18. Let $f$ be a function on the positive real numbers such that $f ( x y ) = f ( x ) + f ( y )$. If $f ( 2024 ) = 2$ then which of the following statement(s) is/ are true?
(a) $f \left( \frac { 1 } { 2024 } \right) = 1$
(b) $f \left( \frac { 1 } { 2024 } \right) = - 1$
(c) $f \left( \frac { 1 } { 2024 } \right) = - 2$
(d) $f \left( \frac { 1 } { 2024 } \right) = 2$

The following description is for questions 19 and 20.

A perfect shuffle of a deck of cards divides the deck into two equal parts and then interleaves the cards from each half, starting with the first card of the first half.
For instance, if we shuffle a deck of cards containing 10 cards arranged $[ 1,2,3,4,5,6,7,8,9,10 ]$ we first create two equal decks with cards $[ 1,2,3,4,5 ]$ and $[ 6,7,8,9,10 ]$ and then interleave them to get a new deck $[ 1,6,2,7,3,8,4,9,5,10 ]$.

When two constants $a , b$ satisfy $\lim _ { x \rightarrow 2 } \frac { x ^ { 2 } - ( a + 2 ) x + 2 a } { x ^ { 2 } - b } = 3$, what is the value of $a + b$? [2 points]
(1) $- 6$
(2) $- 4$
(3) $- 2$
(4) 0
(5) 2

Consider the function $$f ( x ) = \begin{cases} - 3 x + a & ( x \leq 1 ) \\ \frac { x + b } { \sqrt { x + 3 } - 2 } & ( x > 1 ) \end{cases}$$ If $f ( x )$ is continuous on the entire set of real numbers, find the value of $a + b$. (Here, $a$ and $b$ are constants.) [4 points]

It is known that the function $f(x)$ is an odd function defined on $\mathbb{R}$. When $x \in (-\infty, 0)$, $f(x) = 2x^3 + x^2$. Then $f(2) = $ \_\_\_\_

6. Let $f ( x )$ be an odd function, and when $x \geq 0$, $f ( x ) = \mathrm { e } ^ { x } - 1$. Then when $x < 0$, $f ( x ) =$
A. $\mathrm { e } ^ { - x } - 1$
B. $\mathrm { e } ^ { - x } + 1$
C. $- \mathrm { e } ^ { - x } - 1$
D. $- \mathrm { e } ^ { - x } + 1$

If $f ( x ) = \ln \left| a + \frac { 1 } { 1 - x } \right| + b$ is an odd function, then $a = $ $\_\_\_\_$ . $b = $ $\_\_\_\_$ .

107- If $f(x) = 2x + 3$ and $g(f(x)) = 8x^2 + 22x + 20$, what is the rule of $f \circ g$?

• [(1)] $2x^2 - 7x + 3$
• [(2)] $2x^2 - x + 7$
• [(3)] $4x^2 - 2x + 13$
• [(4)] $4x^2 - 4x + 11$

108- If $g(x)=2x-3$ and $(f\circ g)(x)=4(x^2-4x+5)$, what is $f(x)$?
(1) $x^2-4x+3$ (2) $x^2-4x+5$ (3) $x^2-2x+5$ (4) $x^2-2x+3$

110- The figure below shows the graph of the function $y = \sin^{-1}(U(x))$. What is the rule $U(x)$?
[Figure: graph of $y = \sin^{-1}(U(x))$ with a point marked at $x = -1$ and $x = 3$]

p{6cm}} (2) $\dfrac{2}{1-x}$	(1) $\dfrac{2}{x-1}$
[18pt] (4) $\dfrac{1}{2-x}$	(3) $\dfrac{1}{x-2}$

111- What is the value of the expression $169\sin\!\left(2\cos^{-1}\!\left(-\dfrac{5}{13}\right)\right)$?

p{6cm}} (2) $60$	(1) $-120$
[6pt] (4) $120$	(3) $-60$

Let $f : [-1,1] \rightarrow \mathbb{R}$ be a function such that $f\left(\sin\frac{x}{2}\right) = \sin x + \cos x$, for all $x \in [-\pi, \pi]$. The value of $f\left(\frac{3}{5}\right)$ is
(A) $\frac{24}{25}$
(B) $\frac{31}{25}$
(C) $\frac{33}{25}$
(D) $\frac{7}{5}$.

23. If $g ( f ( x ) ) = | \sin x |$ and $f ( g ( x ) ) = ( \sin \sqrt { } x ) 2$, then :
(A) $f ( x ) = \sin 2 x , g ( x ) = \sqrt { } x$
(B) $f ( x ) = \sin x , g ( x ) = | x |$
(C) $f ( x ) = x 2 , g ( x ) = \sin \sqrt { } x$
(D) $f$ and $g$ cannot be determined

25. Let $f ( x ) = a x / ( x + 1 ) , x \neq - 1$. Then for what value of $a$ is $f [ f ( x ) ] = x$ :
(A) $\sqrt { } 2$
(B) $- \sqrt { } 2$
(C) 1
(D) - 1

If $g(x) = x^{2} + x - 2$ and $\frac{1}{2}\,g\circ f(x) = 2x^{2} - 5x + 2$, then $f(x)$ is equal to
(1) $2x-3$
(2) $2x+3$
(3) $2x^{2}+3x+1$
(4) $2x^{2}-3x-1$

