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1. 已知集合 M={2,1,0,1,2},N={xx2x60}M=\{-2,-1,0,1,2\}, N=\left\{x \mid x^{2}-x-6 \geqslant 0\right\}, 则 $...
Sep 23, 2024
Solution by Steps
step 2
Factoring the quadratic gives us (x3)(x+2)=0 (x - 3)(x + 2) = 0 , which yields the roots x=3 x = 3 and x=2 x = -2
step 3
The inequality x2x60 x^2 - x - 6 \geq 0 holds true for x2 x \leq -2 or x3 x \geq 3
step 4
Now we define the sets: M={2,1,0,1} M = \{-2, -1, 0, 1\} and N={xx2 or x3} N = \{x \mid x \leq -2 \text{ or } x \geq 3\}
step 5
The intersection MN M \cap N includes the elements from M M that are also in N N . The only element from M M that satisfies x2 x \leq -2 is 2 -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 M that satisfies the conditions of set N N is 2 -2 .
Generated Graph
Solution by Steps
step 2
From sin(αβ)=13 \sin(\alpha - \beta) = \frac{1}{3} , we can express β \beta in terms of α \alpha : β=αsin1(13)+2πn \beta = \alpha - \sin^{-1}\left(\frac{1}{3}\right) + 2\pi n or β=α+sin1(13)2πn \beta = \alpha + \sin^{-1}\left(\frac{1}{3}\right) - 2\pi n for nZ n \in \mathbb{Z}
step 3
Using cos(α)sin(β)=16 \cos(\alpha) \sin(\beta) = \frac{1}{6} , we can substitute sin(β) \sin(\beta) using the identity sin(β)=sin(αsin1(13)) \sin(\beta) = \sin(\alpha - \sin^{-1}\left(\frac{1}{3}\right)) to find cos(2α+2β) \cos(2\alpha + 2\beta)
step 4
We apply the double angle formula: cos(2α+2β)=cos(2(α+β))=2cos2(α+β)1 \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(2α+2β)=19 \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 p1 p_1 (60 dB) is greater than the maximum of p2 p_2 (60 dB)
step 3
Option B states p_2 > 10p_3 . The maximum of p2 p_2 (60 dB) is not greater than 10×40 10 \times 40 (400 dB), so this is false
step 4
Option C states p3=100p0 p_3 = 100p_0 . Since p3 p_3 is 40 dB, this does not hold true as 40100p0 40 \neq 100p_0
step 5
Option D states p1100p2 p_1 \leq 100p_2 . The maximum of p1 p_1 (90 dB) is not less than or equal to 100×60 100 \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(ωx)1 f(x) = \cos(\omega x) - 1 and set it equal to zero: cos(ωx)1=0 \cos(\omega x) - 1 = 0
step 2
This implies cos(ωx)=1 \cos(\omega x) = 1 , which occurs at ωx=2πn \omega x = 2\pi n for nZ n \in \mathbb{Z}
step 3
Therefore, x=2πnω x = \frac{2\pi n}{\omega} . We need to find the values of n n such that x x is in the interval [0,2π] [0, 2\pi]
step 4
The values of n n can be 0,1,2 0, 1, 2 to ensure x x remains within the interval, leading to three zeros of the function
step 5
Thus, we have n=0 n = 0 gives x=0 x = 0 , n=1 n = 1 gives x=2πω x = \frac{2\pi}{\omega} , and n=2 n = 2 gives x=4πω x = \frac{4\pi}{\omega}
step 6
For 4πω \frac{4\pi}{\omega} to be less than or equal to 2π 2\pi , we require ω2 \omega \geq 2 . Therefore, the range of ω \omega is (0,2) (0, 2)
Answer
ω \omega 的取值范围是 (0,2) (0, 2)
Key Concept
The zeros of the cosine function and their relation to the parameter ω \omega
Explanation
The function cos(ωx)1=0 \cos(\omega x) - 1 = 0 has zeros determined by the values of n n , which must be limited to ensure x x remains within the specified interval. This leads to the conclusion about the range of ω \omega .
Solution by Steps
step 1
We start with the rectangular prism ABCDA1B1C1D1ABCD-A_1B_1C_1D_1 where AB=2AB=2 and AA1=4AA_1=4. The points A2,B2,C2,D2A_2, B_2, C_2, D_2 are on the edges AA1,BB1,CC1,DD1AA_1, BB_1, CC_1, DD_1 respectively, with AA2=1AA_2=1, BB2=DD2=2BB_2=DD_2=2, and CC2=3CC_2=3
step 2
To prove that B2C2A2D2B_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 B2PB_2P when the dihedral angle PA2C2D2P-A_2C_2-D_2 is 150150^{\circ}. We can use the cosine rule in triangle B2A2PB_2A_2P
step 4
The cosine rule states that c2=a2+b22abcos(C)c^2 = a^2 + b^2 - 2ab \cos(C). Here, let B2A2=2B_2A_2 = 2 and A2PA_2P be the unknown we need to find. The angle CC is 150150^{\circ}, so cos(150)=32\cos(150^{\circ}) = -\frac{\sqrt{3}}{2}
step 5
Setting up the equation: B2P2=B2A22+A2P22B2A2A2P(32)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 B2PB_2P
Answer
The length B2PB_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 ABCDA1B1C1D1ABCD-A_1B_1C_1D_1 where AB=2AB=2 and AA1=4AA_1=4. The points A2,B2,C2,D2A_2, B_2, C_2, D_2 are on the edges AA1,BB1,CC1,DD1AA_1, BB_1, CC_1, DD_1 respectively, with AA2=1AA_2=1, BB2=2BB_2=2, and CC2=3CC_2=3, DD2=2DD_2=2
step 2
To prove that B2C2A2D2B_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 B2PB_2P when the dihedral angle PA2C2D2P-A_2C_2-D_2 is 150150^{\circ}. We can use the cosine rule in triangle B2A2D2B_2A_2D_2 to find B2PB_2P
step 4
Applying the cosine rule, we have B2P2=B2A22+A2D222B2A2A2D2cos(150)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 B2PB_2P
Answer
The length B2PB_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.
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