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,
\[\begin{align}
\lim_{x\rightarrow0}\frac{-3\sin(3x)}{2x}
&=\lim_{x\rightarrow0}\frac{(-3\sin(3x))'}{(2x)'}\\
&=\lim_{x\rightarrow0}\frac{-9\cos(3x)}{2}\\
&=\frac{-9\cos(3\cdot0)}{2}\\
&=\frac{-9}{2}.
\end{align}\]