Let $f ( x ) = 2 ^ { 10 } x + 1$ and $g ( x ) = 3 ^ { 10 } x - 1$. If $( f \circ g ) ( x ) = x$, then $x$ is equal to:
(1) $\frac { 2 ^ { 10 } - 1 } { 2 ^ { 10 } - 3 ^ { - 10 } }$
(2) $\frac { 1 - 2 ^ { - 10 } } { 3 ^ { 10 } - 2 ^ { - 10 } }$
(3) $\frac { 3 ^ { 10 } - 1 } { 3 ^ { 10 } - 2 ^ { - 10 } }$
(4) $\frac { 1 - 3 ^ { - 10 } } { 2 ^ { 10 } - 3 ^ { - 10 } }$

If $g(x) = x ^ { 2 } + x - 1$ and $(g \circ f)(x) = 4x ^ { 2 } - 10x + 5$, then $f \left( \frac { 5 } { 4 } \right)$ is equal to
(1) $\frac { 3 } { 2 }$
(2) $- \frac { 1 } { 2 }$
(3) $\frac { 1 } { 2 }$
(4) $- \frac { 3 } { 2 }$

Let $f : R - \left\{ \frac { \alpha } { 6 } \right\} \rightarrow R$ be defined by $f ( x ) = \left( \frac { 5 x + 3 } { 6 x - \alpha } \right)$. Then the value of $\alpha$ for which $( f \circ f ) ( x ) = x$, for all $x \in R - \left\{ \frac { \alpha } { 6 } \right\}$, is
(1) No such $\alpha$ exists
(2) 5
(3) 8
(4) 6

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a differentiable function that satisfies the relation $f ( x + y ) = f ( x ) + f ( y ) - 1 , \forall x , y \in \mathbb { R }$. If $f ^ { \prime } ( 0 ) = 2$, then $| f ( - 2 ) |$ is equal to

Let $f: R - \{0\} \rightarrow R$ be a function satisfying $f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$ for all $x, y$, $f(y) \neq 0$. If $f'(1) = 2024$, then
(1) $xf'(x) - 2024f(x) = 0$
(2) $xf'(x) + 2024f(x) = 0$
(3) $xf'(x) + f(x) = 2024$
(4) $xf'(x) - 2023f(x) = 0$

Let $f : \mathbb{R} - \{0\} \rightarrow \mathbb{R}$ be a function such that $f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$. If $\lim_{x \rightarrow 0}\left(\frac{1}{\alpha x} + f(x)\right) = \beta$; $\alpha, \beta \in \mathbb{R}$, then $\alpha + 2\beta$ is equal to
(1) 5
(2) 3
(3) 4
(4) 6

Let $f : \mathbf { R } - \{ 0 \} \rightarrow ( - \infty , 1 )$ be a polynomial of degree 2, satisfying $f ( x ) f \left( \frac { 1 } { x } \right) = f ( x ) + f \left( \frac { 1 } { x } \right)$. If $f ( K ) = - 2 K$, then the sum of squares of all possible values of $K$ is :
(1) 7
(2) 6
(3) 1
(4) 9

Let $f$ be a function such that $3 f ( x ) + 2 f \left( \frac { m } { 19 x } \right) = 5 x , x \neq 0$ where $m = \sum _ { i = 1 } ^ { 9 } ( i ) ^ { 2 }$, then $f ( 5 ) - f ( 2 )$ is equal to

If $g ( x ) = 3 x ^ { 2 } + 2 x - 3 , f ( 0 ) = - 3, 4 g ( f ( x ) ) = 3 x ^ { 2 } - 32 x + 72$. Then $\mathrm { f } ( \mathrm { g } ( 2 ) )$ is equal to (A) $- \frac { 25 } { 6 }$ (B) $\frac { 25 } { 6 }$ (C) $\frac { 7 } { 2 }$ (D) $\frac { 5 } { 2 }$

2. For ALL APPLICANTS.

(i) Let $k \neq \pm 1$. The function $f ( t )$ satisfies the identity
$$f ( t ) - k f ( 1 - t ) = t$$
for all values of $t$. By replacing $t$ with $1 - t$, determine $f ( t )$.
(ii) Consider the new identity
$$f ( t ) - f ( 1 - t ) = g ( t )$$
(a) Show that no function $f ( t )$ satisfies $( * )$ when $g ( t ) = t$.
(b) What condition must the function $g ( t )$ satisfy for there to be a solution $f ( t )$ to $( * )$ ?
(c) Find a solution $f ( t )$ to $( * )$ when $g ( t ) = ( 2 t - 1 ) ^ { 3 }$.

$$f\left(\frac{x-1}{x+1}\right) = x^{2} - x + 2$$
Given this, what is the value of $f(3)$?
A) 5
B) 6
C) 7
D) 8
E) 11

$$f ( 2 x + 5 ) = \tan \left( \frac { \pi } { 2 } x \right)$$
For the function $f$ given by the equality, what is the value $f ^ { -1 } ( 1 )$?
A) $\frac { \pi } { 2 }$
B) $\frac { \pi } { 4 }$
C) $\pi$
D) $2 \pi$
E) $3 \pi$