Set 1

Given the operation $\triangle(a, b) = ka + \frac{b}{m}$ where $k$ and $m$ are positive integers. If $\triangle(2, 4) = 8$ and $\triangle(3, 5) = 11$ , then the value of $\triangle(5, 9)$ is ....

Given $A$ , $B$ , and $D$ are matrices with

A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} a & 2 \\ b & 1 \end{pmatrix}, \quad D = 3A + B

If $AB = \begin{pmatrix} 1 & 3 \\ 17 & 11 \end{pmatrix}$ , then matrix $D$ is ....

$\begin{pmatrix} 4 & 1 \\ 7 & 9 \end{pmatrix}$

$\begin{pmatrix} 8 & -1 \\ 15 & 10 \end{pmatrix}$

$\begin{pmatrix} 9 & 4 \\ 14 & 10 \end{pmatrix}$

$\begin{pmatrix} 8 & 5 \\ 13 & 6 \end{pmatrix}$

$\begin{pmatrix} 11 & 5 \\ 10 & 6 \end{pmatrix}$

The sequence $\frac{3}{2}, p, 6, q$ is a geometric sequence with a positive ratio.

Which statements are true based on the information above?

$\text{(1)}$ The ratio of the sequence is $2$ .

$\text{(2)}$ The value of $pq$ is $72$ .

$\text{(3)}$ The difference between the sixth and second term is $45$ .

$\text{(4)}$ Every term in the sequence is an even number.

$(1), (2), \text{ and } (3)$

$(1) \text{ and } (3)$

$(2) \text{ and } (4)$

$(4)$

$(1), (2), (3), \text{ and } (4)$

If the size of each small square on the map is $2 \text{ cm} \times 2 \text{ cm}$ , the map scale is $1 : 200,000$ , and an airplane takes $12 \text{ minutes}$ to fly from City B to City A with a map distance of $12 \text{ cm}$ , then the average speed of the airplane is ....

$24 \text{ km/h}$

$48 \text{ km/h}$

$72 \text{ km/h}$

$96 \text{ km/h}$

$120 \text{ km/h}$

Rhombus $PQRS$ has a perimeter of $20$ units. Diagonal $PR$ is perpendicular to diagonal $QS$ . The gradient of $QR$ is ....

Rhombus PQRS

Diagram of rhombus

PQRS

on the coordinate plane.

Rhombus $PQRS$ has a perimeter of $20$ units. Diagonal $PR$ is perpendicular to diagonal $QS$ . The equation of the line that passes through $Q$ and is perpendicular to $PS$ is ....

Rhombus PQRS

Diagram of rhombus

PQRS

on the coordinate plane.

Given the trigonometric ratios of two acute angles $A$ and $B$ with $\sin A = \frac{5}{13}$ and $\cos B = \frac{8}{17}$ .

How many of the following four statements are true based on the information above?

\text{(1)} \quad \tan A + \tan B = \frac{19}{20}

\text{(2)} \quad \cos A \sin B + \cos B \sin A = \frac{220}{221}

\text{(3)} \quad \frac{\tan A}{\tan B} = \frac{25}{32}

\text{(4)} \quad \frac{\sin A \cos B}{\sin B \cos A} = \frac{2}{9}

Explanation

Given $\sin A = \frac{5}{13}$ and $\cos B = \frac{8}{17}$ where $A$ and $B$ are acute angles.

From $\sin A = \frac{5}{13}$ , the opposite side is $5$ and the hypotenuse is $13$ . Then the adjacent side is

\text{Adjacent side} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12

So for angle $A$

\sin A = \frac{5}{13}, \quad \cos A = \frac{12}{13}, \quad \tan A = \frac{5}{12}

Angle A

Visualization of angle

A

with

\sin A = \frac{5}{13}

\cos A = \frac{12}{13}

, and

\tan A = \frac{5}{12}

Sin (22.62°) = 5/13Cos (22.62°) = 12/13Tan (22.62°) = 5/12

22.62°0.39 Radian

From $\cos B = \frac{8}{17}$ , the adjacent side is $8$ and the hypotenuse is $17$ . Then the opposite side is

