All practice questions

Question 1601Question →

For what values of the real constant $k$ does the quadratic equation $(k-2)x^2 - 2kx + (2k - 3) = 0$ have two distinct real roots $r_1$ and $r_2$ such that $r_1 < 1 < r_2$ ?

1 < k < 6

2 < k < 5

k < 2

or

k > 5

1 < k < 5

k > 2

Show answer & explanation

Answer: $2 < k < 5$

Answer

The correct inequality expressing all possible values of

k

is

2 < k < 5

.

The correct inequality

2 < k < 5

is obtained by requiring

(k-2)f(1) < 0

, which ensures

x = 1

falls between the two distinct real roots. Evaluating

f(1) = k - 5

yields

(k-2)(k-5) < 0

, giving

2 < k < 5

. The discriminant condition

\Delta = -4(k-1)(k-6) > 0

gives

1 < k < 6

, which fully encompasses

(2, 5)

.

Step-by-Step Solution

1

Define the quadratic function and state the conditions for

r_1 < 1 < r_2

.

Let

f(x) = (k-2)x^2 - 2kx + (2k - 3)

. For a quadratic function to have two real roots with

x = 1

located between them, the product of the leading coefficient

(k-2)

and

f(1)

must be strictly negative, i.e.,

(k-2)f(1) < 0

.

If a parabola opens upwards (

k-2 > 0

), its value at a point between its roots must be negative (

f(1) < 0

). If it opens downwards (

k-2 < 0

), its value at a point between its roots must be positive (

f(1) > 0

).

2

Evaluate

f(1)

in terms of

k

.

f(1) = (k-2)(1)^2 - 2k(1) + (2k - 3) = k - 2 - 2k + 2k - 3 = k - 5

.

Substitute

x = 1

directly into the expression for

f(x)

.

3

Solve the inequality

(k-2)f(1) < 0

.

(k-2)(k-5) < 0 \implies 2 < k < 5

.

The product of two linear factors

(k-2)

and

(k-5)

is negative between their roots,

k = 2

and

k = 5

.

4

Verify discriminant condition

\Delta > 0

for real roots.

\Delta = (-2k)^2 - 4(k-2)(2k-3) = 4k^2 - 4(2k^2 - 7k + 6) = -4k^2 + 28k - 24 = -4(k-1)(k-6) > 0 \implies 1 < k < 6

.

Since the interval

(2, 5)

is entirely contained within

(1, 6)

, any

k \in (2, 5)

automatically guarantees two distinct real roots.

Key Concept

Location of roots of quadratic equations and sign analysis of quadratic functions.

Estimated Time:2m 0s

Question 1602Question →

An isosceles triangle has two sides of length $10$ units each and a third side of integer length $x$ units. If the area of the triangle is strictly greater than $24$ square units and less than or equal to $48$ square units, which of the following could be the value of $x$ ? Select all such values.

Select all that apply

4

6

14

16

20

Show answer & explanation

Answer: 6; 16

Answer

The values 6 and 16 are the valid side lengths.

The values 6 and 16 produce valid areas of approximately 28.62 and exactly 48 square units respectively, both of which satisfy the given condition that the area must be strictly greater than 24 and less than or equal to 48.

Step-by-Step Solution

1

Express the height and area of the isosceles triangle in terms of

x

.

Height

h = \sqrt{10^2 - (x/2)^2} = \sqrt{100 - \frac{x^2}{4}}

, so Area

A = \frac{1}{2} x \sqrt{100 - \frac{x^2}{4}} = \frac{1}{4} x \sqrt{400 - x^2}

.

The altitude to the base of an isosceles triangle bisects the base into two equal segments of length

\frac{x}{2}

.

2

Set up the inequality for the area constraints

24 < A \le 48

.

24 < \frac{1}{4} x \sqrt{400 - x^2} \le 48 \implies 96 < x \sqrt{400 - x^2} \le 192

.

Multiplying all parts by

4

isolates the radical expression.

3

Square all terms to analyze the function

f(x^2) = x^2(400 - x^2)

.

9216 < x^2(400 - x^2) \le 36864

.

Squaring positive quantities preserves the inequality direction.

4

Evaluate the area function for each given option choice.

For

x=4

:

A \approx 19.6

(too small). For

x=6

:

A \approx 28.6

(valid). For

x=14

:

A \approx 50.0

(too large). For

x=16

:

A = 48

(valid). For

x=20

: degenerate triangle with

A = 0

(invalid).

Direct evaluation identifies which integer choices satisfy

24 < A \le 48

.

Key Concept

Properties of isosceles triangles, Pythagorean theorem for altitude, area bounds, and triangle inequality.

Question 1603Question →

A dataset $D$ consists of $11$ distinct positive integers. The median of $D$ is $40$ , and the arithmetic mean of $D$ is $45$ . A new dataset $D'$ is created by increasing each of the $5$ largest integers in $D$ by $10$ and decreasing each of the $5$ smallest integers in $D$ by a positive integer $x$ . If the median of $D'$ is strictly less than the arithmetic mean of $D'$ , what is the maximum possible integer value of $x$ ?

20

21

19

25

10

Show answer & explanation

Answer: $20$

Answer

The maximum possible integer value of

x

is

20

.

The sum of the original

11

values is

11 \times 45 = 495

. When the

5

largest integers are each increased by

10

, the sum increases by

+50

. When the

5

smallest integers are each decreased by

x

, the sum decreases by

-5x

. The new sum is

545 - 5x

, making the new mean

\frac{545 - 5x}{11}

. Because

D

has

11

elements, the median is the

6^{\text{th}}

element. Changing the smallest

5

and largest

5

elements does not alter the value of the

6^{\text{th}}

element, so the median remains

40

. Requiring the median to be strictly less than the new mean gives

40 < \frac{545 - 5x}{11} \implies 440 < 545 - 5x \implies 5x < 105 \implies x < 21

. The greatest integer less than

21

is

20

.

Step-by-Step Solution

1

Calculate the total sum of the original dataset

D

.

Since

D

has

11

elements with a mean of

45

, the sum of elements is

11 \times 45 = 495

.

The mean formula is

\text{Mean} = \frac{\text{Sum}}{n}

, so

\text{Sum} = n \times \text{Mean}

.

2

Determine the median and sum of the modified dataset

D'

.

The median remains the

6^{\text{th}}

element, which is

40

. The sum of

D'

is

495 + 5(10) - 5(x) = 545 - 5x

.

Modifying only the

5

smallest and

5

largest elements leaves the

6^{\text{th}}

central element unchanged.

3

Set up and solve the inequality comparing the median to the mean of

D'

.

Solve

40 < \frac{545 - 5x}{11} \implies 440 < 545 - 5x \implies 5x < 105 \implies x < 21

.

