What is the value of $\sqrt [ 3 ] { 9 } \times 3 ^ { \frac { 1 } { 3 } }$? [2 points]
(1) 1
(2) $3 ^ { \frac { 1 } { 2 } }$
(3) 3
(4) $3 ^ { \frac { 3 } { 2 } }$
(5) 9

What is the value of $\left( 2 ^ { \sqrt { 3 } } \times 4 \right) ^ { \sqrt { 3 } - 2 }$? [2 points]
(1) $\frac { 1 } { 4 }$
(2) $\frac { 1 } { 2 }$
(3) 1
(4) 2
(5) 4

What is the value of $\left( \frac { 4 } { 2 ^ { \sqrt { 2 } } } \right) ^ { 2 + \sqrt { 2 } }$? [2 points]
(1) $\frac { 1 } { 4 }$
(2) $\frac { 1 } { 2 }$
(3) 1
(4) 2
(5) 4

For a natural number $m$ ($m \geq 2$), let $f ( m )$ be the number of natural numbers $n \geq 2$ such that an integer exists among the $n$-th roots of $m ^ { 12 }$. What is the value of $\sum _ { m = 2 } ^ { 9 } f ( m )$? [4 points]
(1) 37
(2) 42
(3) 47
(4) 52
(5) 57

Find the value of $\sqrt[3]{24} \times 3^{\frac{2}{3}}$. [2 points]
(1) 6
(2) 7
(3) 8
(4) 9
(5) 10

What is the value of $\sqrt[3]{5} \times 25^{\frac{1}{3}}$? [2 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5

What is the value of $9 ^ { \frac { 1 } { 4 } } \times 3 ^ { - \frac { 1 } { 2 } }$? [2 points]
(1) 1
(2) $\sqrt { 3 }$
(3) 3
(4) $3 \sqrt { 3 }$
(5) 9

7. If real numbers $\mathrm { a } , \mathrm { b }$ satisfy $\frac { 1 } { a } + \frac { 2 } { b } = \sqrt { a b }$, then the minimum value of $ab$ is
A. $\sqrt { 2 }$
B. 2
C. $2 \sqrt { 2 }$
D. 4

Among the three numbers $2 ^ { - 3 }$, $3 ^ { \frac { 1 } { 2 } }$, $\log _ { 2 } 5$, the largest is

Exercise I

I-A- $\quad \frac { ( 2 \sqrt { 3 } ) ^ { 2 } \times 12 ^ { 3 } \times 3 ^ { 2 } } { 3 ^ { - 4 } \times ( \sqrt { 2 } ) ^ { 4 } } = 3 ^ { 10 } \times 2 ^ { 8 }$. I-B- $\quad 2 \sqrt { 27 } - ( 2 \sqrt { 3 } - 1 ) ^ { 2 } = 10 \sqrt { 3 } - 13$. I-C- $\quad \ln \left( \frac { e } { 4 } \right) + \ln \left( \frac { 1 } { 9 e } \right) + \ln ( 36 e ) = 1$. I-D- $\quad e ^ { 2 \ln 3 + \ln 5 } + e ^ { - 2 \ln 5 } = 20$. I-E- For every real number $x$ different from $-2$ and from $2$, $\frac { 2 } { x + 2 } - \frac { 1 } { x - 2 } + \frac { 8 } { x ^ { 2 } - 4 } = \frac { 1 } { x - 2 }$. I-F- For every real number $x$, $\frac { e ^ { 2 x } + 2 e ^ { x } + 1 } { e ^ { x } + 1 } = e ^ { x } + 1$.
For each statement, indicate whether it is TRUE or FALSE.

If $x = 1 + \sqrt[5]{2} + \sqrt[5]{4} + \sqrt[5]{8} + \sqrt[5]{16}$, then the value of $\left(1 + \frac{1}{x}\right)^{30}$ is
(A) 2
(B) 5
(C) 32
(D) 64

If $x$ is a solution of the equation $\sqrt { 2 x + 1 } - \sqrt { 2 x - 1 } = 1 , \left( x \geq \frac { 1 } { 2 } \right)$, then $\sqrt { 4 x ^ { 2 } - 1 }$ is equal to :
(1) $\frac { 3 } { 4 }$
(2) $\frac { 1 } { 2 }$
(3) $2 \sqrt { 2 }$
(4) 2

$50^{\text{th}}$ root of a number $x$ is 12 and $50^{\text{th}}$ root of another number $y$ is 18. Then the remainder obtained on dividing $(x + y)$ by 25 is $\_\_\_\_$.

