FORM GR1268 Question 32

Question:
\(\displaystyle\frac{d}{dx}\int_{x^3}^{x^4}d^{t^2}dt=\)

(A) \(\displaystyle e^{x^6}\left(e^{x^8-x^6}-1\right)\)

(B) \(\displaystyle 4x^3e^{x^8}\)

(C) \(\displaystyle \frac{1}{\sqrt{1-e^{x^2}}}\)

(D) \(\displaystyle \frac{e^{x^2}}{x^2}-1\)

(E) \(\displaystyle x^2e^{x^6}\left(4xe^{x^8-x^6}-3\right)\)

Answer:
(E)

Answer Key:
Let \(f(t)=e^{t^2}\) and \(F'(t)=f(t)\).
\[\begin{align}
\frac{d}{dx}\int_{x^3}^{x^4}f(t)dt
&= \frac{d}{dx}\left.F(t)\right|_{x^3}^{x^4}\\
&=\frac{d}{dx}\left(F(x^4)-F(x^3)\right)\\
&=\frac{d}{dx}F(x^4)-\frac{d}{dx}F(x^3)\\
&=\frac{d}{dx^4}F(x^4)\cdot\frac{d}{dx}x^4-\frac{d}{dx^3}F(x^3)\cdot\frac{d}{dx}x^3\\
&=f(x^4)4x^3-f(x^3)3x^2\\
&=4x^3e^{(x^4)^2}-3x^2e^{(x^3)^2}\\
&=4x^3e^{x^8}-3x^2e^{x^6}\\
&=x^2e^{x^6}\left(4xe^{x^8-x^6}-3\right).
\end{align}\]

FORM GR1268 Question 31

Question:
Of the numbers \(2,3,\) and \(5\), which are eigenvalues of the matrix \(\displaystyle\begin{pmatrix}3 & 5 & 3\\ 1 & 7 & 3\\ 1 & 2 & 8\end{pmatrix}\)?

(A) None     (B) \(2\) and \(3\) only     (C) \(2\) and \(5\) only     (D) \(3\) and \(5\) only     (E) \(2,3,\) and \(5\)

Answer:
(C)

Answer Key:
If \(\lambda\) is an eigenvalue of a square matrix \(\mathbf A\), then \(\mathbf{Av}=\lambda \mathbf v\) for some nonzero vector \(\mathbf v\).
Then \(\mathbf{Av}=\lambda \mathbf{Iv}\), so \((\mathbf A-\lambda \mathbf I)\mathbf v=\mathbf 0\).
If \(\mathbf A-\lambda \mathbf I\) is invertible, then \(\mathbf v=(\mathbf A-\lambda \mathbf I)^{-1}\mathbf 0=\mathbf 0\).
But we want \(\mathbf v\) to be a non-zero vector, so \(\mathbf A-\lambda \mathbf I\) needs to be singular.

\[\begin{align}
\begin{pmatrix}3 & 5 & 3\\ 1 & 7 & 3\\ 1 & 2 & 8\end{pmatrix}-2I&=\begin{pmatrix}1 & 5 & 3\\ 1 & 5 & 3\\ 1 & 2 & 6\end{pmatrix}\quad \text{(singular)}\\
\\
\begin{pmatrix}3 & 5 & 3\\ 1 & 7 & 3\\ 1 & 2 & 8\end{pmatrix}-3I&=\begin{pmatrix}0 & 5 & 3\\ 1 & 4 & 3\\ 1 & 2 & 5\end{pmatrix}\quad \text{(non-singular)}\\
\\
\begin{pmatrix}3 & 5 & 3\\ 1 & 7 & 3\\ 1 & 2 & 8\end{pmatrix}-5I&=\begin{pmatrix}-2 & 5 & 3\\ 1 & 2 & 3\\ 1 & 2 & 3\end{pmatrix}\quad \text{(singular)}\\
\end{align}\]

FORM GR1268 Question 30

Question:
For what positive value of \(c\) does the equation \(\log x=cx^4\) have exactly one real solution for \(x\)?

(A) \(\displaystyle\frac{1}{4e}\)     (B) \(\displaystyle\frac{1}{4e^4}\)     (C) \(\displaystyle\frac{e^4}{4}\)     (D) \(\displaystyle\frac{4}{e^{1/4}}\)     (E) \(\displaystyle 4e^{1/4}\)

Answer:
(A)

Answer Key:
Depicted below are graphs of \(f(x)=\log x\) and \(g(x)=cx^4\) with \(c=1\).

We want to adjust \(c\) so that two curves touch.

Suppose the two curves touch at \(x=x_0\).