The problem specifies that the median must be strictly less than the mean.

4

Identify the maximum integer value satisfying the inequality.

The largest integer strictly less than

21

is

20

.

x

must be an integer.

Key Concept

Effect of data modifications on mean and median

Estimated Time:2m 30s

Question 1604Question →

In the right rectangular solid $ABCDEFGH$ , the base $ABCD$ is a rectangle with edge lengths $AB = 6$ and $BC = 6\sqrt{3}$ . The vertical edge $CG = 12$ . Point $M$ is the midpoint of edge $CG$ . What is the perimeter of triangle $BDM$ ?

24 + 6\sqrt{2}

12 + 12\sqrt{2}

24 + 6\sqrt{3}

18 + 6\sqrt{2} + 6\sqrt{3}

12 + 6\sqrt{2} + 6\sqrt{5}

Show answer & explanation

Answer: $24 + 6\sqrt{2}$

Answer

The perimeter of triangle

BDM

is

24 + 6\sqrt{2}

.

The correct answer is obtained by recognizing three right triangles within the 3D figure:

\triangle BCD

has legs

6

and

6\sqrt{3}

giving

BD = 12

;

\triangle BCM

has legs

6\sqrt{3}

and

6

giving

BM = 12

; and

\triangle DCM

has legs

6

and

6

giving

DM = 6\sqrt{2}

. Summing the three sides yields a perimeter of

24 + 6\sqrt{2}

.

Step-by-Step Solution

1

Calculate the length of base diagonal

BD

using the right triangle

\triangle BCD

.

BD = 12

In right triangle

\triangle BCD

, legs are

CD = AB = 6

and

BC = 6\sqrt{3}

. Using the

30^\circ\text{--}60^\circ\text{--}90^\circ

ratio

(1 : \sqrt{3} : 2)

, hypotenuse

BD = 2 \times 6 = 12

(or via Pythagorean theorem:

BD = \sqrt{6^2 + (6\sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12

).

2

Calculate the length of segment

BM

using the right triangle

\triangle BCM

.

BM = 12

Since

M

is the midpoint of vertical edge

CG = 12

,

CM = 6

. Vertical edge

CG

is perpendicular to base

ABCD

, so

\triangle BCM

is a right triangle at

C

. The legs are

CM = 6

and

BC = 6\sqrt{3}

. Applying the

30^\circ\text{--}60^\circ\text{--}90^\circ

ratio

(1 : \sqrt{3} : 2)

, hypotenuse

BM = 2 \times 6 = 12

.

3

Calculate the length of segment

DM

using the right triangle

\triangle DCM

.

DM = 6\sqrt{2}

In right triangle

\triangle DCM

, legs are

CD = 6

and

CM = 6

. Since the legs are equal,

\triangle DCM

is a

45^\circ\text{--}45^\circ\text{--}90^\circ

isosceles right triangle with side ratio

1 : 1 : \sqrt{2}

. Thus, hypotenuse

DM = 6\sqrt{2}

.

4

Sum the three side lengths to find the perimeter of

\triangle BDM

.

Perimeter =

12 + 12 + 6\sqrt{2} = 24 + 6\sqrt{2}

The perimeter of

\triangle BDM

is

BD + BM + DM

.

Key Concept

Applying special right triangle ratios (

30^\circ\text{--}60^\circ\text{--}90^\circ

and

45^\circ\text{--}45^\circ\text{--}90^\circ

) to 3D rectangular solids.

Question 1605Question →

A solid cube has a total surface area of $S$ and a volume of $V$ . A solid right circular cylinder has a height equal to its base diameter. If the total surface area of the cylinder is also equal to $S$ , what is the volume of the cylinder in terms of $V$ ?

\frac{2}{\sqrt{\pi}} V

\frac{\sqrt{\pi}}{2} V

\frac{1}{\sqrt{\pi}} V

\sqrt{\frac{27}{2\pi}} V

\frac{2}{\pi} V

Show answer & explanation

Answer: $\frac{2}{\sqrt{\pi}} V$

Answer

\frac{2}{\sqrt{\pi}} V

The total surface area of a cube with edge

a

is

6a^2

and its volume is

V = a^3

. For a cylinder with base radius

r

and height

h = 2r

, the total surface area is

2\pi r^2 + 2\pi r(2r) = 6\pi r^2

. Setting

6\pi r^2 = 6a^2

yields

r = \frac{a}{\sqrt{\pi}}

and

h = \frac{2a}{\sqrt{\pi}}

. Substituting these into the volume formula

V_{\text{cyl}} = \pi r^2 h

produces

\pi \left(\frac{a^2}{\pi}\right) \left(\frac{2a}{\sqrt{\pi}}\right) = \frac{2a^3}{\sqrt{\pi}} = \frac{2}{\sqrt{\pi}} V

.

Step-by-Step Solution

1

Express the surface area and volume of the cube in terms of its side length

a

.

Surface area

S = 6a^2

and volume

V = a^3

.

A cube with edge length

a

has

6

identical square faces of area

a^2

and volume

a^3

.

2

Set up the total surface area formula for the cylinder with radius

r

and height

h = 2r

, and equate it to

S

.

Total surface area

S_{\text{cyl}} = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r (2r) = 6\pi r^2 = 6a^2

.

The cylinder's height is equal to its base diameter (

2r

). Equating surface areas gives

6\pi r^2 = 6a^2

.

3

Solve for radius

r

in terms of edge length

a

.

r^2 = \frac{a^2}{\pi} \implies r = \frac{a}{\sqrt{\pi}}

.

Dividing both sides by

6\pi

and taking the square root isolates

r

.

4

Calculate the volume of the cylinder in terms of

V

.

V_{\text{cyl}} = \pi r^2 h = \pi \left(\frac{a^2}{\pi}\right) \left(\frac{2a}{\sqrt{\pi}}\right) = \frac{2a^3}{\sqrt{\pi}} = \frac{2}{\sqrt{\pi}} V

.

Substituting

r^2 = \frac{a^2}{\pi}

and

h = \frac{2a}{\sqrt{\pi}}

into the cylinder volume formula

\pi r^2 h

yields the answer in terms of

V = a^3

.

Key Concept

Volume and surface area relationship between geometric solids

Estimated Time:2m 0s

Question 1606Question →

For all real numbers $x$ and $y$ , the custom operation $\odot$ is defined by $x \odot y = x^2 y - y^2 x$ . The function $f$ is defined by $f(t) = t \odot 3$ . If $t$ is a positive real number such that $f(f(t)) = 0$ and $f(t) \neq 0$ , what is the value of $t$ ?