\text{Opposite side} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15

So for angle $B$

\sin B = \frac{15}{17}, \quad \cos B = \frac{8}{17}, \quad \tan B = \frac{15}{8}

Angle B

Visualization of angle

B

with

\sin B = \frac{15}{17}

\cos B = \frac{8}{17}

, and

\tan B = \frac{15}{8}

Sin (61.93°) = 15/17Cos (61.93°) = 8/17Tan (61.93°) = 15/8

61.93°1.08 Radian

Statement $\text{(1)}$ : $\tan A + \tan B = \frac{19}{20}$

\tan A + \tan B = \frac{5}{12} + \frac{15}{8}

= \frac{10}{24} + \frac{45}{24}

= \frac{55}{24}

So $\tan A + \tan B = \frac{55}{24}$ , not $\frac{19}{20}$ . Statement $\text{(1)}$ is false.

Statement $\text{(2)}$ : $\cos A \sin B + \cos B \sin A = \frac{220}{221}$

\cos A \sin B + \cos B \sin A = \frac{12}{13} \cdot \frac{15}{17} + \frac{8}{17} \cdot \frac{5}{13}

= \frac{180}{221} + \frac{40}{221}

= \frac{220}{221}

Statement $\text{(2)}$ is true.

Statement $\text{(3)}$ : $\frac{\tan A}{\tan B} = \frac{25}{32}$

\frac{\tan A}{\tan B} = \frac{\frac{5}{12}}{\frac{15}{8}}

= \frac{5}{12} \cdot \frac{8}{15}

= \frac{40}{180}

= \frac{2}{9}

So $\frac{\tan A}{\tan B} = \frac{2}{9}$ , not $\frac{25}{32}$ . Statement $\text{(3)}$ is false.

Statement $\text{(4)}$ : $\frac{\sin A \cos B}{\sin B \cos A} = \frac{2}{9}$

\frac{\sin A \cos B}{\sin B \cos A} = \frac{\sin A}{\cos A} \cdot \frac{\cos B}{\sin B}

= \tan A \cdot \frac{1}{\tan B}

= \frac{\tan A}{\tan B}

= \frac{2}{9}

Statement $\text{(4)}$ is true.

Therefore, the true statements are $\text{(2)}$ and $\text{(4)}$ . The number of true statements is $2$ .

Given that $a$ and $b$ are natural numbers that satisfy $a = 2 \times b$ and $b$ is divisible by $5$ .

How many of the following four statements are true based on the information above?

$(1)$ $a$ is an even number

$(2)$ The unit digit of $a$ is always $0$

$(3)$ $a + b$ is divisible by $5$

$(4)$ $3a + 2b$ is an odd number

The graph of quadratic function $f(x) = ax^2 + bx + c$ intersects the $y$ -axis at point $(0, 8)$ and has a vertex at $(2, -4)$ .

Based on the information above, the true statements are

$\text{(I)}$ The complete equation of the quadratic function is $f(x) = 3x^2 - 12x + 8$ .

$\text{(II)}$ The point $(3, -1)$ lies on function $f$ .

$\text{(III)}$ The value of $f(-1)$ is $22$ .

$\text{I, II, and III}$

$\text{I and II}$

$\text{II and III}$

$\text{I}$

$\text{III}$

Explanation

Given that the graph of quadratic function $f(x) = ax^2 + bx + c$ intersects the $y$ -axis at point $(0, 8)$ and has a vertex at $(2, -4)$ .

From point $(0, 8)$ , substitute $x = 0$ and $f(0) = 8$ into the function equation

f(0) = a(0)^2 + b(0) + c

8 = c

So $c = 8$ .

From the vertex $(2, -4)$ , the $x$ -coordinate of the vertex is $x_p = -\frac{b}{2a} = 2$ , so

-\frac{b}{2a} = 2

-b = 4a

b = -4a

The $y$ -coordinate of the vertex is $f(2) = -4$ , so

f(2) = a(2)^2 + b(2) + c

-4 = 4a + 2b + 8

-12 = 4a + 2b

Substitute $b = -4a$ into the equation $-12 = 4a + 2b$

-12 = 4a + 2(-4a)

-12 = 4a - 8a

-12 = -4a

a = 3

From $b = -4a$ , we get $b = -4(3) = -12$ .