Then \(f'(x_0)=g'(x_0)\), so
\[\begin{align}
\frac{1}{x_0}&=4cx_0^3\\
1&=4cx_0^4\\
x_0^4&=\frac{1}{4c}\\
x_0&=\left(\frac{1}{4c}\right)^{\frac{1}{4}}.
\end{align}\]
We also have \(f(x_0)=g(x_0)\). Thus,
\[\begin{align}
\log x_0&=cx_0^4\\
\log\left(\frac{1}{4c}\right)^{\frac{1}{4}}&=c\cdot\frac{1}{4c}\\
\frac{1}{4}\log\frac{1}{4c}&=\frac{1}{4}\\
\log\frac{1}{4c}&=1\\
\frac{1}{4c}&=e\\
4c&=\frac{1}{e}\\
c&=\frac{1}{4e}.
\end{align}\]

FORM GR1268 Question 29

Question:
A tree is a connected graph with no cycles. How many nonisomorphic trees with \(5\) vertices exist?

(A) \(1\)     (B) \(2\)     (C) \(3\)     (D) \(4\)     (E) \(5\)

Answer:
(C)

Answer Key:
A tree with \(5\) vertices have \(5-1=4\) edges.
To see this, starting with a node, attach edge + node one at a time to construct a tree.

The degree of a node is the number of edges attached ("incident") to the node.
Every edge is incident to two nodes.
So, the sum of degrees have to be \(4\times2=8\).

Isomorphic trees have the same degree sequence.
A degree sequence lists the degree of nodes in an descending order.
In other words, if \((d_1,d_2,d_3,d_4,d_5)\) is a degree sequence, \(d_i\ge d_j\) when \(i\lt j\) and \(d_1+d_2+d_3+d_4+d_5=8\).

All possible degree sequences are \((4,1,1,1,1)\), \((3,2,1,1,1)\), and \((2,2,2,1,1)\).
Corresponding trees are shown below.

FORM GR1268 Question 28

Question:
Let \(f\) be a one-to-one (injective), positive-valued function defined on \(\mathbb R\). Assume that \(f\) is differentiable at \(x=1\) and that in the \(xy\)-plane the line \(y-4=3(x-1)\) is tangent to the graph of \(f\) at \(x=1\). Let \(g\) be the function defined by \(g(x)=\sqrt x\) for \(x\ge0\). which of the following is FALSE?

(A) \(f'(1)=3\)

(B) \(\displaystyle\left(f^{-1}\right)^{\prime}(4)=\frac{1}{3}\)

(C) \((fg)'(1)=5\)

(D) \((g\circ f)'(1)=\frac{1}{2}\)

(E) \((g\circ f)(1)=2\)

Answer:
(D)

Answer Key:
The line \(y-4=3(x-1)\) has a constant derivative, \(3\).
Because \(f\) is tangent to the line at \(x=1\), we have \(f'(1)=3\).

The line goes through the point \((x,y)=(1,4)\).
The slope of \(f^{-1}\) is the reciprocal of the slope of \(f\) at that point.
Thus, (B) is true.

For (C), since \(\displaystyle g'(x)=\frac{1}{2\sqrt x}\), we have
\[\begin{align}
(fg)'(1)&=f'(1)g(1)+f(1)g'(1)\\
&=3\cdot\sqrt1+4\cdot\frac{1}{2\sqrt1}\\
&=5.
\end{align}\]

For (D),
\[\begin{align}
(g\circ f)'(1)&=(g(f(x))'(1)\\
&=\frac{dg}{df}(f(1))\frac{df}{dx}(1)\\
&=\frac{1}{2\sqrt{f(1)}}\cdot3\\
&=\frac{1}{2\sqrt4}\cdot 3\\
&=\frac{3}{4}.
\end{align}\]

For (E), \((g\circ f)(1)=g(f(1))=g(4)=\sqrt{4}=2\).

FORM GR1268 Question 27

Question:
\((1+i)^{10}=\)

(A) \(1\)     (B) \(i\)     (C) \(32\)     (D) \(32i\)     (E) \(32(i+1)\)

Answer:
(D)

Answer Key:
\[\begin{align}
(1+i)^{10}&=\left(\sqrt2 e^{\frac{\pi}{4}i}\right)^{10}\\
&=2^{\frac{10}{2}}e^{\frac{10\pi}{4}i}\\
&=32e^{\frac{5\pi}{2}i}\\
&=32\left(\cos\frac{5\pi}{2}+i\sin\frac{5\pi}{2}\right)\\
&=32\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\\
&=32(0+i)\\
&=32i.
\end{align}\]

FORM GR1268 Question 26

Question:
\[\begin{align}
3x&\equiv 5\pmod{11}\\
2y&\equiv 7\pmod{11}
\end{align}\]
If \(x\) and \(y\) are integers that satisfy the congruences above, then \(x+y\) is congruent module \(11\) to which of the following?

(A) \(1\)     (B) \(3\)     (C) \(5\)     (D) \(7\)     (E) \(9\)

Answer:
(D)

Answer Key:
Multiply the first equation by \(2\) to obtain \(6x\equiv10\pmod{11}\).
Multiply the second equation by \(3\) to obtain \(6y\equiv21\equiv10\pmod{11}\).
Add the two equations to obtain \(6(x+y)\equiv20\equiv9\pmod{11}\).

