Giải thích các bước giải:
$\begin{array}{l} C1:\\ y = \sqrt 3 \sin x + \cos x + 2\\ \Leftrightarrow \frac{y}{2} = \frac{{\sqrt 3 }}{2}\sin x + \frac{1}{2}\cos x + 1\\ \Leftrightarrow \frac{y}{2} = \sin \left( {x + \frac{\pi }{6}} \right) + 1 \end{array}$
Mà − 1 ≤ sin ( x + π 6 ) ≤ 1 - 1 \le \sin \left( {x + \frac{\pi }{6}} \right) \le 1 − 1 ≤ sin ( x + 6 π ) ≤ 1
$\begin{array}{l} \Leftrightarrow 0 \le \sin \left( {x + \frac{\pi }{6}} \right) + 1 \le 2\\ \Leftrightarrow 0 \le \frac{y}{2} \le 2\\ \Leftrightarrow 0 \le y \le 4 \end{array}$
+) Nếu y = 4 y=4 y = 4
⇔ sin ( x + π 6 ) = 1 ⇔ x + π 6 = π 2 + k 2 π ⇔ x = π 3 + k 2 π ( k ∈ Z ) \Leftrightarrow \sin \left( {x + \frac{\pi }{6}} \right) = 1 \Leftrightarrow x + \frac{\pi }{6} = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{\pi }{3} + k2\pi \left( {k \in Z} \right) ⇔ sin ( x + 6 π ) = 1 ⇔ x + 6 π = 2 π + k 2 π ⇔ x = 3 π + k 2 π ( k ∈ Z )
→ M a x y = 4 ⇔ x = π 3 + k 2 π ( k ∈ Z ) \to Maxy = 4 \Leftrightarrow x = \frac{\pi }{3} + k2\pi \left( {k \in Z} \right) → M a x y = 4 ⇔ x = 3 π + k 2 π ( k ∈ Z )
+) Nếu y = 0 y=0 y = 0
⇔ sin ( x + π 6 ) = − 1 ⇔ x + π 6 = − π 2 + k 2 π ⇔ x = − 2 π 3 + k 2 π ( k ∈ Z ) \Leftrightarrow \sin \left( {x + \frac{\pi }{6}} \right) = - 1 \Leftrightarrow x + \frac{\pi }{6} = - \frac{\pi }{2} + k2\pi \Leftrightarrow x = - \frac{{2\pi }}{3} + k2\pi \left( {k \in Z} \right) ⇔ sin ( x + 6 π ) = − 1 ⇔ x + 6 π = − 2 π + k 2 π ⇔ x = − 3 2 π + k 2 π ( k ∈ Z )
⇒ M i n y = 0 ⇔ x = − 2 π 3 + k 2 π ( k ∈ Z ) \Rightarrow Miny = 0 \Leftrightarrow x = - \frac{{2\pi }}{3} + k2\pi \left( {k \in Z} \right) ⇒ M i n y = 0 ⇔ x = − 3 2 π + k 2 π ( k ∈ Z )
Vậy
M a x y = 4 ⇔ x = π 3 + k 2 π ( k ∈ Z ) Maxy = 4 \Leftrightarrow x = \frac{\pi }{3} + k2\pi \left( {k \in Z} \right) M a x y = 4 ⇔ x = 3 π + k 2 π ( k ∈ Z )
M i n y = 0 ⇔ x = − 2 π 3 + k 2 π ( k ∈ Z ) Miny = 0 \Leftrightarrow x = - \frac{{2\pi }}{3} + k2\pi \left( {k \in Z} \right) M i n y = 0 ⇔ x = − 3 2 π + k 2 π ( k ∈ Z )
C2:
Đường thẳng ( d ) : 2 x + y + 3 = 0 \left( d \right):2x + y + 3 = 0 ( d ) : 2 x + y + 3 = 0
Lấy điểm A ( − 1 , − 1 ) ∈ ( d ) A\left( { - 1, - 1} \right) \in \left( d \right) A ( − 1 , − 1 ) ∈ ( d )
T v → = ( 5 , − 1 ) A = A ′ ⇒ A ′ ( 4 , − 2 ) {T_{\overrightarrow v = \left( {5, - 1} \right)}}A = A' \Rightarrow A'\left( {4, - 2} \right) T v = ( 5 , − 1 ) A = A ′ ⇒ A ′ ( 4 , − 2 )
Và ${T_{\overrightarrow v = \left( {5, - 1} \right)}}\left( d \right) = \left( d \right)' \Rightarrow \left\{ \begin{array}{l} A'\left( {4, - 2} \right) \in \left( d \right)'\\ \left( d \right)'//\left( d \right) \end{array} \right.$
$\begin{array}{l} \Rightarrow \left( d \right)':2\left( {x - 4} \right) + 1\left( {y + 2} \right) = 0\\ \Rightarrow \left( d \right)':2x + y - 6 = 0 \end{array}$
Vậy ( d ) ′ : 2 x + y − 6 = 0 \left( d \right)':2x + y - 6 = 0 ( d ) ′ : 2 x + y − 6 = 0