So the quadratic function equation is $f(x) = 3x^2 - 12x + 8$ .

Statement $\text{(I)}$ : The complete equation of the quadratic function is $f(x) = 3x^2 - 12x + 8$

Based on the calculation above, the quadratic function equation is $f(x) = 3x^2 - 12x + 8$ . Therefore, statement $\text{(I)}$ is true.

Statement $\text{(II)}$ : The point $(3, -1)$ lies on function $f$

Substitute $x = 3$ into the function $f(x) = 3x^2 - 12x + 8$

f(3) = 3(3)^2 - 12(3) + 8

= 3(9) - 36 + 8

= 27 - 36 + 8

= -1

So $f(3) = -1$ , which means the point $(3, -1)$ lies on function $f$ . Therefore, statement $\text{(II)}$ is true.

Statement $\text{(III)}$ : The value of $f(-1)$ is $22$

Substitute $x = -1$ into the function $f(x) = 3x^2 - 12x + 8$

f(-1) = 3(-1)^2 - 12(-1) + 8

= 3(1) + 12 + 8

= 3 + 12 + 8

= 23

So $f(-1) = 23$ , not $22$ . Therefore, statement $\text{(III)}$ is false.

Therefore, the true statements are $\text{(I)}$ and $\text{(II)}$ .

Function $f$ is expressed as $f(x) = \frac{1}{\sqrt{-x^2 + 4x + 21}}$ .

Based on the given information, which relationship between quantities $P$ and $Q$ is correct?

$P$	$Q$
$7$	$x$ such that $f(x)$ is defined as a real number

Quantity $P$ is greater than $Q$

Quantity $P$ is less than $Q$

Quantity $P$ is equal to $Q$

The information is insufficient to determine the relationship

Cannot be determined

The following table presents the mathematics scores of three student groups.

Group 1	Group 2	Group 3
$8$	$6$	$3$
$2$	$9$	$b$
$5$	$7$	$10$
$8$		$8$
$7$

The sum of the mode of Group 1's scores and the median of Group 2's scores is equal to $2$ times the average of Group 3's scores.

Based on the given information, which of the following relationships between quantities $P$ and $Q$ is correct?

$P$	$Q$
$b$	$8$

Quantity $P$ is greater than $Q$

Quantity $P$ is less than $Q$

Quantity $P$ is equal to $Q$

The information is insufficient to determine the relationship

Cannot be determined

Given a regular pyramid $T.ABCD$ with $AB = 4 \text{ cm}$ and point $O$ is the intersection point of lines $AC$ and $BD$ . What is the volume of the pyramid $T.ABCD$ ?

Regular Pyramid T.ABCD

Visualization of regular pyramid

T.ABCD

with square base

ABCD

and center of base at point

O

Decide whether the following statements $(1)$ and $(2)$ are sufficient to answer the question.

(1) \quad AC = 4\sqrt{2} \text{ cm}

(2) \quad AT = \sqrt{17} \text{ cm}

Statement $(1)$ ALONE is sufficient, but statement $(2)$ ALONE is not sufficient

Statement $(2)$ ALONE is sufficient, but statement $(1)$ ALONE is not sufficient

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient

EACH statement ALONE is sufficient

Statements $(1)$ and $(2)$ TOGETHER are NOT sufficient

Explanation

Given a regular pyramid $T.ABCD$ with $AB = 4 \text{ cm}$ and point $O$ is the intersection point of lines $AC$ and $BD$ .

The volume of the pyramid can be calculated using the formula:

V = \frac{1}{3} \times L_{\text{base}} \times h = \frac{1}{3} \times s^2 \times h

where $s$ is the side length of the base and $h$ is the height of the pyramid.

Since $AB = 4 \text{ cm}$ , the base area $L_{\text{base}} = 4^2 = 16 \text{ cm}^2$ .

To calculate the volume, we need to know the height of the pyramid $h = TO$ .