Compare this equation to
\(6\cdot1=6\equiv6\pmod{11}\)
\(6\cdot3=18\equiv7\pmod{11}\)
\(6\cdot5=30\equiv8\pmod{11}\)
\(6\cdot7=42\equiv9\pmod{11}\)
\(6\cdot9=54\equiv10\pmod{11}\)

FORM GR1268 Question 25

Question:

The graph of the derivative \(h'\) is shown above, where \(h\) is a real-valued function. Which of the following open intervals contains a value \(c\) for which the point \((c,h(c))\) is an inflection point of \(h\)?

(A) \((-2,-1)\)     (B) \((-1,0)\)     (C) \((0,1)\)     (D) \((1,2)\)     (E) \((2,3)\)

Answer:
(A)

Answer Key:
An inflection point is where a curve switches from concave to convex, or vice versa.
A curve is concave if \(h^{\prime\prime}\lt0\) and convex if \(h^{\prime\prime}\gt0\).
At an inflection point, \(h^{\prime\prime}=0\).
Thus, we are looking for a point where \(h'\) is flat (the slope is zero).

There is such a point in the interval \((-2,-1)\).
Since \(h^{\prime\prime}\) changes from \(\lt0\) to \(\gt0\), \(h\) changes from being concave to being convex.

Alternatively, you can use integration.
The area between \(h'\) and \(x\)-axis is initially decreasing.
From \(x=-2\) to around \(x=-1.5\), the area decreases at an increasing speed.
From around \(x=-1.5\) to around \(x=-0.5\), the area decreases at a decreasing speed.
From around \(x=-0.5\), the area increases at an increasing speed.

FORM GR1268 Question 24

Question:
Consider the system of linear equations
\[\begin{align}
w&+&3x&+&2y&+&2z&=0\\
w&+&4x&+&y&&&=0\\
3w&+&5x&+&10y&+&14z&=0\\
2w&+&5x&+&5y&+&6z&=0\\
\end{align}\]
with solutions of the form \((w,x,y,z)\), where \(w,x,y,\) and \(z\) are real. Which of the following statements is FALSE?

(A) The system is consistent.
(B) The system has infinitely many solutions.
(C) The sum of any two solutions is a solution.
(D) \((-5,1,1,0)\) is a solution.
(E) Every solution is a scalar multiple of \((-5,1,1,0)\).

Answer:
(E)

Answer Key:
Plug in \((-5,1,1,0)\) and confirm that it is a solution.
Thus, (D) is true.

Since the system has a solution, it is consistent.
Thus, (A) is true.

Let \(\displaystyle \mathbf A= \begin{bmatrix}1 & 3 & 2 & 2\\ 1 & 4 & 1 & 0\\ 3 & 5 & 10 & 14\\ 2 & 5 & 5& 6\end{bmatrix}\).
If \(\mathbf v=\begin{bmatrix}w\\x\\y\\z\end{bmatrix}\) is a solution,then \(\mathbf{Av}=\mathbf 0\).
Then for any scalar \(k\), we have \(k\mathbf{Av}=\mathbf{A}(k\mathbf v)=\mathbf 0\), so \(k\mathbf v\) is also a solution.
Thus, (B) is true.

If \(\mathbf v\) and \(\mathbf u\) are solutions, \(\mathbf{Av}=\mathbf 0\) and \(\mathbf{Au}=\mathbf 0\),
so \(\mathbf{Av}+\mathbf{Au}=\mathbf{A(v+u)}=\mathbf 0\).
In other words, \(\mathbf v+u\) is a solution.
Thus, (C) is true.

Since answer choices (A) through (D) are true, it must be that (E) is false.

FORM GR1268 Question 23

Question:
Let \((\mathbb Z_{10}, +,\cdot)\) be the ring of integers modulo \(10\), and let \(S\) be the subset of \(\mathbb Z_{10}\) represented by \(\{0,2,4,6,8\}\). Which of the following statement is FALSE?

(A) \((S,+,\cdot)\) is closed under addition modulo \(10\).
(B) \((S,+,\cdot)\) is closed under multiplication modulo \(10\).
(C) \((S,+,\cdot)\) has an identity under addition modulo \(10\).
(D) \((S,+,\cdot)\) has no identity under multiplication modulo \(10\).
(E) \((S,+,\cdot)\) is commutative under addition modulo \(10\).

Answer:
(D)

Answer Key:
\(S\) contains all even remainders.

(A) is true because even + even = even, and even number has an even remainder when divided by \(10\).

(B) is true because even \(\cdot\) even = even, and even number has an even remainder when divided by \(10\).

(C) is true because \(0\) is additive identity.

(D) is not true because \(6\) is multiplicative identity.
That is,
\(0\cdot6=0\equiv0\pmod{10}\)
\(2\cdot6=12\equiv2\pmod{10}\)
\(4\cdot6=24\equiv4\pmod{10}\)
\(6\cdot6=36\equiv6\pmod{10}\)
\(8\cdot6=48\equiv8\pmod{10}\)

(E) is true addition of integers are commutative.

FORM GR1268 Question 22

Question:
What is the volume of the solid in \(xyz\)-space bounded by the surfaces \(y=x^2,y=2-x^2,z=0,\) and \(z=y+3\)?

