Question:
Let \(f\) be the function defined by \(\displaystyle f(x)=\sum_{n=1}^\infty\frac{x^n}{n}\) for all \(x\) such that \(-1\lt x\lt 1\). Then \(f'(x)=\)
(A) \(\displaystyle\frac{1}{1-x}\) (B) \(\displaystyle\frac{x}{1-x}\) (C) \(\displaystyle\frac{1}{1+x}\) (D) \(\displaystyle\frac{x}{1+x}\) (E) \(0\)
Answer:
(A)
Answer Key:
We need the following conditions for term-by-term differentiation:
(1) The series converges at some point \(x_0\); and
(2) The series of derivatives uniformly converges on \((-1,1)\).
The first condition is easy to check.
Take \(x_0=0\).
Then \(f(0)=0\), so it converges to \(0\).
To check the second condition, consider
\[\sum_{n=1}^\infty\left(\frac{x^n}{n}\right)'=\sum_{n=1}^\infty\frac{nx^{n-1}}{n}=\sum_{n=1}^\infty x^{n-1}.\]
This is a geometric series with initial term \(1\) and ratio \(x\), so
\[\sum_{n=1}^\infty x^{n-1}=\frac{1}{1-x}.\]
Note on the sum of geometric series:
Let \(\displaystyle S=\sum_{k=1}^n ar^{k-1} = a + ar + ar^2 + \cdots+ar^{n-1}\).
Then \(\displaystyle rS=\sum_{k=1}^n ar^k = ar +ar^2+ar^3+\cdots+ar^n\).
Subtract the second equation from the first equation:
\[\begin{align}
S-rS &= a-ar^n\\
(1-r)S&=a(1-r^n)\\
S&=\frac{a(1-r^n)}{1-r}.
\end{align}\]
If \(|r|\lt1\), let \(n\rightarrow\infty\) to obtain \(\displaystyle S=\frac{a}{1-r}\).
Thursday, August 13, 2015
Wednesday, August 12, 2015
FORM GR1268 Question 17
Question:
Which of the following equations has the greatest number of real solutions?
(A) \(x^3=10-x\)
(B) \(x^2+5x-7=x+8\)
(C) \(7x+5=1-3x\)
(D) \(e^x=x\)
(E) \(\sec x=e^{-x^2}\)
Answer:
(B)
Answer Key:
Graph left-hand-side and right-hand-side functions.
(A) has one intersection.
(B) has two intersections.
(C) has one intersection.
(D) has zero intersection.
(E) has one intersection.
The only tricky one is (E).
Because \(\displaystyle\sec x = \frac{1}{\cos x}\), we have \(\displaystyle\sec(0)=\frac{1}{\cos(0)}=1\).
Because the range of \(\cos x\) is \([-1,1]\), the range of \(\sec x\) is \((-\infty, -1]\cup[1, \infty)\).
\(e^{-x^2}\) is like \(e^{-x}\), so it decays as \(x\) increases.
It also decays as \(x\) decreases because of the squaring.
It takes maximum at \(x=0\) and \(e^{-0^2}=1\).
Thus, the range of \(e^{-x^2}\) is \((0,1]\).
Since it takes the value of \(1\) only at \(x=0\), there is only one intersection with \(\sec x\).
Which of the following equations has the greatest number of real solutions?
(A) \(x^3=10-x\)
(B) \(x^2+5x-7=x+8\)
(C) \(7x+5=1-3x\)
(D) \(e^x=x\)
(E) \(\sec x=e^{-x^2}\)
Answer:
(B)
Answer Key:
Graph left-hand-side and right-hand-side functions.
(A) has one intersection.
(B) has two intersections.
(C) has one intersection.
(D) has zero intersection.
(E) has one intersection.
The only tricky one is (E).
Because \(\displaystyle\sec x = \frac{1}{\cos x}\), we have \(\displaystyle\sec(0)=\frac{1}{\cos(0)}=1\).
Because the range of \(\cos x\) is \([-1,1]\), the range of \(\sec x\) is \((-\infty, -1]\cup[1, \infty)\).
\(e^{-x^2}\) is like \(e^{-x}\), so it decays as \(x\) increases.
It also decays as \(x\) decreases because of the squaring.
It takes maximum at \(x=0\) and \(e^{-0^2}=1\).
Thus, the range of \(e^{-x^2}\) is \((0,1]\).