3

\frac{3 + \sqrt{5}}{2}

\frac{3 + \sqrt{13}}{2}

\frac{3 + \sqrt{13}}{4}

\frac{9 + \sqrt{13}}{2}

Show answer & explanation

Answer: $\frac{3 + \sqrt{13}}{2}$

Answer

The correct value of

t

is

\frac{3 + \sqrt{13}}{2}

.

Applying the custom binary operator gives

f(t) = 3t^2 - 9t

. Substituting

u = f(t)

into

f(u) = 0

yields

3u(u - 3) = 0

, so

u = 0

or

u = 3

. Because

f(t) \neq 0

, it must be that

f(t) = 3

. Setting

3t^2 - 9t = 3

leads to

t^2 - 3t - 1 = 0

. Applying the quadratic formula yields the positive value

\frac{3 + \sqrt{13}}{2}

.

Step-by-Step Solution

1

Express

f(t)

using the custom symbol definition.

f(t) = t \odot 3 = t^2(3) - (3)^2 t = 3t^2 - 9t

.

Apply the rule

x \odot y = x^2 y - y^2 x

with

x = t

and

y = 3

.

2

Analyze the nested function condition

f(f(t)) = 0

.

Let

u = f(t)

. Then

f(u) = 3u^2 - 9u = 3u(u - 3) = 0

, which yields

u = 0

or

u = 3

.

Evaluate the outer function

f

at the argument

u = f(t)

.

3

Apply the problem constraints to determine the exact value of

f(t)

.

Since

f(t) \neq 0

,

u = f(t) = 3

. Thus,

3t^2 - 9t = 3

.

Eliminate

f(t) = 0

based on the explicit condition given in the problem.

4

Solve the quadratic equation for

t > 0

.

Dividing

3t^2 - 9t - 3 = 0

by 3 gives

t^2 - 3t - 1 = 0

. Using the quadratic formula,

t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{13}}{2}

. Since

t > 0

,

t = \frac{3 + \sqrt{13}}{2}

.

Find the positive real root of the simplified quadratic equation.

Key Concept

Nested Function Evaluation and Custom Symbol Operations

Estimated Time:2m 30s

Question 1607Question →

In triangle $ABC$ , the length of side $AB$ is $16$ units and the length of side $AC$ is $13$ units. If the area of triangle $ABC$ is $96$ square units and the altitude from vertex $C$ to side $AB$ intersects the line segment $AB$ at point $D$ , what is the length of segment $AD$ ?

5

11

12

7

1

Show answer & explanation

Answer: $5$

Answer

The length of segment

AD

is

5

units.

The correct answer is

5

. First, find altitude

CD

using the triangle area equation:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \implies 96 = \frac{1}{2} \times 16 \times CD \implies CD = 12

. Next, because

CD \perp AB

, triangle

ADC

is a right triangle with hypotenuse

AC = 13

and leg

CD = 12

. Applying the Pythagorean theorem yields

AD = \sqrt{13^2 - 12^2} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Calculate the length of altitude

CD

using the triangle area formula.

CD = 12

units.

The area of triangle

ABC

is given by

\text{Area} = \frac{1}{2} \times AB \times CD

. Substituting the given values:

96 = \frac{1}{2} \times 16 \times CD \implies 96 = 8 \times CD \implies CD = 12

.

2

Apply the Pythagorean theorem in right triangle

ADC

to find

AD

.

AD = 5

units.

Altitude

CD

is perpendicular to

AB

, forming right triangle

ADC

with hypotenuse

AC = 13

and leg

CD = 12

. By the Pythagorean theorem,

AD^2 + CD^2 = AC^2 \implies AD^2 + 12^2 = 13^2 \implies AD^2 + 144 = 169 \implies AD^2 = 25 \implies AD = 5

.

Key Concept

Triangles: Properties, Perimeter, and Area

Estimated Time:1m 30s

Question 1608Question →

The sum of the measures of all interior angles of a convex polygon, excluding one interior angle $\theta$ , is equal to $2190^\circ$ . If the degree measure of $\theta$ is an integer, what is the perimeter of a regular polygon with $n$ sides, where $n$ is the number of sides of the original polygon and each side length is $\frac{\theta}{10}$ units?

195

210

225

240

255

Show answer & explanation

Answer: 225

Answer

The perimeter of the regular polygon is 225.

The sum of the interior angles of a convex polygon with

n

sides is

(n-2) \times 180^\circ

. Adding the excluded interior angle

\theta

to

2190^\circ

yields the total sum

S

. Because

0^\circ < \theta < 180^\circ

,

S

must fall strictly between

2190^\circ

and

2370^\circ

. The only multiple of

180^\circ

in this interval is

2340^\circ

. Setting

(n-2)\times 180^\circ = 2340^\circ

yields

n = 15

. Solving for

\theta

gives

\theta = 2340^\circ - 2190^\circ = 150^\circ

. The side length is

\frac{150}{10} = 15

, making the perimeter

15 \times 15 = 225

.

Step-by-Step Solution

1

Set up the inequality for the sum of interior angles of a convex polygon.

The total sum of interior angles for a convex

n

-gon is

S = (n-2) \times 180^\circ

. Given

S - \theta = 2190^\circ

, we have

S = 2190^\circ + \theta

.

The sum of interior angles of any convex polygon with

n

sides is

(n-2) \times 180^\circ

.

2

Determine the value of

S

using the bounds for an interior angle of a convex polygon.

Since

0^\circ < \theta < 180^\circ

, it follows that

2190^\circ < S < 2190^\circ + 180^\circ = 2370^\circ

. The only multiple of

180^\circ

in this range is

2340^\circ

.

S

must be an integer multiple of

180^\circ

and

\theta

must be strictly between

0^\circ

and

180^\circ

.

3

Calculate the number of sides

n

and the missing angle

\theta

.

(n-2) \times 180^\circ = 2340^\circ \implies n - 2 = 13 \implies n = 15

. Then

\theta = 2340^\circ - 2190^\circ = 150^\circ

.

Solving the linear equations gives exact values for the number of sides and the excluded angle.

4

Compute the perimeter of the regular regular

n

-gon.

Side length

= \frac{\theta}{10} = \frac{150}{10} = 15

. Perimeter

= n \times \text{side length} = 15 \times 15 = 225

.

The perimeter of a regular polygon is the product of its number of sides and its individual side length.

Key Concept

Sum of Interior Angles of Convex Polygons

Estimated Time:2m 0s

Question 1609Question →

A triangle has side lengths of $x$ , $x + 4$ , and $14$ units, where $x$ is an integer. Which of the following could be the perimeter of the triangle? Select all that apply.

Select all that apply

26

28

30

34

40

Show answer & explanation

Answer: $30$ ; $34$ ; $40$

Answer

The possible perimeters of the triangle are 30, 34, and 40.