(A) \(\displaystyle\frac{8}{3}\)     (B) \(\displaystyle\frac{16}{3}\)     (C) \(\displaystyle\frac{32}{3}\)     (D) \(\displaystyle\frac{104}{105}\)     (E) \(\displaystyle\frac{208}{105}\)

Answer:
(C)

Answer Key:
First, you want to picture this solid.

Take \(xy\)-plane parallel to the ground,
\(y\)-axis pointing to the right, and \(x\)-axis pointing toward you.
Draw \(y=x^2\) and \(y=2-x^2\).
The surrounded area look like eyelid or football.

Take \(z\)-axis pointing to the ceiling.
Extrude the eyelid along the \(z\)-axis.
Now we have a "column" whose cross-section is an eyelid.

Chop the column at \(z=0\) and \(z=y+3\).

We will use disc method for integration.

If we slice along the \(z\)-axis, we need the area of eyelid at each level of \(z\).
We'd rather avoid it, especially toward the top of the column,
where the cross-section is no longer an eyelid.

If we slice along the \(y\)-axis, each slice will be a rectangle.
At \(y=\tilde y\), the height of the rectangle is \(z=\tilde y+3\).

The width of the rectangle is a little tricky to calculate.
First note that \(y=x^2\) and \(y=2-x^2\) intersect at \(x=\pm1\).

For \(0\le \tilde y\le 1\), the width of the rectangle is the distance between the \(x\)-coordinates where \(y=x^2\) and \(y=\tilde y\) intersect.
That is, \(\tilde y=x^2\), so \(x=\pm\sqrt{\tilde y}\), so the width is \(\sqrt{\tilde y}-(-\sqrt{\tilde y})=2\sqrt{\tilde y}\).

For \(1\lt \tilde y\le 2\), the width of the rectangle is the distance between the \(x\)-coordinates where \(y=2-x^2\) and \(y=\tilde y\) intersect.
That is, \(\tilde y=2-x^2\), so \(x=\pm\sqrt{2-\tilde y}\), so the width is \(2\sqrt{2-\tilde y}\).

Then we integrate along the \(y\)-axis:
\[V=\int_0^12\sqrt{\tilde y}(\tilde y+3)d\tilde y+\int_1^2 2\sqrt{2-\tilde y}(\tilde y+3)d\tilde y,\]
which I'd rather not deal with.

If we instead slice the solid along the \(x\)-axis, each slice will be a trapezoid.

At \(x=\tilde x\), the width of the trapezoid is \((2-\tilde x^2)-x^2=2(1-\tilde x^2)\).
The \(y\)-coordinate of the left edge is \(y=\tilde x^2\), so the height is \(z=\tilde x^2+3\).
The \(y\)-coordinate of the right edge is \(y=2-\tilde x^2\), so the height is \(z=(2-\tilde x^2)+3=5-\tilde x^2\).
Thus, the are of the trapezoid is \(\frac{1}{2}[2(1-\tilde x^2)][(\tilde x^2+3)+(5-\tilde x^2)]=8(1-\tilde x^2)\).

Integrating along the \(x\)-axis yields
\[\begin{align}
V&=\int_{-1}^18(1-\tilde x^2)d\tilde x\\
&=8\int_{-1}^1(1-\tilde x^2)d\tilde x\\
&=8\left[x-\frac{\tilde x^3}{3}\right]_{-1}^1\\
&=8\left[\left(1-\frac{1}{3}\right)-\left(-1-\left(\frac{-1}{3}\right)\right)\right]\\
&=8\cdot\frac{4}{3}\\
&=\frac{32}{3}.
\end{align}\]

FORM GR1268 Question 21

Question:
What is the value of \(\displaystyle\int_{-\pi/4}^{\pi/4}\left(\cos t+\sqrt{1+t^2}\sin^3t\cos^3t\right)dt\)?

(A) \(0\)     (B) \(\sqrt2\)     (C) \(\sqrt2-1\)     (D) \(\displaystyle\frac{\sqrt2}{2}\)     (E) \(\displaystyle\frac{\sqrt2-1}{2}\)

Answer:
(B)

Answer Key:
We should realize that there is no easy antiderivative.

Even if we mess around using \(\sin^2ｔ+\cos^2ｔ=1\) and such,
because of the product of \(\sqrt{1+t^2}\) and trigonometric function,
it is impossible to find antiderivative.

If it is a curve we are familiar with, we may be able to solve geometrically.
However, we have no idea what this curve looks like.

So, let's start with the easy part.
Integrating the first term yields
\[\begin{align}
\int_{-\pi/4}^{\pi/4}\cos t
&= \left.\sin t\right|_{-\pi/4}^{\pi/4}\\
&=\sin\left(\frac{\pi}{4}\right)-\sin\left(-\frac{\pi}{4}\right)\\
&=\frac{1}{\sqrt2}-\left(-\frac{1}{\sqrt2}\right)\\
&=\frac{2}{\sqrt2}\\
&=\sqrt2.
\end{align}\]
If the integral of the second term is zero, the answer is (B).