Since it takes the value of \(1\) only at \(x=0\), there is only one intersection with \(\sec x\).
Tuesday, August 11, 2015
FORM GR1268 Question 16
Question:
Suppose \(A, B,\) and \(C\) are statements such that \(C\) is true if exactly one of \(A\) and \(B\) is true. If \(C\) is false, which of the following statements must be true?
(A) If \(A\) is true, then \(B\) is false.
(B) If \(A\) is false, then \(B\) is false.
(C) If \(A\) is false, then \(B\) is true.
(D) Both \(A\) and \(B\) are true.
(E) Both \(A\) and \(B\) are false.
Answer:
(B)
Answer Key:
There are four scenarios:
\(A\) is true and \(B\) is true \(\Rightarrow\) \(C\) is false.
\(A\) is true and \(B\) is false \(\Rightarrow\) \(C\) is true.
\(A\) is false and \(B\) is true \(\Rightarrow\) \(C\) is true.
\(A\) is false and \(B\) is false \(\Rightarrow\) \(C\) is false.
If \(C\) is false, then either (i) both \(A\) and \(B\) are true; or (ii) both \(A\) and \(B\) are false.
Thus, if one is true, the other must be true.
If one is false, the other must be false.
(D) is not necessarily true because both (A) and (B) can be false.
(E) is not necessarily true because both (A) and (B) can be true.
Suppose \(A, B,\) and \(C\) are statements such that \(C\) is true if exactly one of \(A\) and \(B\) is true. If \(C\) is false, which of the following statements must be true?
(A) If \(A\) is true, then \(B\) is false.
(B) If \(A\) is false, then \(B\) is false.
(C) If \(A\) is false, then \(B\) is true.
(D) Both \(A\) and \(B\) are true.
(E) Both \(A\) and \(B\) are false.
Answer:
(B)
Answer Key:
There are four scenarios:
\(A\) is true and \(B\) is true \(\Rightarrow\) \(C\) is false.
\(A\) is true and \(B\) is false \(\Rightarrow\) \(C\) is true.
\(A\) is false and \(B\) is true \(\Rightarrow\) \(C\) is true.
\(A\) is false and \(B\) is false \(\Rightarrow\) \(C\) is false.
If \(C\) is false, then either (i) both \(A\) and \(B\) are true; or (ii) both \(A\) and \(B\) are false.
Thus, if one is true, the other must be true.
If one is false, the other must be false.
(D) is not necessarily true because both (A) and (B) can be false.
(E) is not necessarily true because both (A) and (B) can be true.
Monday, August 10, 2015
FORM GR1268 Question 15
Question:
Let \(S,T,\) and \(U\) be nonempty sets, and let \(f:S\rightarrow T\) and \(g:T\rightarrow U\) be functions such that the function \(g\circ f:S\rightarrow U\) is one-to-one (injective). Which of the following must be true?
(A) \(f\) is one-to-one.
(B) \(f\) is onto.
(C) \(g\) is one-to-one.
(D) \(g\) is onto.
(E) \(g\circ f\) is onto.
Answer:
(A)
Answer Key:
"One-to-one" means no two elements in the domain maps to the same element in the target.
"Onto" means every element in the target is "hit" by at least one element in the domain.
If \(f\) is not one-to-one, more than one elements in \(S\) map to the same element in \(T\) (through \(f\)), and therefore, they necessarily have to map to the same element in \(U\) (through \(g\)).
Then \(g\circ f\) cannot be one-to-one.
Below, I provide a counterexample for (B) through (E).
Let
\[\begin{align}
S &= \{1, 2\}\\
T &= \{a, b, c\}\\
U &= \{x, y, z\}.
\end{align}\]
Define \(f\) as follows:
\[\begin{align}
f(1)&=a\\
f(2)&=b.
\end{align}\]
Define \(g\) as follows:
\[\begin{align}
g(a)&=x\\
g(b)&=y\\
g(c)&=y.
\end{align}\]
Then \(g\circ f\) is defined as follows:
\[\begin{align}
g\circ f(1)&=g(f(1))=g(a)=x\\
g\circ f(2)&=g(f(2))=g(b)=y.
\end{align}\]
Thus, \(g\circ f\) is one-to-one.
\(f\) is not onto because no element in \(S\) maps to \(c\in T\).
\(g\) is not one-to-one because both \(b\) and \(c\) map to \(y\in U\).
\(g\) is not onto because no element in \(T\) maps to \(z\in U\).