The Triangle Inequality Theorem dictates that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the remaining side. For side lengths

x

,

x + 4

, and

14

, setting

x + (x + 4) > 14

yields

2x > 10

, so

x > 5

. Since

x

is specified as an integer,

x

can be any integer greater than or equal to 6. Substituting valid values of

x

into the perimeter formula

P = 2x + 18

gives possible perimeters of 30 (when

x = 6

), 34 (when

x = 8

), and 40 (when

x = 11

). Thus, the values 30, 34, and 40 are all valid perimeters.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem

x > 5

According to the Triangle Inequality Theorem, the sum of any two side lengths must be strictly greater than the third side length. Therefore,

x + (x + 4) > 14

, which simplifies to

2x + 4 > 14 \implies 2x > 10 \implies x > 5

. The other two inequality conditions (

(x + 4) + 14 > x

and

x + 14 > x + 4

) are satisfied for all positive values of

x

.

2

Express the perimeter in terms of

x

P = 2x + 18

The perimeter

P

is the sum of the three side lengths:

P = x + (x + 4) + 14 = 2x + 18

.

3

Evaluate the given choices against the valid bounds of

x

Valid integer values of

x \ge 6

correspond to perimeters

P \ge 30

that are even numbers.

Since

x

is an integer and

x > 5

, the minimum integer value for

x

is 6, which yields a minimum perimeter of

P = 2(6) + 18 = 30

. Testing each option:
- For 26:

2x + 18 = 26 \implies x = 4

(Invalid,

x \le 5

)
- For 28:

2x + 18 = 28 \implies x = 5

(Invalid,

x \le 5

)
- For 30:

2x + 18 = 30 \implies x = 6

(Valid)
- For 34:

2x + 18 = 34 \implies x = 8

(Valid)
- For 40:

2x + 18 = 40 \implies x = 11

(Valid)

Key Concept

Triangle Inequality Theorem and Perimeter Calculation

Question 1610Question →

Dataset $X$ consists of $20$ distinct real numbers with standard deviation $s_X > 0$ and interquartile range $IQR_X > 0$ . Dataset $Y$ is created by multiplying each number in Dataset $X$ by $-3$ and then adding $7$ . Dataset $Z$ is created by adding a single 21st value, equal to the arithmetic mean of Dataset $X$ , to Dataset $X$ . Which of the following statements must be true regarding the measures of dispersion of these datasets?

The standard deviation of Dataset

Y

is

3s_X

, and the standard deviation of Dataset

Z

is strictly less than

s_X

.

The standard deviation of Dataset

Y

is

3s_X

, and the standard deviation of Dataset

Z

is equal to

s_X

.

The standard deviation of Dataset

Y

is

-3s_X + 7

, and the standard deviation of Dataset

Z

is strictly less than

s_X

.

The standard deviation of Dataset

Y

is

3s_X + 7

, and the standard deviation of Dataset

Z

is strictly greater than

s_X

.

The standard deviation of Dataset

Y

is

-3s_X

, and the standard deviation of Dataset

Z

is equal to

s_X

.

Show answer & explanation

Answer: The standard deviation of Dataset $Y$ is $3s_X$ , and the standard deviation of Dataset $Z$ is strictly less than $s_X$ .

Answer

The standard deviation of Dataset

Y

is

3s_X

, and the standard deviation of Dataset

Z

is strictly less than

s_X

.

Linear transformation scales standard deviation by the absolute value of the multiplier, making

s_Y = |-3|s_X = 3s_X

. Constant additions do not affect spread. Adding a data point equal to the mean does not change the total sum of squared deviations, but increases the sample size

n

, thereby reducing the average squared deviation and yielding

s_Z < s_X

.

Step-by-Step Solution

1

Analyze the linear transformation on Dataset X to form Dataset Y.

For any linear transformation of data

y_i = a x_i + b

, the standard deviation transforms according to

s_Y = |a| s_X

. Here

a = -3

and

b = 7

, so

s_Y = |-3| s_X = 3s_X

.

Adding a constant shift

b

shifts all points equally without altering their spread relative to the mean, whereas multiplying by

a

scales distance by

|a|

.

2

Analyze the effect of inserting the mean into Dataset X to form Dataset Z.

Adding

\mu_X

to Dataset

X

yields a new dataset with mean

\mu_Z = \mu_X

. The sum of squared deviations

\sum (z_i - \mu_Z)^2 = \sum (x_i - \mu_X)^2 + (\mu_X - \mu_X)^2 = \sum (x_i - \mu_X)^2

remains identical.

The extra point contributes

0

squared distance from the mean.

3

Compare the standard deviation

s_Z

to

s_X

.

Because the sum of squared deviations remains unchanged while the total count of elements increases from

20

to

21

, the average squared deviation (variance) decreases. Hence,

s_Z < s_X

.

Dividing the same numerator by a larger denominator (

21

or

20

depending on sample/population formula) results in a smaller variance and smaller standard deviation.

Key Concept

Effect of linear transformations and mean-value insertion on standard deviation

Alternative Method

Consider extreme simple cases: if Dataset X has two values

\{ -1, 1 \}

, mean is 0,

s_X = 1

. Dataset Y becomes

\{ 10, 4 \}

, mean is 7,

s_Y = 3 = 3s_X

. Adding 0 to X gives

\{ -1, 0, 1 \}

, mean remains 0, variance becomes

2/3

, so

s_Z = \sqrt{2/3} < 1

.

Estimated Time:2m 0s

Question 1611Question →

In the $xy$ -plane, line $m$ is defined by the equation $ax + by = c$ , where $a$ , $b$ , and $c$ are non-zero real numbers such that $ab < 0$ and $ac > 0$ . Line $k$ is perpendicular to line $m$ and intersects line $m$ at its $x$ -intercept. Which of the following statements must be true? Select all such statements.

Select all that apply

Line

m

has a positive slope.

Line

k

passes through Quadrant II.

The

y

-intercept of line

k

is negative.

The point of intersection of line

m

and line

k

lies in Quadrant I.

Line

m

passes through Quadrant II.

Show answer & explanation

Answer: Line $m$ has a positive slope.; Line $k$ passes through Quadrant II.

Answer

The statements asserting that line

m

has a positive slope and that line

k

passes through Quadrant II must be true.

The statement regarding line

m

having a positive slope is correct because

ab < 0

implies

a

and

b

have opposite signs, making

-\frac{a}{b} > 0

. The statement asserting line

k

passes through Quadrant II is correct because line

k

possesses a negative slope

\frac{b}{a} < 0

and a positive

y

-intercept

-\frac{bc}{a^2} > 0

, ensuring it enters Quadrant II.

Step-by-Step Solution

1

Determine the slope and intercepts of line

m

.