Let $\alpha = \frac { ( 4 ! ) ! } { ( 4 ! ) ^ { 3 ! } }$ and $\beta = \frac { ( 5 ! ) ! } { ( 5 ! ) ^ { 4 ! } }$. Then :
(1) $\alpha \in \mathrm { N }$ and $\beta \notin \mathrm { N }$
(2) $\alpha \notin \mathrm { N }$ and $\beta \in \mathrm { N }$
(3) $\alpha \in \mathrm { N }$ and $\beta \in \mathrm { N }$
(4) $\alpha \notin \mathrm { N }$ and $\beta \notin \mathrm { N }$

Suppose that positive real numbers $a$ and $b$ satisfy
$$a ^ { 2 } = 3 + \sqrt { 5 } , \quad b ^ { 2 } = 3 - \sqrt { 5 } .$$
Let $c$ be the fractional portion of $a + b$. We are to find the value of $\frac { 1 } { c } - c$.
(1) We see that $( a b ) ^ { 2 } = \mathbf { N }$ and $( a + b ) ^ { 2 } = \mathbf { O P }$.
(2) Since $\mathbf { Q }$ $< a + b < \mathbf { Q } + 1$, the value of $c$ is $\sqrt { \mathbf { R S } } - \mathbf { T }$.
Thus we obtain $\frac { 1 } { c } - c = \mathbf { U }$.

Suppose that positive real numbers $a$ and $b$ satisfy
$$a ^ { 2 } = 3 + \sqrt { 5 } , \quad b ^ { 2 } = 3 - \sqrt { 5 } .$$
Let $c$ be the fractional portion of $a + b$. We are to find the value of $\frac { 1 } { c } - c$.
(1) We see that $( a b ) ^ { 2 } = \mathbf { N }$ and $( a + b ) ^ { 2 } = \mathbf { O P }$.
(2) Since $\mathbf { Q }$ $< a + b < \mathbf { Q } + 1$, the value of $c$ is $\sqrt { \mathbf { R S } } - \mathbf { T }$.
Thus we obtain $\frac { 1 } { c } - c = \mathbf { U }$.

For the real numbers $a$ and $b$ satisfying
$$a ^ { 3 } = \frac { 1 } { \sqrt { 5 } - 2 } , \quad b ^ { 3 } = 2 - \sqrt { 5 }$$
we are to find the value of $a + b$.
When we set $x = a + b$, we have
$$x ^ { 3 } = ( a + b ) ^ { 3 } = a ^ { 3 } + b ^ { 3 } + \mathbf { A } a b ( a + b ) .$$
Since $a b = \mathbf { B C }$, we know that this $x$ satisfies
$$x ^ { 3 } + \mathbf { D } x - \mathbf { E } = 0 .$$
The left side of this equation can be factorized as follows:
$$\begin{aligned} x ^ { 3 } + \mathbf { D } x - \mathbf { E } & = \left( x ^ { 3 } - \mathbf { F } \right) + \mathbf { D } \left( x - \frac { \mathbf { F } } { \mathbf { F } } \right) \\ & = ( x - \mathbf { F } ) \left( x ^ { 2 } + x + \mathbf { G } \right) . \end{aligned}$$
Since
$$x ^ { 2 } + x + \mathbf { G } = \left( x + \frac { \mathbf { H } } { \mathbf { I } } \right) ^ { 2 } + \frac { \mathbf { J K } } { \mathbf { L } } > 0 ,$$
we obtain $x = a + b = \mathbf { M }$.

For the real numbers $a$ and $b$ satisfying
$$a ^ { 3 } = \frac { 1 } { \sqrt { 5 } - 2 } , \quad b ^ { 3 } = 2 - \sqrt { 5 }$$
we are to find the value of $a + b$.
When we set $x = a + b$, we have
$$x ^ { 3 } = ( a + b ) ^ { 3 } = a ^ { 3 } + b ^ { 3 } + \mathbf { A } a b ( a + b ) .$$
Since $a b = \mathbf { B C }$, we know that this $x$ satisfies
$$x ^ { 3 } + \mathbf { D } x - \mathbf { E } = 0 .$$
The left side of this equation can be factorized as follows:
$$\begin{aligned} x ^ { 3 } + \mathbf { D } x - \mathbf { E } & = \left( x ^ { 3 } - \mathbf { F } \right) + \mathbf { D } \left( x - \frac { \mathbf { F } } { \mathbf { E } } \right) \\ & = ( x - \mathbf { F } ) \left( x ^ { 2 } + x + \mathbf { G } \right) . \end{aligned}$$
Since
$$x ^ { 2 } + x + \frac { \mathbf { G } } { \mathbf { H } } = \left( x + \frac { \mathbf { H } } { \mathbf { I } } \right) ^ { 2 } + \frac { \mathbf { J K } } { \mathbf { L } } > 0 ,$$
we obtain $x = a + b = \mathbf { M }$.