Is it the case that the integral of the second term is zero?
Fortunately, it is.

Here's why:
\(\sqrt{1+t^2}\) is an even function.
\(\sin^3t\) is an odd function (because cubing just shrinks \(\sin t\)).
\(\cos^3t\) is an even function.
And even \(\times\) odd \(\times\) even \(=\) odd.

The integral of the second term is zero because the function is odd and the limits of the integration is symmetric around zero.

FORM GR1268 Question 20

Question:
Let \(g\) be the function defined by \(g(x)=e^{2x+1}\) for all real \(x\). Then \(\displaystyle\lim_{x\rightarrow0}\frac{g(g(x))-g(e)}{x}=\)

(A) \(2e\)     (B) \(4e^2\)     (C) \(e^{2e+1}\)     (D) \(2e^{2e+1}\)     (E) \(4e^{2e+2}\)

Answer:
(E)

Answer Key:
Numerator evaluated at zero = \(g(g(0))-g(e)=g(e)-g(e)=0\).
Denominator evaluated at zero = \(0\).

Thus, by L'Hospital rule,
\[\begin{align}
\lim_{x\rightarrow0}\frac{g(g(x))-g(e)}{x}
&=\lim_{x\rightarrow0}\frac{(g(g(x))-g(e))'}{x'}\\
&=\lim_{x\rightarrow0}\frac{g'(g(x))g'(x)}{1}\\
&=\lim_{x\rightarrow0}2e^{2g(x)+1}2e^{2x+1}\\
&=2e^{2g(0)+1}2e^{2(0)+1}\\
&=2e^{2e+1}2e\\
&=4e^{2e+2}
\end{align}\]
Note that \((g(e))'=0\) because \(g(e)\) is just a number (i.e., not a function of \(x\)).
If \((g(g(x)))'\) is confusing, let \(u(x)=g(x)\).
Then
\[\begin{align}
(g(g(x)))'&=(g(u(x)))'\\
&=\frac{dg(u)}{du}\cdot\frac{du(x)}{dx}\\
&=\frac{dg(u)}{du}\cdot\frac{dg(x)}{dx}\\
&=2e^{2u+1}\cdot 2e^{2x+1}\\
&=2e^{2g(x)+1}\cdot 2e^{2x+1}.
\end{align}\]

FORM GR1268 Question 19

Question:
If \(z\) is a complex variable and \(\bar z\) denotes the complex conjugate of \(z\), which is \(\displaystyle\lim_{z\rightarrow0}\frac{(\bar z)^2}{z^2}\)?

(A) \(0\)     (B) \(1\)     (C) \(i\)     (D) \(\infty\)     (E) The limit does not exist.

Answer:
(E)

Answer Key:
Let \(z=x+iy\).
Then \(\bar z = x-iy\).

So, \(z^2=(x+iy)^2=x^2+2xyi+(iy)^2=x^2-y^2+2xyi\)
and \((\bar z)^2=(x-iy)^2=x^2-2xyi+(iy)^2=x^2-y^2-2xyi\).

Let \(z\) approach zero along the \(x\)-axis.
In other words, set \(y=0\).
Then
\[\lim_{x\rightarrow0}\frac{x^2-0^2-2x(0)i}{x^2-0^2+2x(0)i}
=\lim_{x\rightarrow0}\frac{x^2}{x^2}
=\lim_{x\rightarrow0}1=1.\]
Let \(z\) approach zero along the line \(y=x\).
In other words, set \(y=x\).
Then
\[\lim_{x\rightarrow0}\frac{x^2-x^2-2x^2i}{x^2-x^2+2x^2i}
=\lim_{x\rightarrow0}\frac{-2x^2i}{2x^2i}
=\lim_{x\rightarrow0}(-1)=-1.\]
Because the limits don't agree, the limit does not exist.

Alternatively, you can use polar coordinate.
Let \(z=re^{i\theta}\).
Then \(\bar z=re^{-i\theta}\).
\[\lim_{r\rightarrow0}\frac{(\bar z )^2}{z^2}
=\lim_{r\rightarrow0}\frac{r^2e^{-i2\theta}}{r^2e^{i2\theta}}
=\lim_{r\rightarrow0}e^{-i4\theta}
=e^{-i4\theta}.\]
The limit does not exist because it depends on \(\theta\).

\(z\) approaching zero along \(x\)-axis corresponds to \(\theta=0\),
in which case, the limit is \(e^{-i4(0)}=e^0=1\).

\(z\) approaching zero along the line \(y=x\) corresponds to \(\theta=\frac{\pi}{4}\),
in which case, the limit is \(e^{-i4\left(\frac{\pi}{4}\right)}=e^{-\pi i}=-1\).

FORM GR1268 Question 18

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 \(\displaystyle S=\frac{a}{1-r}\).

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 \(\sec x\).

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.

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 \(z\in U\).

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, \(\displaystyle\frac{d}{dx}\int_c^xg(t)dt=\frac{d}{dx}[G(x)-G(c)]=g(x)\).