\(g\circ f\) is not onto because no element in \(S\) maps to \(z\in U\).
Let \(S,T,\) and \(U\) be nonempty sets, and let \(f:S\rightarrow T\) and \(g:T\rightarrow U\) be functions such that the function \(g\circ f:S\rightarrow U\) is one-to-one (injective). Which of the following must be true?
(A) \(f\) is one-to-one.
(B) \(f\) is onto.
(C) \(g\) is one-to-one.
(D) \(g\) is onto.
(E) \(g\circ f\) is onto.
Answer:
(A)
Answer Key:
"One-to-one" means no two elements in the domain maps to the same element in the target.
"Onto" means every element in the target is "hit" by at least one element in the domain.
If \(f\) is not one-to-one, more than one elements in \(S\) map to the same element in \(T\) (through \(f\)), and therefore, they necessarily have to map to the same element in \(U\) (through \(g\)).
Then \(g\circ f\) cannot be one-to-one.
Below, I provide a counterexample for (B) through (E).
Let
\[\begin{align}
S &= \{1, 2\}\\
T &= \{a, b, c\}\\
U &= \{x, y, z\}.
\end{align}\]
Define \(f\) as follows:
\[\begin{align}
f(1)&=a\\
f(2)&=b.
\end{align}\]
Define \(g\) as follows:
\[\begin{align}
g(a)&=x\\
g(b)&=y\\
g(c)&=y.
\end{align}\]
Then \(g\circ f\) is defined as follows:
\[\begin{align}
g\circ f(1)&=g(f(1))=g(a)=x\\
g\circ f(2)&=g(f(2))=g(b)=y.
\end{align}\]
Thus, \(g\circ f\) is one-to-one.
\(f\) is not onto because no element in \(S\) maps to \(c\in T\).
\(g\) is not one-to-one because both \(b\) and \(c\) map to \(y\in U\).
\(g\) is not onto because no element in \(T\) maps to \(z\in U\).
\(g\circ f\) is not onto because no element in \(S\) maps to \(z\in U\).
Saturday, August 8, 2015
FORM GR1268 Question 14
Question:
Suppose \(g\) is a continuous real-valued function such that \(\displaystyle3x^5+96=\int_c^x g(t)dt\) for each \(x\in\mathbb R\), where \(c\) is a constant. What is the value of \(c\)?
(A) \(-96\) (B) \(-2\) (C) \(4\) (D) \(15\) (E) \(32\)
Answer:
(B)
Answer Key:
Method 1
Because the equation is true for any real number, it is true for \(x=c\).
Then \(\displaystyle 3c^5+96=\int_c^c g(t)dt=0\).
Thus, \(3c^5=-96\), so \(c^5=-32\), so \(c=(-32)^{1/5}=-2\).
Method 2
Differentiate both sides with respect to \(x\) to obtain \(15x^4=g(x)\).
(See note below if you are not sure about this step.)
Then
\[\begin{align}
3x^5+96&=\int_c^x g(t)dt\\
&= \int_c^x15t^4dt\\
&=\left.3t^5\right|_c^x\\
&=3x^5-3c^5.
\end{align}\]
Thus, \(-3c^5=96\), so \(c^5=-32\), so \(c=(-32)^{1/5}=-2\).
Note on differentiating integral:
Suppose the antiderivative of \(g(t)\) is \(G(t)\).
Then \(\displaystyle\int_c^xg(t)dt=\left.G(t)\right|_c^x=G(x)-G(c)\).
Since \(G(c)\) is a constant, \(\displaystyle\frac{d}{dx}\int_c^xg(t)dt=\frac{d}{dx}[G(x)-G(c)]=g(x)\).
Suppose \(g\) is a continuous real-valued function such that \(\displaystyle3x^5+96=\int_c^x g(t)dt\) for each \(x\in\mathbb R\), where \(c\) is a constant. What is the value of \(c\)?
(A) \(-96\) (B) \(-2\) (C) \(4\) (D) \(15\) (E) \(32\)
Answer:
(B)
Answer Key:
Method 1
Because the equation is true for any real number, it is true for \(x=c\).
Then \(\displaystyle 3c^5+96=\int_c^c g(t)dt=0\).
Thus, \(3c^5=-96\), so \(c^5=-32\), so \(c=(-32)^{1/5}=-2\).
Method 2
Differentiate both sides with respect to \(x\) to obtain \(15x^4=g(x)\).