Analysis of Statement $(1)$ : $AC = 4\sqrt{2} \text{ cm}$

The diagonal of a square with side $s$ is $s\sqrt{2}$ .

If $AC = 4\sqrt{2} \text{ cm}$ , then:

s\sqrt{2} = 4\sqrt{2}

s = 4

This is consistent with the given information ( $AB = 4 \text{ cm}$ ).

Point $O$ is the midpoint of the diagonal, so:

AO = OC = \frac{AC}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \text{ cm}

However, statement $(1)$ does not provide information about the height of the pyramid $TO$ . Without the height, the volume cannot be determined.

Therefore, statement $(1)$ ALONE is not sufficient.

Analysis of Statement $(2)$ : $AT = \sqrt{17} \text{ cm}$

Since $AB = 4 \text{ cm}$ , the base diagonal $AC = 4\sqrt{2} \text{ cm}$ .

Point $O$ is the midpoint of the diagonal, so:

AO = \frac{AC}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \text{ cm}

Consider the right triangle $AOT$ with a right angle at $O$ (since $TO$ is perpendicular to the base).

Using the Pythagorean theorem:

AT^2 = AO^2 + TO^2

(\sqrt{17})^2 = (2\sqrt{2})^2 + TO^2

17 = 8 + TO^2

TO^2 = 9

TO = 3 \text{ cm}

With height $TO = 3 \text{ cm}$ and base area $L_{\text{base}} = 16 \text{ cm}^2$ , the volume of the pyramid can be calculated:

V = \frac{1}{3} \times 16 \times 3 = 16 \text{ cm}^3

Therefore, statement $(2)$ ALONE is sufficient to answer the question.

Regular Pyramid T.ABCD with Side Length Labels

Visualization of regular pyramid

T.ABCD

with labels for side lengths.

Conclusion:

Statement $(1)$ ALONE is not sufficient, but statement $(2)$ ALONE is sufficient to answer the question.

Functions $f$ and $g$ are defined as $f(x) = \frac{x-a}{x+a}$ for $x \neq -a$ and $g(x) = bx^2 - x + 2$ . Is $\frac{g(-2)}{f(-2)} > 0$ ?

Decide whether statements $(1)$ and $(2)$ below are sufficient to answer the question.

(1) \quad a < 5 \text{ and } b < 5

(2) \quad a > 2 \text{ and } b > 0

Statement $(1)$ ALONE is sufficient, but statement $(2)$ ALONE is not sufficient

Statement $(2)$ ALONE is sufficient, but statement $(1)$ ALONE is not sufficient

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient

EACH statement ALONE is sufficient

Statements $(1)$ and $(2)$ TOGETHER are NOT sufficient

Explanation

Given functions $f(x) = \frac{x-a}{x+a}$ for $x \neq -a$ and $g(x) = bx^2 - x + 2$ .

The question asks whether $\frac{g(-2)}{f(-2)} > 0$ .

Calculate Function Values

First, calculate the values of $g(-2)$ and $f(-2)$ .

For $g(-2)$ , substitute $x = -2$ into function $g(x)$ :

g(-2) = b(-2)^2 - (-2) + 2

g(-2) = 4b + 2 + 2

g(-2) = 4b + 4

For $f(-2)$ , substitute $x = -2$ into function $f(x)$ :

f(-2) = \frac{-2-a}{-2+a}

f(-2) = \frac{-(2+a)}{-(2-a)}

f(-2) = \frac{2+a}{2-a}

Therefore, the ratio we are looking for is:

\frac{g(-2)}{f(-2)} = \frac{4b + 4}{\frac{2+a}{2-a}} = \frac{(4b + 4)(2-a)}{2+a}

We can simplify by factoring the numerator:

\frac{g(-2)}{f(-2)} = \frac{4(b + 1)(2-a)}{2+a}

For $\frac{g(-2)}{f(-2)} > 0$ , we need $\frac{4(b + 1)(2-a)}{2+a} > 0$ .

Since $4$ is always positive, we need to analyze the sign of $\frac{(b + 1)(2-a)}{2+a}$ .