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 \((0,3)\),
\[\begin{align}
f(3)&=f(0)-1(3-0)\\
5&=f(0)-3\\
8&=f(0).
\end{align}\]

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.

FORM GR1268 Question 11

Question:
The region bounded by the curves \(y=x\) and \(y=x^2\) in the first quadrant of the \(xy\)-plane is rotated about the \(y\)-axis. The volume of the resulting solid of revolution is

(A) \(\displaystyle\frac{\pi}{12}\)     (B) \(\displaystyle\frac{\pi}{6}\)     (C) \(\displaystyle\frac{\pi}{3}\)     (D) \(\displaystyle\frac{2\pi}{3}\)     (E) \(\displaystyle\frac{3\pi}{2}\)

Answer:
(B)

Answer Key:
We will use the disc method for integration.
That is, we slice the solid perpendicular to the \(y\)-axis.

The cross sections of the solid are rings.
The diameter of the outer and inner circle depends on where on the \(y\)-axis you are at.
So, we want to express \(x\) in terms of \(y\):
The diameter of the inner circle is \(x=y\), derived from \(y=x\).
The diameter of the outer circle is \(x=\sqrt{y}\), derived from \(y=x^2\).

Thus, the area of the cross section at \(y\)-coordinate \(y\) is
\[A(y)=\pi(\sqrt{y}^2-y^2)=\pi(y-y^2).\]
To get the volume, integrate over \(0\le y\le 1\).
(Because the curves intersect at \((0,0)\) and \((1,1)\).)
\[\begin{align}
\int_0^1A(y)dy&=\int_0^1\pi(y-y^2)dy\\
&=\left.\pi\left(\frac{y^2}{2}-\frac{y^3}{3}\right)\right|_0^1\\
&=\pi\left(\frac{1}{2}-\frac{1}{3}\right)\\
&=\frac{\pi}{6}.
\end{align}\]

FORM GR1268 Question 10

Question:
\[\sqrt{(x+3)^2+(y-2)^2}=\sqrt{(x-3)^2+y^2}\]
In the \(xy\)-plane, the set of points whose coordinates satisfy the equation above is
(A) a line      (B) a circle     (C) an ellipse     (D) a parabola     (E) one branch of a hyperbola

Answer:
(A)

Answer Key:
Square both sides:
\[\begin{align}
(x+3)^2+(y-2)^2&=(x-3)^2+y^2\\
x^2+6x+9+y^2-4y+4&=x^2-6x+9+y^2\\
12x-4y+4=0
\end{align}\]
This is a line.


Alternatively, we can think this problem geometrically.

\(\sqrt{(x+3)^2+(y-2)^2}=r\) is a circle with radius \(r\) centered at \((-3,2)\).

\(\sqrt{(x-3)^2+y^2}=r\) is a circle with radius \(r\) centered at \((3,0)\).

The distance between the centers is \(\sqrt{(-3-3)^2+(2-0)^2}=\sqrt{40}=2\sqrt{10}\).

When \(r\lt\sqrt{10}\), the circles do not intersect.

When \(r=\sqrt{10}\), the circles intersect at one point.

When \(r\gt\sqrt{10}\), the circles intersect at two points.

The collection of the intersections is a line that is equidistant from the centers of the two circles.

FORM GR1268 Question 9

Question:

Let \(g\) be a continuous real-valued function defined on \(\mathbb R\) with the following properties.

\[g'(0)=0\]

\[g^{\prime\prime}(-1)\gt0\]

\[g^{\prime\prime}(x)\lt0\text{   if   }0\lt x\lt 2.\]

Which of the following could be part of the graph of \(g\)?

Answer:

(A)

Answer Key:

Since \(g'(0)=0\), the slope of \(g\) at \(x=0\) is zero.

Since \(g^{\prime\prime}(-1)\gt0\), \(g\) is convex at \(x=-1\).

Since \(g^{\prime\prime}(x)\lt0\) for \(0\lt x\lt 2\), \(g\) is concave over \(0\lt x\lt 2\).

Think about this.

When \(g^{\prime\prime}\gt0\):

For \(g'\lt0\), the slope is getting less negative.

For \(g'\gt0\), the slope is getting more positive (steeper).

So, \(g\) is convex.

When \(g^{\prime\prime}\lt0\):

For \(g'\gt0\), the slope is decreasing (less steep).

For \(g'\lt0\), the slope is getting more negative.

So, \(g\) is concave.

Or, you can use memory trick like \(+\,\,_\cup+\) for convexity and \(-\,\,_\cap-\) for concavity.

FORM GR1268 Question 8

Question:
Which of the following is NOT a group?
(A) The integers under addition
(B) The nonzero integers under multiplication
(C) The nonzero real numbers under multiplication
(D) The complex numbers under addition
(E) The nonzero complex numbers under multiplication

Answer:
(B)

Answer Key:
Since all the answer choices deal with numbers, associativity holds.

For (A):
Identity is \(0\).
Inverse of \(x\in\mathbb Z\) is \(-x\in\mathbb Z\).
Integer plus integer is integer (closure).

