$\lim_{n \rightarrow \infty} \left\{\left(2^{\frac{1}{2}} - 2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}} - 2^{\frac{1}{5}}\right) \ldots \left(2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}}\right)\right\}$ is equal to
(1) 1
(2) 0
(3) $\sqrt{2}$
(4) $\frac{1}{\sqrt{2}}$

If the total maximum value of the function $f ( x ) = \left( \frac { \sqrt { 3 e } } { 2 \sin x } \right) ^ { \sin ^ { 2 } x } , x \in \left( 0 , \frac { \pi } { 2 } \right)$, is $\frac { k } { e }$, then $\left( \frac { k } { e } \right) ^ { 8 } + \frac { k ^ { 8 } } { e ^ { 5 } } + k ^ { 8 }$ is equal to
(1) $e ^ { 3 } + e ^ { 6 } + e ^ { 11 }$
(2) $e ^ { 5 } + e ^ { 6 } + e ^ { 11 }$
(3) $e ^ { 3 } + e ^ { 6 } + e ^ { 10 }$
(4) $e ^ { 3 } + e ^ { 5 } + e ^ { 11 }$

The sum of all the solutions of the equation $( 8 ) ^ { 2 x } - 16 \cdot ( 8 ) ^ { x } + 48 = 0$ is :
(1) $1 + \log _ { 8 } ( 6 )$
(2) $1 + \log _ { 6 } ( 8 )$
(3) $\log _ { 8 } ( 6 )$
(4) $\log _ { 8 } ( 4 )$

$\lim _ { x \rightarrow 0 } \frac { e - ( 1 + 2 x ) ^ { \frac { 1 } { 2 x } } } { x }$ is equal to
(1) 0
(2) $\frac { - 2 } { e }$
(3) e
(4) $e - e ^ { 2 }$

Let $f : \mathbb{R} \rightarrow (0, \infty)$ be strictly increasing function such that $\lim_{x \rightarrow \infty} \dfrac{f(7x)}{f(x)} = 1$. Then, the value of $\lim_{x \rightarrow \infty} \left(\dfrac{f(5x)}{f(x)} - 1\right)$ is equal to
(1) 4
(2) 0
(3) $\dfrac{7}{5}$
(4) 1

If $\alpha = \lim _ { x \rightarrow 0 ^ { + } } \left( \frac { \mathrm { e } ^ { \sqrt { \tan x } } - \mathrm { e } ^ { \sqrt { x } } } { \sqrt { \tan x } - \sqrt { x } } \right)$ and $\beta = \lim _ { x \rightarrow 0 } ( 1 + \sin x ) ^ { \frac { 1 } { 2 } \cot x }$ are the roots of the quadratic equation $a x ^ { 2 } + b x - \sqrt { \mathrm { e } } = 0$, then $12 \log _ { \mathrm { e } } ( \mathrm { a } + \mathrm { b } )$ is equal to $\_\_\_\_$

If $f ( x ) = \frac { 2 ^ { x } } { 2 ^ { x } + \sqrt { 2 } } , \mathrm { x } \in \mathbb { R }$, then $\sum _ { \mathrm { k } = 1 } ^ { 81 } f \left( \frac { \mathrm { k } } { 82 } \right)$ is equal to
(1) $1.81 \sqrt { 2 }$
(2) 41
(3) 82
(4) $\frac { 81 } { 2 }$

Q68. $\lim _ { x \rightarrow 0 } \frac { e - ( 1 + 2 x ) ^ { \frac { 1 } { 2 x } } } { x }$ is equal to
(1) 0
(2) $\frac { - 2 } { e }$
(3) e
(4) $e - e ^ { 2 }$
Q69. If the variance of the frequency distribution

$x$	$c$	$2 c$	$3 c$	$4 c$	$5 c$	$6 c$
$f$	2	1	1	1	1	1

is 160 , then the value of $c \in N$ is
(1) 7
(2) 8
(3) 5
(4) 6

Q85. If $\alpha = \lim _ { x \rightarrow 0 ^ { + } } \left( \frac { \mathrm { e } ^ { \sqrt { \tan x } } - \mathrm { e } ^ { \sqrt { x } } } { \sqrt { \tan x } - \sqrt { x } } \right)$ and $\beta = \lim _ { x \rightarrow 0 } ( 1 + \sin x ) ^ { \frac { 1 } { 2 } \cot x }$ are the roots of the quadratic equation $a x ^ { 2 } + b x - \sqrt { \mathrm { e } } = 0$, then $12 \log _ { \mathrm { e } } ( \mathrm { a } + \mathrm { b } )$ is equal to $\_\_\_\_$