Statement Analysis

Analysis of Statement $(1)$ : $a < 5$ and $b < 5$

From this statement, we only know that both $a$ and $b$ are less than $5$ . However, there is no definite information about the signs of the factors involved.

Let's check some possibilities:

If $b = 0$ , then $b + 1 = 1 > 0$
If $b = -2$ , then $b + 1 = -1 < 0$
For $2-a$ , if $a = 1$ , then $2-a = 1 > 0$
For $2-a$ , if $a = 3$ , then $2-a = -1 < 0$
For $2+a$ , if $a = 0$ , then $2+a = 2 > 0$
For $2+a$ , if $a = -3$ , then $2+a = -1 < 0$

Since the signs of these factors can vary with many possible combinations, we cannot determine whether $\frac{g(-2)}{f(-2)} > 0$ .

Therefore, statement $(1)$ ALONE is not sufficient.

Analysis of Statement $(2)$ : $a > 2$ and $b > 0$

From this statement, let's analyze the sign of each factor:

Since $b > 0$ , then $b + 1 > 1 > 0$ (positive)
Since $a > 2$ , then $2 - a < 0$ (negative)
Since $a > 2$ , then $2 + a > 4 > 0$ (positive)

Thus, we have:

\frac{(b + 1)(2-a)}{2+a} = \frac{(+)(-)}{(+)} = \frac{(-)}{(+)} = (-)

Since the result is negative, $\frac{g(-2)}{f(-2)} < 0$ , so the answer is no, $\frac{g(-2)}{f(-2)}$ is not greater than $0$ .

Therefore, statement $(2)$ ALONE is sufficient to answer the question definitively (the answer is no).

If $N$ satisfies $7 - N \times (-3) = 52$ , the value of $N$ is ....

Triangle $ABC$ is similar to triangle $ADE$ .

Triangle ABC and ADE

Visualization of two similar triangles with marked vertices.

The area of triangle $ADE$ is ....

The result of

\frac{x + 2y}{3} + \frac{2x - y}{4} - \frac{3x + 2y}{6}

is ....

$\frac{x + y}{12}$

$\frac{2x + y}{12}$

$\frac{3x + y}{12}$

$\frac{4x + y}{12}$

$\frac{5x + y}{12}$

Given the function $f(x) = (x+3)(x^2+6x+8)^4$ and $F(x) = \int f(x) \, dx$ . If $F(-3) = 2$ , then the function $F(x)$ is ....

$\frac{1}{10}(x^2+6x+8)^5 + \frac{1}{10}$

$\frac{1}{10}(x^2+6x+8)^5 + \frac{11}{10}$

$\frac{1}{10}(x^2+6x+8)^5 + \frac{21}{10}$

$\frac{1}{10}(x^2+6x+8)^5 + \frac{31}{10}$

$\frac{1}{10}(x^2+6x+8)^5 + \frac{41}{10}$

Explanation

Given the function $f(x) = (x+3)(x^2+6x+8)^4$ and $F(x) = \int f(x) \, dx$ with the condition $F(-3) = 2$ .

U-Substitution

To solve this integral, we use the substitution method.

Let $u = x^2 + 6x + 8$ .

The derivative of $u$ with respect to $x$ :

\frac{du}{dx} = 2x + 6

Factor:

\frac{du}{dx} = 2(x + 3)

Multiply both sides by $dx$ :

du = 2(x + 3) \, dx

Divide both sides by $2$ :

\frac{du}{2} = (x + 3) \, dx

Solving the Integral

Substitute into the integral:

F(x) = \int f(x) \, dx = \int (x+3)(x^2+6x+8)^4 \, dx

With substitution $u = x^2 + 6x + 8$ and $(x+3) \, dx = \frac{du}{2}$ :

F(x) = \int u^4 \cdot \frac{du}{2}

= \frac{1}{2} \int u^4 \, du

= \frac{1}{2} \cdot \frac{u^5}{5} + C

= \frac{1}{10} u^5 + C

Substitute back $u = x^2 + 6x + 8$ :

F(x) = \frac{1}{10}(x^2 + 6x + 8)^5 + C

Determining the Constant C

Use the condition $F(-3) = 2$ :