For (B):
Identity is \(1\).
Inverse of \(x\in\mathbb Z\setminus\{0\}\) is \(\frac{1}{x}\notin\mathbb Z\setminus\{0\}\).
Integer times integer is integer (closure).

For (C):
Identity is \(1\).
Inverse of \(x\in\mathbb R\setminus\{0\}\) is \(\frac{1}{x}\in\mathbb R\setminus\{0\}\).
Non-zero real number times non-zero real number is non-zero real number (closure).

For (D):
Identity is \(0\).
Inverse of \(x+iy\in\mathbb C\) is \(-x+i(-y)\in\mathbb C\).
Complex number plus complex number is complex number (closure).

For (E):
Identity is \(1\).
Inverse of \(x+iy\in\mathbb C\setminus\{0\}\) is
\[\displaystyle\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}+i\left(-\frac{y}{x^2+y^2}\right)\in\mathbb C\setminus\{0\}.\]
Non-zero complex number times non-zero complex number is non-zero complex number (closure).

FORM GR1268 Question 7

Question:

The figure above shows the graph of the derivative \(f'\) of a function \(f\), where \(f\) is continuous on the interval \([0,4]\) and differentiable on the interval \((0,4)\). Which of the following gives the correct ordering of the values \(f(0), f(2),\) and \(f(4)\)?

(A) \(f(0)\lt f(2)\lt f(4)\)

(B) \(f(0)\lt f(4)=f(2)\)

(C) \(f(0)\lt f(4)\lt f(2)\)

(D) \(f(4)=f(2)\lt f(0)\)

(E) \(f(4)\lt f(0)\lt f(2)\)

Answer:

(C)

Answer Key:

Since \(f'(x)\gt0\) for \(0\lt x\lt 2\), \(f\) is increasing on the interval \((0,2)\).

Since \(f'(x)\lt0\) for \(2\lt x\lt 4\), \(f\) is decreasing on the interval \((2,4)\).

So, \(f(2)\) is the largest.

Now, the question is, which is larger, \(f(0)\) or \(f(4)\)?

Intuitively, since the rate of increase on the interval \((0,2)\) is larger than the rate of decrease on the interval \((2,4)\), \(f(0)\) is "further away" from \(f(2)\).

Hence, \(f(0)\lt f(4)\).

Formally, since \(\displaystyle\int_0^2 f'(x)dx=f(2)-f(0)\), we have \(\displaystyle f(0)=f(2)-\int_0^2f'(x)dx\).

Similarly, since \(\displaystyle\int_2^4f'(x)dx=f(4)-f(2)\), we have \(\displaystyle f(4)=f(2)+\int_2^4f'(x)dx\).

Because \(\displaystyle\int_0^2f'(x)dx\gt -\int_2^4f'(x)dx\), we have \(f(0)\lt f(4)\).

FORM GR1268 Question 6

Question:
Which of the following shows the numbers \(2^{1/2}, 3^{1/3},\) and \(6^{1/6}\) in increasing order?

(A) \(2^{1/2}\lt3^{1/3}\lt6^{1/6}\)

(A) \(6^{1/6}\lt3^{1/3}\lt2^{1/2}\)

(A) \(6^{1/6}\lt2^{1/2}\lt3^{1/3}\)

(A) \(3^{1/3}\lt2^{1/2}\lt6^{1/6}\)

(A) \(3^{1/3}\lt6^{1/6}\lt2^{1/2}\)

Answer:
(C)

Answer Key:
\(2^{1/2}=2^{3/6}=8^{1/6}\)
\(3^{1/3}=3^{2/6}=9^{1/6}\)

FORM GR1268 Question 5

Question:
Sofia and Tess will each randomly choose one of the 10 integers from 1 to 10. What is the probability that neither integer chosen will be the square of the other?

(A) 0.64     (B) 0.72     (C) 0.81     (D) 0.90     (E) 0.95

Answer:
(E)

Answer Key:
There are \(10^2=100\) possible ordered pairs.
That is, the cardinality of the set \(\{(S,T)\mid S,T\in\{1,...,10\}\}\) is 100.

There are five ordered pairs we don't want: \((1,1),(2,4),(4,2),(3,9),(9,3)\).
The probability that one integer will be the square of the other is \(\frac{5}{100}=0.05\).

The probability that neither integer will be the square of the other is \(1-0.05=0.95\).

FORM GR1268 Question 4

Question:
Let \(V\) and \(W\) be 4-dimensional subspaces of a 7-dimensional vector space \(X\). Which of the following CANNOT be the dimension of the subspace \(V\cap W\)?

(A) 0     (B) 1     (C) 2      (D) 3     (E) 4

Answer:
(A)

Answer Key:
A basis of \(V\) has four vectors, and these four vectors are linearly independent.
A basis of \(W\) has four vectors, and these four vectors are linearly independent.
Because there cannot be more than seven linearly independent vectors in a 7-dimensional vector space, these eight vectors are not linearly independent.
That is, \(V\) and \(W\) has common dimension.

