AMC12 2018 A

AMC12 2018 A · Q9

AMC12 2018 A · Q9. It mainly tests Inequalities (AM-GM etc. basic), Trigonometry (basic).

Which of the following describes the largest subset of values of $y$ within the closed interval $[0, \pi]$ for which $\sin(x + y) \le \sin(x) + \sin(y)$ for every $x$ between 0 and $\pi$, inclusive?

以下哪项描述了在闭区间 $[0, \pi]$ 内最大的 $y$ 值子集，使得对于所有 $x \in [0, \pi]$，有 $\sin(x + y) \le \sin(x) + \sin(y)$？

(A) $y = 0$ $y = 0$

(B) $0 \le y \le \frac{\pi}{4}$ $0 \le y \le \frac{\pi}{4}$

(D) $0 \le y \le \frac{3\pi}{4}$ $0 \le y \le \frac{3\pi}{4}$

(E) $0 \le y \le \pi$ $0 \le y \le \pi$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): If $0 \le x \le \pi$ and $0 \le y \le \pi$, then $\sin(x) \ge 0$, $\sin(y) \ge 0$, $\cos(x) \le 1$, and $\cos(y) \le 1$. Therefore $$ \sin(x+y)=\sin(x)\cdot\cos(y)+\cos(x)\cdot\sin(y)\le \sin(x)+\sin(y). $$ The given inequality holds for all $y$ such that $0 \le y \le \pi$.

答案（E）：若 $0 \le x \le \pi$ 且 $0 \le y \le \pi$，则 $\sin(x) \ge 0$、$\sin(y) \ge 0$、$\cos(x) \le 1$、$\cos(y) \le 1$。因此 $$ \sin(x+y)=\sin(x)\cdot\cos(y)+\cos(x)\cdot\sin(y)\le \sin(x)+\sin(y)。 $$ 所给不等式对所有满足 $0 \le y \le \pi$ 的 $y$ 都成立。

Topics