Line

m

:

y = -\frac{a}{b}x + \frac{c}{b}

. Slope is

-\frac{a}{b} > 0

because

ab < 0

.

x

-intercept is

\left(\frac{c}{a}, 0\right)

where

\frac{c}{a} > 0

because

ac > 0

.

y

-intercept is

\left(0, \frac{c}{b}\right)

where

\frac{c}{b} < 0

because

b

and

c

have opposite signs.

Converting standard line equations to slope-intercept form exposes the signs of slopes and intercepts based on coefficient products.

2

Determine the slope, equation, and properties of line

k

.

Since line

k

is perpendicular to line

m

, its slope is the negative reciprocal of

-\frac{a}{b}

, which is

\frac{b}{a} < 0

. Line

k

passes through

\left(\frac{c}{a}, 0\right)

, giving equation

y = \frac{b}{a}\left(x - \frac{c}{a}\right) = \frac{b}{a}x - \frac{bc}{a^2}

.

Perpendicular lines have slopes whose product is

-1

.

3

Analyze quadrant coverage for both lines and verify statements.

Line

m

has positive slope and negative

y

-intercept

\implies

passes through Quadrants I, III, IV. Line

k

has negative slope and positive

y

-intercept

-\frac{bc}{a^2} > 0 \implies

passes through Quadrants I, II, IV. Intersection is at

\left(\frac{c}{a}, 0\right)

on the positive

x

-axis.

Systematic sign analysis determines quadrant trajectory and exact axis locations.

Key Concept

Properties of lines, perpendicular slopes, and sign analysis of intercepts in coordinate geometry.

Question 1612Question →

In the $xy$ -plane, line $L_1$ is defined by the equation $2x - 3y = 6$ . Line $L_2$ is perpendicular to line $L_1$ and passes through the point $P(4, -1)$ . Point $Q(a, b)$ lies on line $L_2$ such that the distance between point $P$ and point $Q$ is $\sqrt{13}$ . Which of the following statements regarding point $Q$ or line $L_2$ could be true? Select all such statements.

Select all that apply

Point

Q

lies in Quadrant I.

The sum of the coordinates of point

Q

,

a + b

, is equal to

4

.

The distance from point

Q

to the origin is

2\sqrt{13}

.

The

y

-intercept of line

L_2

is

(0, -6)

.

Point

Q

could have coordinates

(4, 2)

.

Show answer & explanation

Answer: Point $Q$ lies in Quadrant I.; The sum of the coordinates of point $Q$ , $a + b$ , is equal to $4$ .; The distance from point $Q$ to the origin is $2\sqrt{13}$ .

Answer

The correct statements are that point Q can lie in Quadrant I, the sum of the coordinates of point Q can equal 4, and the distance from point Q to the origin can be 2√13.

Solving the system formed by line

L_2

(

3x + 2y = 10

) and the distance constraint

(a-4)^2 + (b+1)^2 = 13

yields two possible points:

(2, 2)

and

(6, -4)

. The point

(2, 2)

lies in Quadrant I and has a coordinate sum of

2 + 2 = 4

. The point

(6, -4)

has a distance to the origin of

\sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13}

.

Step-by-Step Solution

1

Determine the slope and equation of line

L_2

.

Line

L_1

has slope

m_1 = \frac{2}{3}

. Therefore, perpendicular line

L_2

has slope

m_2 = -\frac{3}{2}

. Using point

P(4, -1)

, the equation of

L_2

is

y - (-1) = -\frac{3}{2}(x - 4)

, which simplifies to

y = -\frac{3}{2}x + 5

or

3x + 2y = 10

.

Perpendicular lines in the coordinate plane have negative reciprocal slopes.

2

Express the distance constraint between

P(4, -1)

and

Q(a, b)

algebraically.

Since

Q(a, b)

lies on

L_2

,

b = -\frac{3}{2}a + 5

. The distance squared is

(a - 4)^2 + (b + 1)^2 = 13

. Substituting

b + 1 = -\frac{3}{2}(a - 4)

yields

(a - 4)^2 + \left(-\frac{3}{2}(a - 4)\right)^2 = 13

, which simplifies to

\frac{13}{4}(a - 4)^2 = 13

, so

(a - 4)^2 = 4

.

Applying the distance formula and substituting the line equation reduces the problem to a quadratic equation in one variable.

3

Solve for the possible coordinates of point

Q

.

Taking square roots gives

a - 4 = 2

or

a - 4 = -2

, resulting in

a = 6

or

a = 2

. Correspondingly,

b = -4

or

b = 2

. Thus,

Q

can be

(6, -4)

or

(2, 2)

.

Quadratic equations of the form

(x-h)^2 = k

have two real solutions.

4

Evaluate each statement against the possible coordinates

Q(6, -4)

and

Q(2, 2)

.

For

Q(2, 2)

: it lies in Quadrant I (valid), its coordinate sum is

2 + 2 = 4

(valid). For

Q(6, -4)

: its distance to the origin is

\sqrt{6^2 + (-4)^2} = \sqrt{52} = 2\sqrt{13}

(valid). The

y

-intercept of

L_2

is

(0, 5)

, and

(4, 2)

is not on

L_2

.

Direct verification confirms which properties hold for the two solved points.

Key Concept

Perpendicular Slopes and Distance Formula in Coordinate Geometry

Question 1613Question →

Courier A departs from Warehouse X heading toward Warehouse Y at 8:00 AM traveling at a constant speed of $40$ miles per hour. Courier B departs from Warehouse Y heading toward Warehouse X along the same straight route at 9:00 AM traveling at a constant speed of $60$ miles per hour. If the total distance between Warehouse X and Warehouse Y is $190$ miles, at what time will the two couriers meet?

9:54 AM

10:00 AM

10:30 AM

11:00 AM

11:15 AM

Show answer & explanation

Answer: 10:30 AM

Answer

10:30 AM

The correct answer is 10:30 AM. Between 8:00 AM and 9:00 AM, Courier A travels 40 miles alone. At 9:00 AM, the distance remaining between them is 150 miles. Because they travel toward each other, their speeds combine to 100 mph (40 + 60). Dividing 150 miles by 100 mph gives 1.5 hours (1 hour and 30 minutes). Adding 1 hour and 30 minutes to 9:00 AM gives a meeting time of 10:30 AM.

Step-by-Step Solution

1

Calculate the distance traveled by Courier A before Courier B starts moving.

From 8:00 AM to 9:00 AM (1 hour), Courier A travels

40 \text{ mph} \times 1 \text{ hour} = 40 \text{ miles}

.

Courier A has a 1-hour head start.

2

Determine the remaining distance to be covered between the two couriers at 9:00 AM.

Remaining distance

= 190 - 40 = 150 \text{ miles}

.