Consider the function $y = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } }$, where $x \geqq 0$.
(1) We are to find the $x$ at which $y$ is minimized.
When we differentiate $y$, we have
$$\frac { d y } { d x } = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } \left( 2 x \log _ { e } \mathbf { A } - \mathbf { B } \log _ { e } \mathbf { C } \right) .$$
Hence, when we express the value of $x$ at which $y$ is minimized using the common logarithm, we have
$$x = \frac { \mathbf { D } } { \mathbf { F } \left( 1 - \log _ { 10 } \mathbf { E } \right) } .$$
(2) We are to find the smallest positive integer $x$ satisfying $\frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } > 1000$.
From the inequality $y > 1000$, we obtain
$$x ^ { \mathbf { H } } \log _ { 10 } \mathbf { I } - \mathbf { J } x \log _ { 10 } \mathbf { L } - \mathbf { K } > 0 .$$
When we solve the inequality using $0.3$ as an approximate value for $\log _ { 10 } 2 = 0.301 \cdots$, the smallest positive integer $x$ satisfying $y > 1000$ is $\mathbf{Q}$.

We are to find the minimum value of the function
$$f ( x ) = 8 ^ { x } + 8 ^ { - x } - 3 \left( 4 ^ { 1 + x } + 4 ^ { 1 - x } - 2 ^ { 4 + x } - 2 ^ { 4 - x } \right) - 24$$
and the value of $x$ at which the function takes this minimum value.
First, let us set $2 ^ { x } + 2 ^ { - x } = t$. Then, since
$$4 ^ { x } + 4 ^ { - x } = t ^ { 2 } - \mathbf { A } \quad \text { and } \quad 8 ^ { x } + 8 ^ { - x } = t ^ { 3 } - \mathbf { B } t ,$$
$f ( x )$ can be expressed as
$$f ( x ) = t ^ { 3 } - \mathbf { C D } t ^ { 2 } + \mathbf { E F } t$$
When we consider the right side as a function of $t$ and denote it by $g ( t )$, its derivative is
$$g ^ { \prime } ( t ) = \mathbf { G } ( t - \mathbf { H } ) ( t - \mathbf { I } ) \text {, }$$
where $\mathrm { H } < \mathrm { I }$. Here, since $2 ^ { x } + 2 ^ { - x } = t$, the range of the values which $t$ takes is
$$t \geqq \mathbf { J }$$
When $t = \mathbf { J }$, we see that $g ( \mathbf { J } ) = \mathbf { K L }$. When $t > \mathbf { J }$, $g ( t )$ is locally maximized at $t = \mathbf { M }$, and its local maximum is $\mathbf { N O }$, and furthermore, it is locally minimized at $t = \mathbf { P }$, and its local minimum is $\mathbf { Q R }$.
Thus, the minimum value of $f ( x )$ is $\mathbf { S T }$, which is taken at
$$x = \mathbf { U } \quad \text { and } \quad x = \log _ { 2 } ( \mathbf { V } \pm \sqrt { \mathbf { W X } } ) - \mathbf { Y } .$$

3. (i) Find the co-ordinates of the turning points of
$$f ( x ) = e ^ { x } \left( 2 x ^ { 2 } - x - 1 \right)$$
(ii) Sketch the graph of $y = f ( x )$ on the axes below for the range $- 4 \leqslant x \leqslant 2$.
(iii) Now consider
$$g ( x ) = \left\{ \begin{array} { c c } e ^ { x } \left( 2 x ^ { 2 } - x - 1 \right) & \text { if } x < 1 ; \\ \sin ( x - 1 ) & \text { if } x \geqslant 1 . \end{array} \right.$$
Determine, with explanations, the maximum and minimum values of $g ( x )$ as $x$ varies over the real numbers. [Figure]

3. For APPLICANTS IN $\left\{ \begin{array} { l } \text { MATHEMATICS } \\ \text { MATHEMATICS \& STATISTICS } \\ \text { MATHEMATICS \& PHILOSOPHY } \\ \text { MATHEMATICS \& COMPUTER SCIENCE } \end{array} \right\}$ ONLY.