We are to find the positive number $a$ satisfying $a ^ { 3 } = 9 + \sqrt { 80 }$.
Let us consider the positive number $b$ which satisfies $b ^ { 3 } = 9 - \sqrt { 80 }$. Then
$$\left\{ \begin{aligned} a ^ { 3 } + b ^ { 3 } & = \mathbf { A B } \\ a b & = \mathbf { C } \end{aligned} \right.$$
holds.
First, using (2), (1) can be transformed into
$$( a + b ) ^ { 3 } - \mathbf { D } ( a + b ) = \mathbf { A B } .$$
Then, setting $x = a + b$, we have
$$x ^ { 3 } - \mathrm { D } x = \mathrm { AB } .$$
Transforming this equation, we obtain
$$x ^ { 3 } - 27 = \mathbf { D } ( x - \mathbf { E } ) \text {, }$$
which gives
$$( x - \mathbf { F } ) \left( x ^ { 2 } + \mathbf { G } x + \mathbf{H} \right) = 0 .$$
From that we have $x = \mathbf{I}$ and hence
$$a + b = \mathbf{I} \text {. }$$
Thus, from (2), (3) and $a > b$, we have
$$a = \frac { \mathbf { J } + \sqrt { \mathbf { K } } } { \mathbf{L} } .$$

Let $x = \frac { \sqrt { 3 } + 1 } { \sqrt { 3 } - 1 }$ and $y = \frac { \sqrt { 6 } - \sqrt { 2 } } { \sqrt { 6 } + \sqrt { 2 } }$.
(1) We have $x = \mathbf { A } + \sqrt { \mathbf { B } }$ and $y = \mathbf { C } - \sqrt { \mathbf { C } }$. Hence we have
$$x + y = \mathbf { E } , \quad x y = \mathbf { F } , \quad \frac { 1 } { x ^ { 2 } } + \frac { 1 } { y ^ { 2 } } = \mathbf { G H } .$$
Also we have
$$5 \left( x ^ { 2 } - 4 x \right) + 3 \left( y ^ { 2 } - 4 y + 1 \right) = \square \mathbf { I J } .$$
(2) The values of the integers $m$ and $n$ such that $\frac { m } { x } + \frac { n } { y } = 4 + 4 \sqrt { 3 }$ are
$$m = \mathbf { K L } , \quad n = \mathbf { M } .$$

Q1 Let $a = \sqrt { 5 } + \sqrt { 3 }$ and $b = \sqrt { 5 } - \sqrt { 3 }$. We are to find the integers $x$ satisfying
$$2 \left| x - \frac { a } { b } \right| + x < 10$$
(1) We see that $\frac { a } { b } = \mathbf { A } + \sqrt { \mathbf { BC } }$. Hence the largest integer less than $\frac { a } { b }$ is $\mathbf{D}$.
(2) For $\mathbf { F }$ and $\mathbf { H }$ in the following sentence, choose the correct answer from among choices (0) $\sim$ (7) below, and for $\mathbf { E }$ and $\mathbf { G }$, enter the correct numbers.
When $x$ is an integer, the left side of the inequality can be expressed without using the absolute value symbol as follows:
$$\left\{ \begin{array} { l } \text { if } x \leqq \mathbf { E } , \text { then } 2 \left| x - \frac { a } { b } \right| + x = \mathbf { F } , \\ \text { if } x \geqq \mathbf { G } , \text { then } 2 \left| x - \frac { a } { b } \right| + x = \mathbf { H } . \end{array} \right.$$
(0) $x - 6 - 2 \sqrt { 10 }$
(1) $x + 8 + 2 \sqrt { 15 }$
(2) $- x + 8 + 2 \sqrt { 15 }$
(3) $- x + 6 + 2 \sqrt { 10 }$
(4) $3 x - 6 - 2 \sqrt { 10 }$
(5) $3 x - 8 - 2 \sqrt { 15 }$ (6) $- 3 x + 8 + 2 \sqrt { 15 }$ (7) $- 3 x + 6 + 2 \sqrt { 10 }$
(3) Thus, the integers $x$ satisfying inequality $2 \left| x - \frac { a } { b } \right| + x < 10$ are those greater than or equal to $\mathbf { I }$ and less than or equal to $\mathbf { J }$.