Here's the intuition:
We need four vectors for a basis of \(V\) and four vectors for a basis of \(W\).
But there are only seven vectors in a basis of \(X\).
So, at least one of them has to be used for both \(V\) and \(W\).

For visual learners:
Consider the usual 3-dimensional Euclidean space.
Can you make non-overlapping two 2-dimensional subspaces?
That is, can you choose two planes that go through the origin and do not share a line?
The answer is no.

FORM GR1268 Question 3

Question:
\(\displaystyle \int_{e^{-3}}^{e^{-2}}\frac{1}{x\log x}dx=\)

(A) \(1\)     (B) \(\displaystyle\frac{2}{3}\)     (C) \(\displaystyle\frac{3}{2}\)     (D) \(\displaystyle\log\left(\frac{2}{3}\right)\)     (E) \(\displaystyle\log\left(\frac{3}{2}\right)\)

Answer:
(D)

Answer Key:
\[\begin{align}
\int_{e^{-3}}^{e^{-2}}\frac{1}{x\log x}dx
&= \left.\log\left(\log x\right)\right|_{e^{-3}}^{e^{-2}}\\
&=\log\left(\log(e^{-2})\right) - \log\left(\log(e^{-3})\right)\\
&=\log(-2)-\log(-3)\\
&=\log\left(\frac{-2}{-3}\right)\\
&=\log\left(\frac{2}{3}\right)
\end{align}\]

I implicitly used the substitution rule.

Write
\[\int_{e^{-3}}^{e^{-2}}\frac{1}{x\log x}dx
=\int_{e^{-3}}^{e^{-2}}\frac{1}{x}\cdot\frac{1}{\log x}dx
=\int_{e^{-3}}^{e^{-2}}(\log x)'\cdot\frac{1}{\log x}dx.\]

Let \(u=\log x\). Then \(du=\frac{1}{x}dx\).
The limits of integration in terms of \(u\) are \(\log(e^{-3})=-3\) and \(\log(e^{-2})=-2\).

Thus,
\[\begin{align}
\int_{e^{-3}}^{e^{-2}}\frac{1}{x\log x}dx
&=\int_{e^{-3}}^{e^{-2}}\frac{1}{\log x}\frac{1}{x}dx\\
&=\int_{-3}^{-2}\frac{1}{u}du\\
&=\left.\log u\right|_{-3}^{-2}\\
&=\log(-2)-\log(-3)\\
&=\log\left(\frac{-2}{-3}\right)\\
&=\log\left(\frac{2}{3}\right).
\end{align}\]

FORM GR1268 Question 2

Question:
What is the area of an equilateral triangle whose inscribed circle has radius 2?

(A) 12     (B) 16     (C) \(12\sqrt3\)     (D) \(16\sqrt3\)     (E) \(4(3+2\sqrt2)\)

Answer:
(C)

Answer Key:

An equilateral triangle contains six right angled triangles as shown above.
Each of the right angled triangle has \(90^\circ\)-\(30^\circ\)-\(60^\circ\) angles.
The ratio of the corresponding sides is \(2:1:\sqrt3\).

Since we know the radius of the circle is 2, we can calculate the length of the remaining two sides.
Finally, \(\displaystyle A=\frac{bh}{2}=\frac{4\sqrt3\cdot6}{2}=12\sqrt3\).

If you don't remember the ratio of the sides for \(90^\circ\)-\(30^\circ\)-\(60^\circ\) triangle,
start with an equilateral triangle with side length 2, and cut it into half.
Then use the Pythagorean theorem: \(1^2+s^2=2^2\) and \(s=\sqrt{2^2-1^2}=\sqrt3\).

FORM GR1268 Question 1

Question:
\(\displaystyle\lim_{x\rightarrow0}\frac{\cos(3x)-1}{x^2}=\)

(A) \(\displaystyle\frac{9}{2}\)     (B) \(\displaystyle\frac{3}{2}\)     (C) \(\displaystyle-\frac{2}{3}\)     (D) \(\displaystyle-\frac{3}{2}\)     (E) \(\displaystyle-\frac{9}{2}\)

Answer:
(E)

Answer Key:
Since \(\cos(3\cdot0)-1=1-1=0\) and \(0^2=0\), by L'Hospital rule,
\[\lim_{x\rightarrow0}\frac{\cos(3x)-1}{x^2}
=\lim_{x\rightarrow0}\frac{(\cos(3x)-1)'}{(x^2)'}
=\lim_{x\rightarrow0}\frac{-3\sin(3x)}{2x}.\]

 Again, since \(-3\sin(3\cdot0)=0\) and \(2\cdot0=0\), by L'Hospital rule,
\[\begin{align}
\lim_{x\rightarrow0}\frac{-3\sin(3x)}{2x}
&=\lim_{x\rightarrow0}\frac{(-3\sin(3x))'}{(2x)'}\\
&=\lim_{x\rightarrow0}\frac{-9\cos(3x)}{2}\\
&=\frac{-9\cos(3\cdot0)}{2}\\
&=\frac{-9}{2}.
\end{align}\]