, 1, 0)
= ( 1, 2, 0)
.
′
44. D (Calculus) Use logarithmic differentiation to compute f
(x); one finds that
′
(x) = f (x) ·
1
f
(1 + ln x). Although f (x) > 0, note that 1 + ln x < 0 (and thus
2
′
(x) < 0) for x ∈ (0, e
1
f
).
45. A (Calculus) When in doubt, check the units of the formulas.
46. C (Linear Algebra) Compute the eigenvalues of the matrix to be λ = cos t ± i sin t.
π
Since we want their sum to equal 1, this yields t =
.
3
47. E (Probability) Sketch the region inside the unit square [0, 1] × [0, 1]; the restriction
|x
·
·
1
1
1
1
1
y| <
removes triangles with area
=
from the upper left and lower
2
2
2
2
8
2 ·
1
3
right corners of the square. So, the probability (area) equals 1
=
.
8
4
48. A (Calculus) See where the boundary curves map. Letting x = 0, we see that
u = y, v = 1 + y, and thus v = u + 1. Letting x = 1, we see that u = y = v. Letting
y = 0, we see that v = 1. Letting y = 1, we see that v = 2.
49. E (Calculus) These are all true; for I and III, use a u-substitutions (u = x
3 and
u = 3x, respectively) to rewrite the right sides of each equality (to be) as their left
sides. As for II, this is evident if you think of areas (or simply rewrite it as
∫
∫
∫
3
b
3
f =
f +
f .
a
a
b
50. D (Calculus) There are four such functions: f (x) = ±x, ±|x|. That there are no
others follows from f having to be continuous.
51. D (Calculus) Using the Ratio Test, we need r = |2x + y| < 1.
52. D (Linear Algebra) Being homogeneous, (0, 0, 0) is always a solution. Via row
1 0
1
. So, the system has one solution or
reduction, this system reduces to
0 1
2
0 0 b
5
infinitely many solutions whether b ̸ = 5.
53. D (Complex Analysis) Note that z = 1 is the only singularity of the integrand
inside C. So, rewriting this to apply Cauchy’s Integral Formula, we obtain
∫
2
dz/(z+3)
= 2πi ·
1
πi
=
. (Alternately, you can use the Residue Theorem,
2
C
z 1
(z+3)
8
z=1
1
noting that the integrand has residue
at the simple pole z = 1.)
16
2
54. A (Differential Equations) Since V = 10
h = 100h, we obtain the differential
′
0.25t
equation V
+ 0.25V = 1, which has solution V = Ce
+ 400. Regardless of the
value of C, we conclude that lim
V (t) = 400.
t→∞
′′
′
55. D (Calculus) Sketch the graphs of f
, f
, and then f in this order. Alternately,
thinking of these graphs as acceleration, velocity, and position may be useful.
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education