F(-3) = \frac{1}{10}((-3)^2 + 6(-3) + 8)^5 + C = 2

Calculate the value inside the parentheses:

(-3)^2 + 6(-3) + 8 = 9 - 18 + 8 = -1

Substitute:

\frac{1}{10}(-1)^5 + C = 2

\frac{1}{10}(-1) + C = 2

-\frac{1}{10} + C = 2

C = 2 + \frac{1}{10}

C = \frac{20}{10} + \frac{1}{10}

C = \frac{21}{10}

Function F(x)

Substitute the value of $C$ into $F(x)$ :

F(x) = \frac{1}{10}(x^2 + 6x + 8)^5 + \frac{21}{10}

Therefore, the function $F(x)$ is $\frac{1}{10}(x^2 + 6x + 8)^5 + \frac{21}{10}$ .

The number sequence $5, 8, 11, \ldots$ is an arithmetic sequence where $a$ is the first term, $b$ is the common difference, $U_n$ is the $n$ -th term, and $S_n$ is the sum of the first $n$ terms.

Which statements are true based on the information above?

(1) \quad \text{The value of } a \text{ is greater than the value of } b

(2) \quad U_{11} = a \times 7

(3) \quad S_6 = b \times 25

(4) \quad U_{18} = S_5

$(1), (2), \text{ and } (3)$ ONLY are true

$(1) \text{ and } (3)$ ONLY are true

$(2) \text{ and } (4)$ ONLY are true

ALL statements are true

ALL statements are false

Explanation

Given the arithmetic sequence $5, 8, 11, \ldots$ .

From the sequence:

First term $a = 5$
Common difference $b = 8 - 5 = 3$

Formula for the $n$ -th term of an arithmetic sequence:

U_n = a + (n-1)b

Formula for the sum of the first $n$ terms:

S_n = \frac{n}{2}(2a + (n-1)b)

Evaluating Statement (1)

Statement: The value of $a$ is greater than the value of $b$ .

a = 5 \text{ and } b = 3

5 > 3

Therefore, statement $(1)$ is true.

Evaluating Statement (2)

Statement: $U_{11} = a \times 7$ .

Calculate $U_{11}$ :

U_{11} = a + (11-1)b

= 5 + 10 \times 3

= 5 + 30

= 35

Calculate $a \times 7$ :

a \times 7 = 5 \times 7 = 35

Since $U_{11} = 35$ and $a \times 7 = 35$ , then $U_{11} = a \times 7$ .

Therefore, statement $(2)$ is true.

Evaluating Statement (3)

Statement: $S_6 = b \times 25$ .

Calculate $S_6$ :

S_6 = \frac{6}{2}(2a + (6-1)b)

= 3(2 \times 5 + 5 \times 3)

= 3(10 + 15)

= 3 \times 25

= 75

Calculate $b \times 25$ :

b \times 25 = 3 \times 25 = 75

Since $S_6 = 75$ and $b \times 25 = 75$ , then $S_6 = b \times 25$ .

Therefore, statement $(3)$ is true.

Evaluating Statement (4)

Statement: $U_{18} = S_5$ .

Calculate $U_{18}$ :

U_{18} = a + (18-1)b

= 5 + 17 \times 3

= 5 + 51

= 56

Calculate $S_5$ :

S_5 = \frac{5}{2}(2a + (5-1)b)

= \frac{5}{2}(2 \times 5 + 4 \times 3)

= \frac{5}{2}(10 + 12)

= \frac{5}{2} \times 22

= 55

Since $U_{18} = 56$ and $S_5 = 55$ , then $U_{18} \neq S_5$ .

Therefore, statement $(4)$ is false.

Given line $g_1$ passes through point $A(-1, 1)$ and $B(2, 3)$ . Also given the equation of line $g_2$ : $mx + ny - 7 = 0$ , where $m$ and $n$ are real numbers.

Which statements are true based on the information above?

$(1)$ Line $g_1$ and $g_2$ are parallel if $m = 4$ and $n = 6$ .

