帮我解答第一题---1. 已知集合 $A = \{x | -5 < x^3 < 5\}$, $B = \{-3, -1, 0, 2, 3\}$ , 则 $A \cap B = (\cdots\cdots)$
A. $\{-1, 0\}$
B. $\{2, 3\}$
C. $\{-3, -1, 0\}$
D. $\{-1, 0, 2\}$
2. 若 $\frac{z}{z-1} = 1+i$ , 则 $z = (\cdots\cdots)$
A. $-1-i$
B. $-1+i$
C. $1-i$
D. $1+i$
3. 已知向量 $\vec{a} = (0,1)$, $\vec{b} = (2,x)$, 若 $\vec{b} \perp (\vec{b} - 4\vec{a})$, 则 $x = (\cdots\cdots)$
A. $-2$
B. $-1$
C. $1$
D. $2$
4. 已知 $\cos(\alpha + \beta) = m$, $\tan \alpha \tan \beta = 2$, 则 $\cos(\alpha - \beta) = (\cdots\cdots)$
A. $-3m$
B. $-\frac{m}{3}$
C. $\frac{m}{3}$
D. $3m$
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Let's solve this step by step. We have set A defined by the inequality negative 5 less than x cubed less than 5, and set B containing the elements negative 3, negative 1, 0, 2, and 3. To find their intersection, we need to check which elements of B satisfy the condition for set A.
Now let's check each element of set B. For x equals negative 3, we calculate negative 3 cubed which equals negative 27. Since negative 27 is less than negative 5, negative 3 is not in set A. For x equals negative 1, we get negative 1 cubed equals negative 1. Since negative 5 is less than negative 1 which is less than 5, negative 1 is in set A. For x equals 0, we have 0 cubed equals 0, and since negative 5 is less than 0 which is less than 5, 0 is also in set A.
Let's continue with the remaining elements. For x equals 2, we calculate 2 cubed which equals 8. Since 8 is greater than 5, 2 is not in set A. For x equals 3, we get 3 cubed equals 27. Since 27 is greater than 5, 3 is also not in set A. Therefore, the elements that belong to both sets A and B are negative 1 and 0.
Therefore, the intersection of sets A and B is the set containing negative 1 and 0. Comparing this with the given options, we can see that option A, which is the set containing negative 1 and 0, matches our result exactly. The correct answer is A.
To summarize, we solved this set intersection problem by systematically checking each element of set B against the condition for set A. We calculated the cube of each number and verified whether it falls within the range from negative 5 to 5. Only negative 1 and 0 satisfied this condition, giving us the final answer of A.