Computer Science and Computer Science \& Philosophy applicants should turn to page 14.
The function $f ( x )$ is defined for all real numbers and has the following properties, valid for all $x$ and $y$ :
(A) $\quad f ( x + y ) = f ( x ) f ( y )$.
(B) $\quad \mathrm { d } f / \mathrm { d } x = f ( x )$.
(C) $\quad f ( x ) > 0$.
Throughout this question, these should be the only properties of $f$ that you use; no marks will be awarded for any use of the exponential function.
Let $a = f ( 1 )$.
(i) Show that $f ( 0 ) = 1$.
(ii) Let
$$I = \int _ { 0 } ^ { 1 } f ( x ) \mathrm { d } x$$
Show that $I = a - 1$.
(iii) The trapezium rule with $n$ steps is used to produce an estimate $I _ { n }$ for the integral $I$. Show that
$$I _ { n } = \frac { 1 } { 2 n } \left( \frac { b + 1 } { b - 1 } \right) ( a - 1 )$$
where $b = f ( 1 / n )$.
(iv) Given that $I _ { n } \geqslant I$ for all $n$, show that
$$a \leqslant \left( 1 + \frac { 2 } { 2 n - 1 } \right) ^ { n }$$

5. With adequate nutrients, the number of bacteria grows exponentially. Suppose the quantity of bacteria A doubles every two hours, and the quantity of bacteria B triples every three hours. If nutrients are adequate and the initial quantities of both bacteria are equal, approximately how many hours later will the quantity of bacteria B divided by the quantity of bacteria A be closest to 10?
(1) 24 hours.
(2) 48 hours.
(3) 69 hours.
(4) 96 hours.
(5) 117 hours.

II. Multiple-Choice Questions (30 points)

Instructions: For questions 6 to 11, each of the five options is independent, and at least one option is correct. Select the correct options and mark them on the "Answer Section" of the answer sheet. No points are deducted for wrong answers. Five points are awarded for all five options correct, 2.5 points for only one wrong option, and no points for two or more wrong options.

10. Let $a$ be a real number greater than 1. Consider the functions $f(x) = a^x$ and $g(x) = \log_a x$. Which of the following options are correct?
(1) If $f(3) = 6$, then $g(36) = 6$
(2) $\frac{f(238)}{f(219)} = \frac{f(38)}{f(19)}$
(3) $g(238) - g(219) = g(38) - g(19)$
(4) If $P, Q$ are two distinct points on the graph of $y = g(x)$, then the slope of line $PQ$ must be positive
(5) If the line $y = 5x$ and the graph of $y = f(x)$ have two intersection points, then the line $y = \frac{1}{5}x$ and the graph of $y = g(x)$ also have two intersection points

1. For any real number $x$, the minimum value of $27 ^ { \left( x ^ { 2 } + \frac { 2 } { 3 } \right) }$ is
(1) 3
(2) $3 \sqrt { 3 }$
(3) 9
(4) 27
(5) $81 \sqrt { 3 }$

5. In a sealed laboratory, initially there are 1000 bacteria of a certain species, and they reproduce at a rate of 8\% per hour. If reproduction continues at this rate, which of the following options best approximates the number of bacteria after 100 hours?
(1) 9 thousand
(2) 108 thousand
(3) 2200 thousand
(4) 3200 thousand
(5) 32000 thousand

In the early stages of an infectious disease outbreak, since most people have not been infected and have no antibodies, the total number of infected people usually grows exponentially. Under the premise that ``the initial number of infected people is $P _ { 0 }$ , and each infected person on average infects $r$ uninfected people per day'', the total number of people infected with the disease after $n$ days, $P _ { n }$ , can be expressed as
$$P _ { n } = P _ { 0 } ( 1 + r ) ^ { n } \text {, where } P _ { 0 } \geq 1 \text { and } r > 0 \text { . }$$
Answer the following questions:
(1) Given that $A = \frac { \log P _ { 5 } - \log P _ { 2 } } { 3 } , B = \frac { \log P _ { 8 } - \log P _ { 6 } } { 2 }$ , show that $A = B$ . (4 points)
(2) Given that a certain infectious disease in its early stages follows the above mathematical model and the total number of infected people increases tenfold every 16 days, find the value of $\frac { P _ { 20 } } { P _ { 17 } } \times \frac { P _ { 8 } } { P _ { 6 } } \times \frac { P _ { 5 } } { P _ { 2 } }$ . (5 points)
(3) Based on (2), find the value of $\frac { \log P _ { 20 } - \log P _ { 17 } } { 3 }$ . (4 points)

On the coordinate plane, $\Gamma$ is a square with side length 4, centered at the point $(1,1)$, with sides parallel to the coordinate axes. The graph of the function $y = a \times 2 ^ { x }$ intersects $\Gamma$, where $a$ is a real number. The maximum possible range of $a$ is (22)(23) $\leq a \leq$ (24).