(See note below if you are not sure about this step.)
Then
\[\begin{align}
3x^5+96&=\int_c^x g(t)dt\\
&= \int_c^x15t^4dt\\
&=\left.3t^5\right|_c^x\\
&=3x^5-3c^5.
\end{align}\]
Thus, \(-3c^5=96\), so \(c^5=-32\), so \(c=(-32)^{1/5}=-2\).
Note on differentiating integral:
Suppose the antiderivative of \(g(t)\) is \(G(t)\).
Then \(\displaystyle\int_c^xg(t)dt=\left.G(t)\right|_c^x=G(x)-G(c)\).
Since \(G(c)\) is a constant, \(\displaystyle\frac{d}{dx}\int_c^xg(t)dt=\frac{d}{dx}[G(x)-G(c)]=g(x)\).
Friday, August 7, 2015
FORM GR1268 Question 13
Question:
If \(f\) is a continuously differentiable real-valued function defined on the open interval \((-1,4)\) such that \(f(3)=5\) and \(f'(x)\ge-1\) for all \(x\), what is the greatest possible value of \(f(0)\)?
(A) \(3\) (B) \(4\) (C) \(5\) (D) \(8\) (E) \(11\)
Answer:
(D)
Answer Key:
Since the question is a asking for the greatest possible value at \(x=0\),
we want the function to be increasing as much as possible from \(x=3\) to \(x=0\).
Equivalently, we want the function to be decreasing as much as possible from \(x=0\) to \(x=3\).
The largest negative slope we can have is \(f'(x)=-1\).
Assuming \(f'(x)=-1\) over the interval \((0,3)\),
\[\begin{align}
f(3)&=f(0)-1(3-0)\\
5&=f(0)-3\\
8&=f(0).
\end{align}\]
If \(f\) is a continuously differentiable real-valued function defined on the open interval \((-1,4)\) such that \(f(3)=5\) and \(f'(x)\ge-1\) for all \(x\), what is the greatest possible value of \(f(0)\)?
(A) \(3\) (B) \(4\) (C) \(5\) (D) \(8\) (E) \(11\)
Answer:
(D)
Answer Key:
Since the question is a asking for the greatest possible value at \(x=0\),
we want the function to be increasing as much as possible from \(x=3\) to \(x=0\).
Equivalently, we want the function to be decreasing as much as possible from \(x=0\) to \(x=3\).
The largest negative slope we can have is \(f'(x)=-1\).
Assuming \(f'(x)=-1\) over the interval \((0,3)\),
\[\begin{align}
f(3)&=f(0)-1(3-0)\\
5&=f(0)-3\\
8&=f(0).
\end{align}\]
Thursday, August 6, 2015
FORM GR1268 Question 12
Question:
For which integers \(n\) such that \(3\le n\le 11\) is there only one group of order \(n\) (up to isomorphism)?
(A) For no such integer \(n\)
(B) For \(3,5,7,\) and \(11\) only
(C) For \(3,5,7,9,\) and \(11\) only
(D) For \(4,6,8,\) and \(10\) only
(E) For all such integers \(n\)
Answer:
(B)
Answer Key:
There is only one group of order \(3\), namely, \(C_3\), a cyclic group of order \(3\).
To see this, consider a set \(\{1,a,b\}\).
A partially filled multiplication table looks like this:
\[\begin{array}{|c|c|c|c|}
\hline
\times &\textbf{1} & \textbf{a} & \textbf{b}\\ \hline
\textbf{1} & 1 & a & b\\ \hline
\textbf{a} & a & ? &\\ \hline
\textbf{b} & b & & \\ \hline
\end{array}\]
Since no row or column can have any element more than once,
it must be that \(a\times a = 1\) or \(a\times a = b\).
If \(a\times a = 1\),then \(a\times b = b\), which implies \(a=1\), which is not the case.
Thus, it must be that \(a\times a = b\), which leaves us with \(a\times b = 1\).
\[\begin{array}{|c|c|c|c|}
\hline
\times &\textbf{1} & \textbf{a} & \textbf{b}\\ \hline
\textbf{1} & 1 & a & b\\ \hline
\textbf{a} & a & b & 1\\ \hline
\textbf{b} & b & & \\ \hline
\end{array}\]
There is only one way to complete the last row:
\[\begin{array}{|c|c|c|c|}
\hline
\times &\textbf{1} & \textbf{a} & \textbf{b}\\ \hline
\textbf{1} & 1 & a & b\\ \hline
\textbf{a} & a & b & 1\\ \hline
\textbf{b} & b & 1 & a\\ \hline
\end{array}\]
This is a cyclic group of order \(3\).