Where $m$ and $n$ are positive integers, consider the rational number
$$r = \frac { m } { 3 } + \frac { n } { 7 } .$$
We are to find $m$ and $n$ such that among all $r$'s satisfying $r < \sqrt { 2 }$, $r$ is closest to $\sqrt { 2 }$.
It is sufficient that among all $m$'s and $n$'s which satisfy the inequality
$$\mathbf { A } m + \mathbf { B } n < \mathbf { CD } \sqrt { 2 }$$
we find the $m$ and $n$ such that $\mathrm { A } m + \mathrm { B } n$ is closest to $\mathrm { CD } \sqrt { 2 }$.
Squaring both sides of (1), we have
$$(\mathrm { A } m + \mathrm { B } n ) ^ { 2 } < \mathbf { EFG } .$$
Here, the greatest square number which is smaller than EFG is $\mathbf{HIJ} = \mathbf{KL}^{2}$. So, let us consider the equation
$$\mathrm { A } m + \mathrm { B } n = \mathrm { KL } .$$
Transforming this equation, we have
$$n = \frac { \mathbf { MN } - \mathbf { O } m } { \mathbf { P } } .$$
Since $n$ is an integer, $\mathbf { MN } - \mathbf { O } m$ is a multiple of $\mathbf { Q }$. Thus, we obtain
$$m = \mathbf { R } , \quad n = \mathbf { S } .$$

Let $m$ and $n$ be positive integers satisfying $0 < m - n\sqrt{2} < 1$. Denote the integral part of $(m + n\sqrt{2})^3$ by $a$ and the fractional part by $b$.
(1) We are to prove that $a$ is an odd number and $(m - n\sqrt{2})^3 = 1 - b$.
If $(m + n\sqrt{2})^3 = p + q\sqrt{2}$, where $p$ and $q$ are integers, then $$p = m^3 + \mathbf{A}mn^2, \quad q = \mathbf{B}m^2n + \mathbf{C}n^3.$$ Thus, we see that $(m - n\sqrt{2})^3 = p - q\sqrt{2}$.
Furthermore, the integral part of $(m - n\sqrt{2})^3$ is $\mathbf{D}$. When we denote its fractional part by $c$, the following two equations hold: $$\left\{\begin{array}{l} p + q\sqrt{2} = a + b \\ p - q\sqrt{2} = c \end{array}\right.$$
From these we obtain $$\mathbf{E} \quad p - a = b + c.$$
Here, since the left side is an integer and the range of values which the right side takes is $\mathbf{F} < b + c < \mathbf{G}$, we see that $$b + c = \mathbf{H}$$
Hence we see that $a = \mathbf{E}p - \mathbf{H}$, which shows that $a$ is an odd number and that $(m - n\sqrt{2})^3 = 1 - b$.
(2) Let us find the values of $m$ and $n$ when $a = 197$.
Since $a = 197$, we see that $p = \mathbf{IJ}$, that is, $m^3 + \mathbf{A}mn^2 = \mathbf{IJ}$. The positive integers $m$ and $n$ satisfying this equation are $$m = \mathbf{K}, \quad n = \mathbf{L}.$$

Answer the following questions.
(1) The positive integers $m$ and $n$ which simultaneously satisfy the following two inequalities
$$\frac { m } { 3 } < \sqrt { 3 } < \frac { n } { 4 } , \quad \frac { n } { 3 } < \sqrt { 6 } < \frac { m } { 2 }$$
are
$$m = \mathbf { A } , \quad n = \mathbf { B } .$$
(2) Using the results of (1), let us compare the sizes of numbers (1) $\sim$ (5).
(1) $( \sqrt { ( - 3 ) ( - 4 ) } ) ^ { 3 }$
(2) $6 \sqrt { ( - 2 ) ^ { 3 } ( - 3 ) }$
(3) $\sqrt { \left\{ ( - 4 ) ( - 3 ) ^ { 2 } \right\} ^ { 2 } }$
(4) $( - 1 ) ^ { 3 } \sqrt { \left\{ ( - 2 ) ^ { 5 } \right\} ^ { 2 } }$
(5) $\left( \frac { 5 \sqrt { 3 } } { 1 - \sqrt { 6 } } \right) ^ { 2 }$
When the denominator of (5) is rationalized, we have
$$\left( \frac { 5 \sqrt { 3 } } { 1 - \sqrt { 6 } } \right) ^ { 2 } = \mathbf { C D } + \mathbf { E } \sqrt { \mathbf { F } }$$
Of the five numbers, there are $\mathbf { G }$ number(s) greater than 35 and $\mathbf { H }$ negative number(s).
When we arrange the five numbers in the ascending order of their size using the numbers (1) $\sim$ (5), we have
$$\mathbf { I } < \mathbf { J } < \mathbf { K } < \mathbf { L } < \mathbf { M } .$$

If a positive integer $N$ is entered into a calculator, and then the ``$\sqrt{ }$'' key (taking the positive square root) is pressed 3 times consecutively, the display shows the answer as 2. Then $N$ equals which of the following options?
(1) $2^{3}$
(2) $2^{4}$
(3) $2^{6}$
(4) $2^{8}$
(5) $2^{12}$