Subtract Courier A's distance from the total distance of 190 miles.

3

Calculate the combined rate of both couriers and solve for elapsed time after 9:00 AM.

Combined rate

= 40 + 60 = 100 \text{ mph}

. Elapsed time

t = \frac{150}{100} = 1.5 \text{ hours} = 1 \text{ hour } 30 \text{ minutes}

.

Since they move toward each other, their speeds add up to close the gap.

4

Add the elapsed combined time to 9:00 AM to find the meeting time.

9:00 AM

+ 1 \text{ hour } 30 \text{ minutes} = 10:30 \text{ AM}

.

The combined movement began at 9:00 AM when Courier B started traveling.

Key Concept

Distance-rate-time relationship with staggered start times

Estimated Time:1m 30s

Question 1614Question →

A survey of $250$ registered voters was conducted to analyze their primary news sources: Television ( $T$ ), the Internet ( $I$ ), and Print newspapers ( $P$ ). The survey revealed the following results:

- $140$ voters get news from Television.
- $150$ voters get news from the Internet.
- $80$ voters get news from Print newspapers.
- $45$ voters get news from both Television and Print newspapers.
- $60$ voters get news from both the Internet and Print newspapers.
- $30$ voters get news from all three sources.
- $20$ voters do not get news from any of these three sources.

How many of the surveyed voters get news from Television and the Internet, but NOT from Print newspapers?

Show answer & explanation

Answer: 35

Answer

35 voters get news from Television and the Internet, but not from Print newspapers.

Using the principle of inclusion-exclusion for three sets, the total union size is 230 voters (250 total minus 20 who use none). Setting up the formula 230 = 140 + 150 + 80 - 45 - 60 - |T ∩ I| + 30 allows us to solve for |T ∩ I| = 65. To find those who use Television and Internet but NOT Print newspapers, we subtract the 30 voters who use all three sources from 65, resulting in 35 voters.

Step-by-Step Solution

1

Find the size of the union of all three sets

|T ∪ I ∪ P| = 250 - 20 = 230

Subtracting the 20 voters who use none of the three news sources from the total sample of 250 gives the total number of voters in at least one category.

2

Set up the Principle of Inclusion-Exclusion for three sets

230 = 140 + 150 + 80 - 45 - 60 - |T ∩ I| + 30

The formula sums individual set sizes, subtracts pairwise intersections, and adds back the triple intersection.

3

Solve for the total intersection of Television and Internet

|T ∩ I| = 65

Simplifying the equation gives 230 = 295 - |T ∩ I|, which yields |T ∩ I| = 65.

4

Exclude those who also read Print newspapers

|(T ∩ I) \ P| = 65 - 30 = 35

Subtracting the 30 voters who use all three sources leaves only those who use Television and Internet without Print newspapers.

Key Concept

Three-Set Principle of Inclusion-Exclusion

Question 1615Question →

Two automated assembly robots, Robot P and Robot Q, produce identical components. Robot P operates at a constant rate of $p$ components per hour, and Robot Q operates at a constant rate of $q$ components per hour, where $p > q > 0$ . During a shift, Robot P worked for $4$ hours and Robot Q worked for $6$ hours to produce a combined total of $360$ components. If $T$ represents the total number of components produced when Robot P works for $7$ hours and Robot Q works for $3$ hours, which of the following statements must be true? Select all such statements.

Select all that apply

The rate of Robot P,

p

, must be greater than

36

components per hour.

The rate of Robot Q,

q

, must be less than

36

components per hour.

The total number of components

T

must be greater than

360

and less than

630

.

The rate of Robot P,

p

, must be less than

60

components per hour.

The total number of components

T

could be equal to

320

.

Show answer & explanation

Answer: The rate of Robot P, $p$ , must be greater than $36$ components per hour.; The rate of Robot Q, $q$ , must be less than $36$ components per hour.; The total number of components $T$ must be greater than $360$ and less than $630$ .

Answer

The correct statements are: the rate of Robot P,

p

, must be greater than

36

components per hour; the rate of Robot Q,

q

, must be less than

36

components per hour; and the total number of components

T

must be greater than

360

and less than

630

.

The system of equations

2p + 3q = 180

combined with

p > q > 0

strictly bounds

p

between

36

and

90

, and

q

between

0

and

36

. Substituting these boundary conditions into the total expression

T = 5p + 180

yields the strict range

360 < T < 630

. Consequently, the three statements asserting

p > 36

,

q < 36

, and

360 < T < 630

are all mathematically required.

Step-by-Step Solution

1

Set up the linear equation from the initial production shift and simplify.

4p + 6q = 360 \implies 2p + 3q = 180 \implies q = 60 - \frac{2}{3}p

This establishes the exact relationship between the production rates

p

and

q

.

2

Apply the given constraints

p > q > 0

to determine the domain bounds for

p

and

q

.

p > 60 - \frac{2}{3}p \implies \frac{5}{3}p > 60 \implies p > 36

. Also,

q > 0 \implies 60 - \frac{2}{3}p > 0 \implies p < 90

. Thus,

36 < p < 90

and

0 < q < 36

.

Determining extreme bounds for

p

automatically constrains both individual rates.

3

Formulate

T

in terms of

p

and evaluate its numerical boundaries.

T = 7p + 3q = 7p + (180 - 2p) = 5p + 180

. Substituting

36 < p < 90

gives

360 < T < 630

.

Substituting

3q = 180 - 2p

simplifies

T

into a single-variable linear modeling equation.

Key Concept

Linear word problem modeling, variable elimination, and system inequality constraint analysis

Estimated Time:1m 45s

Question 1616Question →

If $x$ is a real number that satisfies the inequality $|3x - 3| \le 9$ , which of the following statements must be true? Select all that apply.

Select all that apply

-4 \le 2x \le 8

|x - 1| \le 3

x^2 \le 16

x \ge 0

1 - x \le 2

Show answer & explanation

Answer: $-4 \le 2x \le 8$ ; $|x - 1| \le 3$ ; $x^2 \le 16$

Answer

The statements that must be true are

-4 \le 2x \le 8

,

|x - 1| \le 3

, and

x^2 \le 16

.

Solving

|3x - 3| \le 9

gives

-9 \le 3x - 3 \le 9

, which simplifies to

-2 \le x \le 4

. Multiplying this range by 2 yields

-4 \le 2x \le 8

. Subtracting 1 gives

-3 \le x - 1 \le 3

, which is

|x - 1| \le 3

. Squaring values in

[-2, 4]

yields non-negative numbers up to 16, so

x^2 \le 16

is also true.

Step-by-Step Solution

1

Unfold the absolute value inequality into a compound inequality.

