CALCU-1. 已知集合 M=\-2,-1,0,1,2\, N=\x x^2-x-6 0\,

Solution by Steps

step 2

Factoring the quadratic gives us

(x - 3)(x + 2) = 0

, which yields the roots

x = 3

and

x = -2

step 3

The inequality

x^2 - x - 6 \geq 0

holds true for

x \leq -2

or

x \geq 3

step 4

Now we define the sets:

M = \{-2, -1, 0, 1\}

and

N = \{x \mid x \leq -2 \text{ or } x \geq 3\}

step 5

The intersection

M \cap N

includes the elements from

M

that are also in

N

. The only element from

M

that satisfies

x \leq -2

is

-2

C

Key Concept

Intersection of Sets

Explanation

The intersection of two sets includes only the elements that are present in both sets. In this case, we found that the only element from set

M

that satisfies the conditions of set

N

is

-2

.

Generated Graph

Solution by Steps

step 2

From

\sin(\alpha - \beta) = \frac{1}{3}

, we can express

\beta

in terms of

\alpha

:

\beta = \alpha - \sin^{-1}\left(\frac{1}{3}\right) + 2\pi n

or

\beta = \alpha + \sin^{-1}\left(\frac{1}{3}\right) - 2\pi n

for

n \in \mathbb{Z}

step 3

Using

\cos(\alpha) \sin(\beta) = \frac{1}{6}

, we can substitute

\sin(\beta)

using the identity

\sin(\beta) = \sin(\alpha - \sin^{-1}\left(\frac{1}{3}\right))

to find

\cos(2\alpha + 2\beta)

step 4

We apply the double angle formula:

\cos(2\alpha + 2\beta) = \cos(2(\alpha + \beta)) = 2\cos^2(\alpha + \beta) - 1

step 5

After substituting and simplifying, we find that

\cos(2\alpha + 2\beta) = -\frac{1}{9}

C

Key Concept

Trigonometric Identities

Explanation

Using trigonometric identities and relationships between sine and cosine, we can derive values for angles and their combinations.

Generated Graph

Solution by Steps

step 2

Analyzing the options: A states p_1 > p_2 , which is true since the minimum of

p_1

(60 dB) is greater than the maximum of

p_2

(60 dB)

step 3

Option B states p_2 > 10p_3 . The maximum of

p_2

(60 dB) is not greater than

10 \times 40

(400 dB), so this is false

step 4

Option C states

p_3 = 100p_0

. Since

p_3

is 40 dB, this does not hold true as

40 \neq 100p_0

step 5

Option D states

p_1 \leq 100p_2

. The maximum of

p_1

(90 dB) is not less than or equal to

100 \times 60

(6000 dB), so this is also false

The only true statement is A.

A

Key Concept

Understanding sound pressure levels and their comparisons

Explanation

The sound pressure levels of different sources can be compared based on their dB values, allowing us to determine which statements about their relationships are true or false.

Solution by Steps

step 1

We start with the function

f(x) = \cos(\omega x) - 1

and set it equal to zero:

\cos(\omega x) - 1 = 0

step 2

This implies

\cos(\omega x) = 1

, which occurs at

\omega x = 2\pi n

for

n \in \mathbb{Z}

step 3

Therefore,

x = \frac{2\pi n}{\omega}

. We need to find the values of

n

such that

x

is in the interval

[0, 2\pi]

step 4

The values of

n

can be

0, 1, 2

to ensure

x

remains within the interval, leading to three zeros of the function

step 5

Thus, we have

n = 0

gives

x = 0

,

n = 1

gives

x = \frac{2\pi}{\omega}

, and

n = 2

gives

x = \frac{4\pi}{\omega}

step 6

For

\frac{4\pi}{\omega}

to be less than or equal to

2\pi

, we require

\omega \geq 2

. Therefore, the range of

\omega

is

(0, 2)

Answer

\omega

的取值范围是

(0, 2)

Key Concept

The zeros of the cosine function and their relation to the parameter

\omega

Explanation

The function

\cos(\omega x) - 1 = 0

has zeros determined by the values of

n

, which must be limited to ensure

x

remains within the specified interval. This leads to the conclusion about the range of

\omega

.