Let $x _ { 0 }$、$y _ { 0 }$ be positive real numbers. If the point $\left( 10 x _ { 0 } , 100 y _ { 0 } \right)$ on the coordinate plane lies on the graph of the function $y = 10 ^ { x }$ , then the point $\left( x _ { 0 } , \log y _ { 0 } \right)$ will lie on the graph of the line $y = a x + b$ , where $a$、$b$ are real numbers. What is the value of $2 a - b$?
(1) 4
(2) 9
(3) 15
(4) 18
(5) 22

A water pumping station found that its electricity consumption (unit: kilowatt-hours) is directly proportional to the cube of the pump motor speed (unit: rpm). Based on this, which of the following five graphs best describes the relationship between the electricity consumption $y$ (kilowatt-hours) and the pump motor speed $X$ (rpm) of this water pumping station?
(1) [Graph 1] (2) [Graph 2] (3) [Graph 3] (4) [Graph 4] (5) [Graph 5]

An organization introduced two different nutrients into culture dishes A and B at 12 o'clock. At this time, the bacterial counts in dishes A and B are $X$ and $Y$ respectively. The quantity in dish A doubles every 3 hours; for example, at 3 PM the quantity in A is $2X$. The quantity in dish B doubles every 2 hours; for example, at 2 PM the quantity in B is $2Y$, and at 4 PM the quantity in B is $4Y$. Part of the measurement results are recorded in the table below. At 6 PM, the organization measured that the quantities in dishes A and B are the same. To estimate the bacterial quantities in dishes A and B from 12 o'clock to 12 midnight using an exponential growth model, select the correct options.

Time (o'clock)	12	13	14	15	16	17	18	19	20	21	22	23	24
Quantity in A	$X$			$2X$
Quantity in B	$Y$		$2Y$		$4Y$

(1) $X > Y$ (2) At 1 PM, the quantity in A is $\frac{4}{3}X$ (3) At 3 PM, the quantity in B is $3Y$ (4) At 7 PM, the quantity in B is 1.5 times that of A (5) At 12 midnight, the quantity in B is 2 times that of A

Research shows that the residual amount of a certain drug in a user's body decreases exponentially over time after taking the drug. It is known that 2 hours after taking the drug, half of the drug dose remains in the body. Which of the following options is correct?
(1) After 3 hours, the body still retains $\frac{1}{3}$ of the drug dose
(2) After 4 hours, the body still retains $\frac{1}{4}$ of the drug dose
(3) After 6 hours, the body still retains $\frac{1}{6}$ of the drug dose
(4) After 8 hours, the body still retains $\frac{1}{8}$ of the drug dose
(5) After 10 hours, the body still retains $\frac{1}{10}$ of the drug dose

Over the past five years, a country's total carbon emissions decreased from $X$ billion metric tons of CO2 equivalent (CO2e) in year 1 to $Y$ billion metric tons of CO2 equivalent (CO2e) in year 5, achieving an average annual carbon reduction of 5\%, that is, $Y = ( 1 - 0.05 ) ^ { 4 } X$. The five-year carbon emission totals and annual growth rates are recorded in the following table, where Year $n$ carbon emission growth rate $= \frac { \text {(Year } n \text { carbon emission total)} - \text {(Year } n - 1 \text { carbon emission total)} } { \text {Year } n - 1 \text { carbon emission total} }$, $n = 2,3,4,5$.

	Year 1	Year 2	Year 3	Year 4	Year 5
\begin{tabular}{ c } Carbon Emission Total
$($ billion metric tons $\mathrm { CO } 2 \mathrm { e } )$

& $X$ & $A$ & $B$ & $C$ & $Y$ \hline Annual Carbon Emission Growth Rate & & - 0.07 & $p$ & $q$ & $r$ \hline \end{tabular}
Select the correct options.
(1) $A = 0.93 X$
(2) $Y \leq 0.8 X$
(3) $\frac { - 0.07 + p + q + r } { 4 } = - 0.05$
(4) $\sqrt [ 4 ] { \frac { Y } { X } } - 1 = - 0.05$
(5) $0.93 ( 1 + p ) ( 1 + q ) ( 1 + r ) = ( 0.95 ) ^ { 4 }$

Let the exponential function $f(x) = 1.2^{x}$. Select the correct options.
(1) $f(0) > 0$
(2) $f(10) > 10$
(3) On the coordinate plane, the graph of $y = 1.2^{x}$ intersects the line $y = x$
(4) On the coordinate plane, the graphs of $y = 1.2^{x}$ and $y = \log(1.2^{x})$ are symmetric about the line $y = x$
(5) For any positive real number $b$, $\log_{1.2} b \neq 1.2^{b}$