-9 \le 3x - 3 \le 9

By definition,

|u| \le k

(where

k \ge 0

) is equivalent to

-k \le u \le k

.

2

Add 3 to all parts of the compound inequality.

-6 \le 3x \le 12

Isolating the variable term

3x

.

3

Divide all parts by 3 to solve for

x

.

-2 \le x \le 4

Dividing by a positive constant preserves the direction of the inequality signs.

4

Test each proposed statement against the interval

[-2, 4]

.

-4 \le 2x \le 8

is true;

|x - 1| \le 3

is true;

x^2 \le 16

is true;

x \ge 0

fails for

x = -1

;

1 - x \le 2

fails for

x = -2

.

Determining which properties hold for every real number in the solution set.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:1m 30s

Question 1617Question →

Dataset $P$ consists of $n$ numerical values with a mean of $40$ and a standard deviation of $6$ . Dataset $Q$ consists of $n$ numerical values with a mean of $60$ and a standard deviation of $6$ . Dataset $R$ is created by combining all $n$ values from Dataset $P$ and all $n$ values from Dataset $Q$ into a single dataset of $2n$ values. If $\sigma_R$ represents the standard deviation of Dataset $R$ , which of the following is the exact value of $\sigma_R$ ?

6

6\sqrt{2}

2\sqrt{34}

12

16

Show answer & explanation

Answer: $2\sqrt{34}$

Answer

The exact value of

\sigma_R

is

2\sqrt{34}

.

The total variance of a combined dataset is given by the law of total variance:

\sigma_R^2 = \text{Mean}(\sigma_P^2, \sigma_Q^2) + \text{Var}(\mu_P, \mu_Q)

. Since

\sigma_P = \sigma_Q = 6

, the average within-group variance is

6^2 = 36

. The combined mean is

50

, and both group means (

40

and

60

) lie

10

units away from

50

, giving a between-group variance of

10^2 = 100

. Combining these yields

\sigma_R^2 = 36 + 100 = 136

, so

\sigma_R = \sqrt{136} = 2\sqrt{34}

.

Step-by-Step Solution

1

Calculate the mean of the combined dataset

R

.

\mu_R = \frac{n(40) + n(60)}{2n} = \frac{100n}{2n} = 50

Since both datasets have equal size

n

, the combined mean is the arithmetic average of the two group means.

2

Express the sum of squared deviations for Dataset

P

around the combined mean

\mu_R = 50

.

\sum_{i=1}^n (p_i - 50)^2 = \sum_{i=1}^n ((p_i - 40) - 10)^2 = \sum_{i=1}^n (p_i - 40)^2 - 20\sum_{i=1}^n (p_i - 40) + 100n = 36n - 0 + 100n = 136n

The variance of

P

gives

\sum (p_i - 40)^2 = 6^2 n = 36n

, and

\sum (p_i - 40) = 0

by the definition of the mean.

3

Express the sum of squared deviations for Dataset

Q

around the combined mean

\mu_R = 50

.

\sum_{j=1}^n (q_j - 50)^2 = \sum_{j=1}^n ((q_j - 60) + 10)^2 = \sum_{j=1}^n (q_j - 60)^2 + 20\sum_{j=1}^n (q_j - 60) + 100n = 36n + 0 + 100n = 136n

The variance of

Q

gives

\sum (q_j - 60)^2 = 6^2 n = 36n

, and

\sum (q_j - 60) = 0

.

4

Compute the combined variance

\sigma_R^2

and take the square root to find

\sigma_R

.

\sigma_R^2 = \frac{136n + 136n}{2n} = \frac{272n}{2n} = 136 \implies \sigma_R = \sqrt{136} = 2\sqrt{34}

The total variance of a combined dataset equals the average within-group variance (

36

) plus the between-group variance around the combined mean (

(40-50)^2 = 100

). Thus

\sigma_R^2 = 36 + 100 = 136

.

Key Concept

Pooled standard deviation for combined datasets with differing means

Estimated Time:2m 0s

Question 1618Question →

Dataset $A$ consists of 100 distinct positive numbers. Dataset $B$ is created by replacing every number in Dataset $A$ that is strictly greater than the 75th percentile of Dataset $A$ with the value of the 75th percentile of Dataset $A$ . All numbers less than or equal to the 75th percentile remain unchanged. Which of the following statements comparing Dataset $B$ to Dataset $A$ must be true? Select all such statements.

Select all that apply

The 25th percentile of Dataset

B

is equal to the 25th percentile of Dataset

A

.

The interquartile range of Dataset

B

is equal to the interquartile range of Dataset

A

.

The standard deviation of Dataset

B

is greater than the standard deviation of Dataset

A

.

The range of Dataset

B

is strictly less than the range of Dataset

A

.

The median of Dataset

B

is strictly greater than the median of Dataset

A

.

Show answer & explanation

Answer: The 25th percentile of Dataset $B$ is equal to the 25th percentile of Dataset $A$ .; The interquartile range of Dataset $B$ is equal to the interquartile range of Dataset $A$ .; The range of Dataset $B$ is strictly less than the range of Dataset $A$ .

Answer

The statements asserting that the 25th percentile remains equal, the interquartile range remains equal, and the range is strictly smaller for Dataset B compared to Dataset A are all correct.

The statements confirming that the 25th percentile remains equal, the interquartile range remains equal, and the range decreases are correct. The lower 75% of the ordered dataset is untouched, leaving

P_{25}

and

P_{75}

unchanged, which preserves the IQR. Furthermore, replacing the top 25 distinct values with

P_{75}

lowers the maximum value while keeping the minimum value the same, strictly decreasing the range.

Step-by-Step Solution

1

Analyze how Dataset B is constructed from Dataset A

Since all 100 values in Dataset A are distinct, exactly 25 values are strictly greater than the 75th percentile (

P_{75}

). In Dataset B, these top 25 values are replaced by

P_{75}

, while the bottom 75 values remain unchanged.

Understanding which specific data points change allows us to determine positional and dispersion metrics.

2

Evaluate positional statistics (25th percentile, median, 75th percentile)

The lower 75% of the ordered dataset is identical in Dataset A and Dataset B. Thus,

P_{25}(B) = P_{25}(A)

,

\text{Median}(B) = \text{Median}(A)

, and

P_{75}(B) = P_{75}(A)

.

Percentiles at or below the 75th percentile depend only on the values at or below those percentile ranks.

3

Evaluate Interquartile Range (IQR) and Range

\text{IQR}(B) = P_{75}(B) - P_{25}(B) = P_{75}(A) - P_{25}(A) = \text{IQR}(A)

. For range,

\text{Min}(B) = \text{Min}(A)

, but

\text{Max}(B) = P_{75}(A) < \text{Max}(A)

. Therefore,

\text{Range}(B) < \text{Range}(A)

.