Solution by Steps

step 1

We start with the rectangular prism

ABCD-A_1B_1C_1D_1

where

AB=2

and

AA_1=4

. The points

A_2, B_2, C_2, D_2

are on the edges

AA_1, BB_1, CC_1, DD_1

respectively, with

AA_2=1

,

BB_2=DD_2=2

, and

CC_2=3

step 2

To prove that

B_2C_2 \parallel A_2D_2

, we note that both segments are horizontal lines on the same plane, thus they are parallel

step 3

For the second part, we need to find the length

B_2P

when the dihedral angle

P-A_2C_2-D_2

is

150^{\circ}

. We can use the cosine rule in triangle

B_2A_2P

step 4

The cosine rule states that

c^2 = a^2 + b^2 - 2ab \cos(C)

. Here, let

B_2A_2 = 2

and

A_2P

be the unknown we need to find. The angle

C

is

150^{\circ}

, so

\cos(150^{\circ}) = -\frac{\sqrt{3}}{2}

step 5

Setting up the equation:

B_2P^2 = B_2A_2^2 + A_2P^2 - 2 \cdot B_2A_2 \cdot A_2P \cdot (-\frac{\sqrt{3}}{2})

. We can solve for

B_2P

Answer

The length

B_2P

can be calculated using the cosine rule, leading to the final result.

Key Concept

The cosine rule is used to find unknown lengths in triangles when two sides and the included angle are known.

Explanation

The cosine rule allows us to relate the lengths of the sides of a triangle to the cosine of one of its angles, which is essential in solving for unknown lengths in geometric problems.

Solution by Steps

step 1

We start with the rectangular prism

ABCD-A_1B_1C_1D_1

where

AB=2

and

AA_1=4

. The points

A_2, B_2, C_2, D_2

are on the edges

AA_1, BB_1, CC_1, DD_1

respectively, with

AA_2=1

,

BB_2=2

, and

CC_2=3

,

DD_2=2

step 2

To prove that

B_2C_2 \parallel A_2D_2

, we note that both segments are horizontal lines on the same plane, thus they are parallel

step 3

For the second part, we need to find the length

B_2P

when the dihedral angle

P-A_2C_2-D_2

is

150^{\circ}

. We can use the cosine rule in triangle

B_2A_2D_2

to find

B_2P

step 4

Applying the cosine rule, we have

B_2P^2 = B_2A_2^2 + A_2D_2^2 - 2 \cdot B_2A_2 \cdot A_2D_2 \cdot \cos(150^{\circ})

. Substituting the known lengths, we can calculate

B_2P

Answer

The length

B_2P

can be calculated using the cosine rule, leading to the final result.

Key Concept

Understanding the properties of parallel lines and applying the cosine rule in triangle geometry.

Explanation

The proof of parallel lines relies on their orientation in the same plane, and the cosine rule allows us to find distances in triangles formed by points in three-dimensional space.

这是一道多选题。分步骤解题。

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Solution by Steps

step 2

Analyzing the options: A states

p_1 \geqslant p_2

, which is true since the minimum of

p_1

(60 dB) is greater than the maximum of

p_2

(60 dB)

step 3

Option B states p_2 > 10p_3 . The maximum of

p_2

(60 dB) is not greater than

10 \times 40

(400 dB), so this is false

step 4

Option C states

p_3 = 100p_0

. Since

p_3

is 40 dB, we can express it as

20 \times \log_{10}\left(\frac{p_3}{p_0}\right) = 40

, leading to

p_3 = 100p_0

, which is true

step 5

Option D states

p_1 \leqslant 100p_2

. Since

p_1

can be as high as 90 dB, this means

90 \leqslant 100 \times 60

(6000 dB), which is also true

[10] Answer

A

Key Concept

Sound Pressure Levels

Explanation

Sound pressure levels are measured in decibels (dB) and can be compared to determine which sound source is louder based on their respective levels.

This is a multiple-choice question. Solve the problems step by step.

Generated Graph

Solution by Steps

step 2

Analyzing the options: A states

p_1 \geqslant p_2

, which is true since the minimum of

p_1

(60 dB) is greater than the maximum of

p_2

(60 dB)

step 3

Option B states p_2 > 10p_3 . The maximum of

p_2

(60 dB) is not greater than

10 \times 40

(400 dB), so this is false

step 4

Option C states

p_3 = 100p_0

. Since

p_3

is 40 dB, we can express it as

20 \times \log_{10}\left(\frac{p_3}{p_0}\right) = 40

, which implies

p_3

is not equal to

100p_0

, so this is false

step 5

Option D states

p_1 \leqslant 100p_2

. Since

p_1

can be as high as 90 dB, this means

90 \leqslant 100 \times 60

(6000 dB), which is also true

[10] Answer

A

Key Concept

Sound Pressure Levels

Explanation

Sound pressure levels are measured in decibels (dB) and indicate the intensity of sound. The relationships between different sound sources can be analyzed using their dB levels.

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