$(2)$ Point $B$ passes through line $g_2$ if $m = 2$ and $n = -1$ .

$(3)$ If $g_1$ and $g_2$ are parallel, then $\frac{m}{n} = -\frac{3}{2}$ .

$(4)$ If $m = 3$ and $n = 2$ , then $g_1$ and $g_2$ are perpendicular.

$(1), (2), \text{ and } (3)$ ONLY are true

$(1) \text{ and } (3)$ ONLY are true

$(2) \text{ and } (4)$ ONLY are true

$(4)$ ONLY is true

ALL statements are true

Explanation

Given line $g_1$ passes through point $A(-1, 1)$ and $B(2, 3)$ , and the equation of line $g_2$ : $mx + ny - 7 = 0$ .

Calculating the Gradient of Line g₁

The gradient of a line passing through two points can be calculated as:

m = \frac{y_2 - y_1}{x_2 - x_1}

For line $g_1$ passing through $A(-1, 1)$ and $B(2, 3)$ :

m_1 = \frac{3 - 1}{2 - (-1)}

= \frac{2}{3}

Therefore, the gradient of line $g_1$ is $m_1 = \frac{2}{3}$ .

Determining the Gradient of Line g₂

The equation of line $g_2$ : $mx + ny - 7 = 0$ .

Convert to the form $y = mx + c$ :

mx + ny - 7 = 0

ny = -mx + 7

y = -\frac{m}{n}x + \frac{7}{n}

Therefore, the gradient of line $g_2$ is $m_2 = -\frac{m}{n}$ .

Evaluating Statement (1)

Statement: Line $g_1$ and $g_2$ are parallel if $m = 4$ and $n = 6$ .

Two lines are parallel if their gradients are equal.

If $m = 4$ and $n = 6$ , then:

m_2 = -\frac{m}{n} = -\frac{4}{6} = -\frac{2}{3}

Since $m_1 = \frac{2}{3}$ and $m_2 = -\frac{2}{3}$ , then $m_1 \neq m_2$ .

Therefore, statement $(1)$ is false.

Evaluating Statement (2)

Statement: Point $B$ passes through line $g_2$ if $m = 2$ and $n = -1$ .

Point $B(2, 3)$ passes through line $g_2$ if it satisfies the equation $mx + ny - 7 = 0$ .

Substitute $m = 2$ , $n = -1$ , and $B(2, 3)$ :

mx + ny - 7 = 2(2) + (-1)(3) - 7

= 4 - 3 - 7

= -6

\neq 0

Since the result is not equal to $0$ , point $B$ does not pass through line $g_2$ .

Therefore, statement $(2)$ is false.

Evaluating Statement (3)

Statement: If $g_1$ and $g_2$ are parallel, then $\frac{m}{n} = -\frac{3}{2}$ .

If both lines are parallel, then $m_1 = m_2$ :

\frac{2}{3} = -\frac{m}{n}

\frac{m}{n} = -\frac{2}{3}

Therefore, if $g_1$ and $g_2$ are parallel, then $\frac{m}{n} = -\frac{2}{3}$ , not $-\frac{3}{2}$ .

Therefore, statement $(3)$ is false.

Evaluating Statement (4)

Statement: If $m = 3$ and $n = 2$ , then $g_1$ and $g_2$ are perpendicular.

Two lines are perpendicular if the product of their gradients equals $-1$ , i.e., $m_1 \times m_2 = -1$ .

If $m = 3$ and $n = 2$ , then:

m_2 = -\frac{m}{n} = -\frac{3}{2}

Check if $m_1 \times m_2 = -1$ :

m_1 \times m_2 = \frac{2}{3} \times \left(-\frac{3}{2}\right)

= -\frac{6}{6}

= -1

Since $m_1 \times m_2 = -1$ , then $g_1$ and $g_2$ are perpendicular.

Therefore, statement $(4)$ is true.

Given two real numbers with the following conditions:

Two times the first number minus three times the second number results in $14$ .

If the first number multiplied by $k$ plus two times the second number, then it results in $4$ , where $k$ is a real number.

If the sum of the first and second numbers results in $-3$ , the value of $k$ is ....

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