IQR depends on

P_{75}

and

P_{25}

, which are unchanged. Range depends on Max and Min; decreasing the maximum value decreases the range.

4

Evaluate Standard Deviation

All altered values were in the upper tail and were moved closer to the center of the distribution. Reducing extreme values decreases variance and standard deviation.

Standard deviation measures average squared distance from the mean; pulling upper extreme values inward reduces standard deviation.

Key Concept

Effect of upper-tail data transformation on measures of position (percentiles, median) and dispersion (range, IQR, standard deviation).

Question 1619Question →

In the $xy$ -plane, line $k$ passes through the point $(2, -1)$ and is perpendicular to the line $3x - 2y = 6$ . Line $m$ is parallel to line $k$ . If the distance between line $k$ and line $m$ is $\sqrt{13}$ units and line $m$ has a positive $y$ -intercept, what is the $y$ -intercept of line $m$ ?

\frac{14}{3}

\frac{17}{2}

\frac{5}{2}

4

7

Show answer & explanation

Answer: $\frac{14}{3}$

Answer

\frac{14}{3}

First, find the slope of the given line

3x - 2y = 6

, which is

\frac{3}{2}

. Because line

k

is perpendicular, its slope is

-\frac{2}{3}

. Using the point

(2, -1)

, line

k

has equation

2x + 3y - 1 = 0

. Line

m

is parallel, so it has equation

2x + 3y + C = 0

. The distance between two parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

. Setting

\frac{|C - (-1)|}{\sqrt{2^2 + 3^2}} = \sqrt{13}

yields

|C + 1| = 13

, which gives

C = 12

or

C = -14

. The

y

-intercept of line

m

is

-\frac{C}{3}

. Since the

y

-intercept must be positive,

C

must be negative, so

C = -14

. Therefore, the

y

-intercept is

-\frac{-14}{3} = \frac{14}{3}

.

Step-by-Step Solution

1

Determine the slope of line

k

.

The line

3x - 2y = 6

has slope

m_1 = \frac{3}{2}

. Because line

k

is perpendicular to it, the slope of line

k

is

m_k = -\frac{2}{3}

.

Perpendicular lines have negative reciprocal slopes.

2

Write the standard form equation of line

k

.

Using point-slope form with

(2, -1)

:

y - (-1) = -\frac{2}{3}(x - 2) \implies 2x + 3y - 1 = 0

.

Standard form

Ax + By + C_1 = 0

is required to apply the distance formula between parallel lines.

3

Set up the equation for line

m

and use the distance formula between parallel lines.

Since line

m

is parallel to line

k

, its equation is

2x + 3y + C = 0

. The distance between line

k

and line

m

is

d = \frac{|C - (-1)|}{\sqrt{2^2 + 3^2}} = \frac{|C + 1|}{\sqrt{13}} = \sqrt{13}

.

The distance between parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

.

4

Solve for constant

C

and calculate the

y

-intercept of line

m

.

|C + 1| = 13 \implies C + 1 = 13

or

C + 1 = -13

, giving

C = 12

or

C = -14

. The

y

-intercept of

2x + 3y + C = 0

is

-\frac{C}{3}

. For a positive

y

-intercept,

C

must be negative, so

C = -14

. Thus, the

y

-intercept is

-\frac{-14}{3} = \frac{14}{3}

.

The problem states that line

m

has a positive

y

-intercept.

Key Concept

Distance between parallel lines and perpendicular slope relationships in coordinate geometry

Estimated Time:3m 0s

Question 1620Question →

Let $P(x) = x^2 - mx + n$ be a quadratic polynomial with real coefficients $m$ and $n$ , having two distinct real roots $\alpha$ and \beta. If the roots satisfy the system of equations $\alpha^3 + \beta^3 = m(n + 7)$ and $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{10}{n^2}$ , which of the following statements MUST be true? Select all such statements.

Select all that apply

The product of the roots,

n

, is equal to

1.5

.

The sum of the squares of the roots,

\alpha^2 + \beta^2

, is equal to

10

.

The discriminant of the polynomial

P(x)

is equal to

7

.

The absolute value of the sum of the roots,

|\alpha + \beta|

, is equal to

5

.

The roots

\alpha

and

\beta

have opposite signs.

Show answer & explanation

Answer: The product of the roots, $n$ , is equal to $1.5$ .; The sum of the squares of the roots, $\alpha^2 + \beta^2$ , is equal to $10$ .; The discriminant of the polynomial $P(x)$ is equal to $7$ .

Answer

The correct statements are those asserting that the product of the roots is

1.5

, the sum of the squares of the roots is

10

, and the discriminant of

P(x)

is

7

.

Using Vieta's formulas and algebraic identity expansions for

\alpha^3 + \beta^3

and

\frac{1}{\alpha^2} + \frac{1}{\beta^2}

establishes the system of equations

m^2 = 4n + 7

and

m^2 = 2n + 10

. Solving this system gives

n = 1.5

,

m^2 = 13

, and a discriminant

\Delta = m^2 - 4n = 7

. Thus, the product of roots is

1.5

, the sum of squares

\alpha^2 + \beta^2 = m^2 - 2n = 10

, and the discriminant is

7

.

Step-by-Step Solution

1

Apply Vieta's formulas to express sum and product of roots.

\alpha + \beta = m \quad \text{and} \quad \alpha\beta = n

Vieta's relations link polynomial coefficients directly to symmetrical root expressions.

2

Expand

\alpha^3 + \beta^3

in terms of

m

and

n

.

\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = m^3 - 3mn = m(m^2 - 3n)

Using algebraic identities converts root powers into functions of

m

and

n

.

3

Equate the expression from Step 2 to

m(n + 7)

to find an equation for

m^2

.

m(m^2 - 3n) = m(n + 7) \implies m^2 - 3n = n + 7 \implies m^2 = 4n + 7

Since the roots are distinct,

m \neq 0

, allowing division by

m

.

4

Simplify the second given equation

\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{10}{n^2}

.

\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{m^2 - 2n}{n^2} = \frac{10}{n^2} \implies m^2 - 2n = 10 \implies m^2 = 2n + 10

Combining fractions over a common denominator

(\alpha\beta)^2 = n^2

isolates

m^2 - 2n

.

5

Solve for

n

,

m^2

, and the discriminant

\Delta

.

4n + 7 = 2n + 10 \implies 2n = 3 \implies n = 1.5; \quad m^2 = 13; \quad \Delta = m^2 - 4n = 13 - 6 = 7

Equating the two expressions for

m^2

yields unique values for

n

,

m^2

, and

\Delta

.

Key Concept

Quadratic Equations, Vieta's Formulas, and Symmetric Polynomial Expressions

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