GRE Practice

GR9768.66: Green's Theorem

66. In the xy-plane, if C is the circle $x^2 + y^2 = 9$, oriented counterclockwise, then $\oint \limits_C -2y \; dx + x^2 \; dy =$

(A) $0$

(B) $6\pi$

(D) $12\pi$

(E) $18\pi$.

Solution:
The hypothesis for Green's Theorem is satisfied since $C$ is a positively oriented, piecewise smooth, simple closed curve. So
$I = \oint_C -2y dx + x^2 dy = \oint_C P dx + Q dy = \iint_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} dA$,
where $D$ is the region enclosed by the circle $C$ centered at the origin and radius 3.
$I = \iint_D 2x + 2 dA = \int_0^{2\pi} \int_0^3 2(r \cos \theta + 2) r dr d\theta$.
$I = \int_0^{2\pi} \int_0^3 2r^2 \cos \theta + 2r dr d\theta$
$I = \int_0^{2\pi} \frac{2r^3}{3} \cos \theta + r^2 |_0^3 d\theta$
$I = \int_0^{2\pi} 18 \cos \theta + 9 d\theta$
$I = 18 \sin \theta + 9 \theta \vert_0^{2\pi} = 18 \sin (2\pi) + 9(2\pi) - 18 \sin (0) - 9(0) = 18 \pi$.
The answer is E.

GR9768.65: Polynomial

65. Let p(x) be the polynomial $x^3 + ax^2 + bx + c$, where a, b, c are real constants. If p(-3) = p(2) = 0 and p'(-3) < 0, which of the following is a possible value of c?

(A) -27

(B) -18

(D) -3

(E) -1/2

Solution:
Since $p^`(x) = 3x^2 + 2ax + b$, we have $p^`(-3) = 27 - 6a + b < 0$.
$p(-3) = -27 + 9a -3b + c = 0$ and $p(2) = 8 + 4a + 2b + c = 0$.
From these two equations, we see that $c = 27 - 9a + 3b = -8 - 4a - 2b$.
So $35 = 5a - 5b$, or $7 = a - b$. Thus, $b = a -7$.
But since $27 - 6a + b < 0$, we see that $27 - 6a + a - 7 = 20 -5a < 0$.
So $a < 4$. This implies $b = -7 +a < -7 + 4 = -3$.
Therefore,
$c = 27 - 9a + 3b < 27 - 9(4) + 3(-3) = 27 - 36 - 9 = -18$.
$c = -8 - 4a - 2b < -8 - 4(4) -2(-3) = -8 +4 + 6 = 2$.
Thus, $c$ must be strictly less than $-18$ so the possible value of $c$ is $-27$.
The answer is A.
-sg-

GR9768.64: Continous Functions

64. Suppose that f is a continuous real-valued function defined on the closed interval [0,1]. Which of the following must be true?

I. There is a constant C > 0 such that | f(x) - f(y) | ≤ C for all x and y in [0,1].
II. There is a constant D > 0 such that | f(x) - f(y) | ≤ 1 for all x and y in [0,1] that satisfy |x-y| ≤ D.
III. There is a constant E > 0 such that | f(x) - f(y) | ≤ E |x-y| for all x and y in [0,1].

(A) I only

(B) III only

(D) II and III only

(E) I, II, and III

Solution:
The definition of continuity: $f$ is continuous at $x = a$ if for all $\epsilon > 0$, there exists $\delta > 0$ such that $0 < |x-a| < \delta $ implies $|f(x) - f(a)| < \epsilon$.
Since $f$ is continuous on a closed interval $[0,1]$, we see that $f$ is also uniformly continuous on $[0,1]$.
The definition of uniformly continuous: $f$ is uniformly continuous on $E = [a,b]$ if given $\epsilon > 0$ and $x, y \in E$, there exists $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ whenever $|x-y| < \delta$.
I is true, with $C = \epsilon > 0$.
II must be true, with $\delta = D > 0$ and $\epsilon = 1$.
III is actually false. Yes, if we have this, then it implies $f$ is uniformly continuous. But what we have is actually the opposite. And we know the converse of a true statement is not necessarily still true. As for a counter example, take $f(x) = x^n$ for really large $n$.
The answer is C.
-sg-

