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Solution

Then $2020 + x = 2020 + 4k + 3 = 4k + 2023$ 

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Solution

Given $|\alpha^2| = 4$ and $-3 \le \lambda \le 2$. Then $|\lambda \alpha^2| = 4|\lambda|$. Minimum value at $\lambda = 0$ → 0. Maximum value at $\lambda = -3$ → $4 \times 3 = 12$. Hence, range is $[0, 12]$. $\boxed{\text{Answer: (C) } [0, 12]}$

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Solution

Case 1: $x \ge 1 \Rightarrow |x|=x,\ |1-x|=x-1$ $\Rightarrow x + (x-1) = 6 \Rightarrow 2x = 7 \Rightarrow x=\tfrac{7}{2}$ (not integer). Case 2: $x < 1 \Rightarrow |x|=-x,\ |1-x|=1-x$ $\Rightarrow -x + (1-x) = 6 \Rightarrow -2x = 5 \Rightarrow x=-\tfrac{5}{2}$ (not integer). Hence, there are **no** non-zero integral solutions.

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Solution

Area of rhombus $= \dfrac{1}{2} \times 8 \times 6 = 24$.