Note that it is isomorphic to \(\mathbb Z/3\mathbb Z\).
So, the correct answer is (A), (B), or (E).
Is there a non-cyclic group of order \(9\)?
The answer is yes.
Consider two cyclic groups \(\{1,a,a^2\}\) and \(\{1,b,b^2\}\).
The Cartesian product forms a new group:
\[\{1,a,a^2\}\times\{1,b,b^2\}=\{1, b, b^2, a, ab, ab^2, a^2, a^2b, a^2b^2\}.\]
The operation is inherited from the original cyclic groups.
For example, \(ab\times a^2b=a^3b^2=1b^2=b^2\).
Convince yourself this set is closed.
Is it a cyclic group?
We know \(1\) is not a generator.
Neither \(a\) nor \(a^2\) is a generator because it cannot generate \(b\).
Neither \(b\) nor \(b^2\) is a generator because it cannot generate \(a\).
Neither \(ab\) nor \(a^2b^2\) is a generator because it cannot generate \(a\).
Neither \(a^2b\) nor \(ab^2\) is a generator because it cannot generate \(a\).
Thus, the group is non-cyclic.
For which integers \(n\) such that \(3\le n\le 11\) is there only one group of order \(n\) (up to isomorphism)?
(A) For no such integer \(n\)
(B) For \(3,5,7,\) and \(11\) only
(C) For \(3,5,7,9,\) and \(11\) only
(D) For \(4,6,8,\) and \(10\) only
(E) For all such integers \(n\)
Answer:
(B)
Answer Key:
There is only one group of order \(3\), namely, \(C_3\), a cyclic group of order \(3\).
To see this, consider a set \(\{1,a,b\}\).
A partially filled multiplication table looks like this:
\[\begin{array}{|c|c|c|c|}
\hline
\times &\textbf{1} & \textbf{a} & \textbf{b}\\ \hline
\textbf{1} & 1 & a & b\\ \hline
\textbf{a} & a & ? &\\ \hline
\textbf{b} & b & & \\ \hline
\end{array}\]
Since no row or column can have any element more than once,
it must be that \(a\times a = 1\) or \(a\times a = b\).
If \(a\times a = 1\),then \(a\times b = b\), which implies \(a=1\), which is not the case.
Thus, it must be that \(a\times a = b\), which leaves us with \(a\times b = 1\).
\[\begin{array}{|c|c|c|c|}
\hline
\times &\textbf{1} & \textbf{a} & \textbf{b}\\ \hline
\textbf{1} & 1 & a & b\\ \hline
\textbf{a} & a & b & 1\\ \hline
\textbf{b} & b & & \\ \hline
\end{array}\]
There is only one way to complete the last row:
\[\begin{array}{|c|c|c|c|}
\hline
\times &\textbf{1} & \textbf{a} & \textbf{b}\\ \hline
\textbf{1} & 1 & a & b\\ \hline
\textbf{a} & a & b & 1\\ \hline
\textbf{b} & b & 1 & a\\ \hline
\end{array}\]
This is a cyclic group of order \(3\).
Note that it is isomorphic to \(\mathbb Z/3\mathbb Z\).
So, the correct answer is (A), (B), or (E).
Is there a non-cyclic group of order \(9\)?
The answer is yes.
Consider two cyclic groups \(\{1,a,a^2\}\) and \(\{1,b,b^2\}\).
The Cartesian product forms a new group:
\[\{1,a,a^2\}\times\{1,b,b^2\}=\{1, b, b^2, a, ab, ab^2, a^2, a^2b, a^2b^2\}.\]
The operation is inherited from the original cyclic groups.
For example, \(ab\times a^2b=a^3b^2=1b^2=b^2\).
Convince yourself this set is closed.
Is it a cyclic group?
We know \(1\) is not a generator.
Neither \(a\) nor \(a^2\) is a generator because it cannot generate \(b\).
Neither \(b\) nor \(b^2\) is a generator because it cannot generate \(a\).
Neither \(ab\) nor \(a^2b^2\) is a generator because it cannot generate \(a\).
Neither \(a^2b\) nor \(ab^2\) is a generator because it cannot generate \(a\).
Thus, the group is non-cyclic.
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