GR9768.63: Intersections

63. At how many points in the xy-plane do the graphs of $y = x^{12}$ and $y=2^x$ intersect?

(A) None

(B) One

(D) Three

(E) Four

Solution:
We know what the graph of $2^x$ looks like: always positive, the $y$-intercept is at $(0,1)$. For $y = x^{12}$, it passes through the vertex $(0,0)$ and it concaves up. There are at most 2 intersections. Now, the two graph intersect exactly once on $(-\infty, 0)$. On the other half, $[0, \infty)$, we'll look at $f(x) = 2^x - x^{12}$ and note that $f(1) = 2^1 - 1^{12} = 1 > 0$, where as $f(2) = 2^2 - 2^{12} < 0$. Thus, by the Rolle's Theorem, we see that $f$ has a zero on $(1,2)$. Thus, $x^{12}$ intersects $2^x$ on this interval. So there are two points of intersection.
The answer is C.
-sg-

Note: the answer provided in the Answer Sheet is D, 3 points of intersection.
result from wolframalpha

GR9768.62: Algebraic Expansion

62. The coefficient of $x^3$ in the expansion of $(1+x)^3(2+x^2)^{10}$ is

(A) $2^{14}$

(B) $31$

(D) $3 \choose 3$ + $2{10\choose 1}$

(E) $3\choose 3$$10\choose 1$$2^9$

Solution:
First, observe that $(1+x)^3 = 1 + 3x + 3x^2 + x^3$.
And $(2+x^2)^{10} = {10\choose0}2^{10}(x^2)^0 + {10\choose1}2^{9} (x^2)^1 + {10\choose2} 2^8 (x^2)^2 \ldots$
$= {10\choose0}2^{10} + {10\choose1}2^{9} x^2 + {10\choose2} 2^8 x^4 \ldots$
$= 2^{10} + 10 \cdot 2^9 x^2 + 45 \cdot 2^8 x^4 + \dots$
Multiplying the two, then collect the terms with $x^3$:
$x^3 \cdot 2^{10}$, $3x \cdot 10 \cdot 2^9 x^2$.
The coefficient is:
$2^{10} + 30 \cdot 2^9 = 2^{10}(1 + 15) = 2^{10} \cdot 16 = 2^{10} \cdot 2^4 = 2^{14}$.
The answer is A.
-sg-

GR9768.61: Greatest Integer, Prime

61. What is the greatest integer that divides $p^4 -1$ for every prime number p greater than 5?

(A) $12$

(B) $30$

(D) $120$

(E) $240$

Solution:
$p^4 -1 = (p-1)(p^3 + p^2 + p + 1)$.
And we see that $(p-1)$ and $(p^3 + p^2 + p + 1)$ are the divisors of $p^4 -1$.
Guess and check:
$p = 13$ works so A is a divisor.
$p = 31$ works so B is a divisor.
$\vdots$
$p = 241$ works so E is a divisor.
And $p-1 = 240$ is the greatest integer provided.
The answer is E.
-sg-

GR9768.60: Ring

60. If S is a ring with the property that $s = s^2$ for each $s \in S$, which of the following must be true?

I. $s+s = 0$ for each $s \in S$.
II. $(s+t)^2 = s^2 + t^2$ for each $s, t \in S$.
III. $S$ is commutative.

(A) III only.

(B) I and II only.

(D) II and III only.

(E) I, II, and III.

Solution:
Consider the ring $\mathbb{Z}_2 = \{0,1\}$. And observe that $0 = 0^2$ and $1 = 1^2$.
Then
$0+0 = 0$ and $1 + 1 = 0$ so I is true.
$(0+1)^2 = 1^2 = 1$ and $0^2 + 1^2 = 1$ so II is true.
$0 = 0 \cdot 1 = 1 \cdot 0 = 0$ so III is true.
The answer is E.
-sg-

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Sunday, May 23, 2010