Giải
Bài 1.
a) Hàm số xác định khi:
$\left\{\begin{matrix}\sin{\left (\dfrac{x}{3} - \dfrac{\pi}{4} \right )} \neq 0\\\tan{4x} \neq -1\\\cos{4x} \neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}x \neq \dfrac{3\pi}{4} + 3k\pi\\x \neq \dfrac{-\pi}{16} + \dfrac{k\pi}{4}\\x \neq \dfrac{\pi}{8} + k\dfrac{\pi}{4}\end{matrix}\right. \, (k \in Z)$
b) Hàm số xác định khi:
$\left\{\begin{matrix}\cos{\left (2x + \dfrac{\pi}{7} \right )} \neq 0\\\cos{4x} \neq 0\\\tan{4x} \neq \cos{2x}\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}x \neq \dfrac{5\pi}{28} + \dfrac{k\pi}{2}\\x \neq \dfrac{\pi}{8} + k\dfrac{\pi}{4}\\\tan{4x} \neq \cos{2x} \, (1)\end{matrix}\right. \, (k \in Z)$
Ta có:
$(1) \Leftrightarrow \cos{2x}\left ( \dfrac{2\sin{2x}}{\cos{4x}} - 1\right ) \neq 0 \Leftrightarrow \left\{\begin{matrix}\cos{2x} \neq 0\\2\sin{2x} - (1 - 2\sin^2{2x}) \neq 0\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}x \neq \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\\sin{2x} \neq \dfrac{-1 + \sqrt{3}}{2}\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}x \neq \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x \neq \dfrac{\arcsin{\dfrac{\sqrt{3} - 1}{2}}}{2} + k\pi\\x \neq \dfrac{\pi}{2} - \dfrac{\arcsin{\dfrac{\sqrt{3} - 1}{2}}}{2} + k\pi\end{matrix}\right. \, (k \in Z)$
Bài 2.
a) $y = \dfrac{5}{2} - \dfrac{1}{2}\cos{4x} - \dfrac{2\tan{x}}{1 + \tan^2{x}}$
$y = \dfrac{5}{2} - \dfrac{1}{2}(1 - 2\sin^2{2x}) - \sin{2x} = 2 + \sin^2{2x} - \sin{2x}$
$y = \sin^2{2x} - 1 - (\sin{2x} + 1) + 4 \leq 4$
Dấu "=" xảy ra khi $\sin{2x} = -1$
b) $y = \sin^2{\left ( \dfrac{15\pi}{8} - 4x\right )} - \sin^2{\left ( \dfrac{17\pi}{8} - 4x\right )}$
$= \dfrac{1 - \cos{\left ( \dfrac{15\pi}{4} - 8x\right )}}{2} - \dfrac{1 - \cos{\left ( \dfrac{17\pi}{4} - 8x\right )}}{2}$
$= \dfrac{1}{2}\left [ \cos{\left ( \dfrac{17\pi}{4} - 8x \right )} - \cos{\left ( \dfrac{15\pi}{4} - 8x\right )} \right ]$
$= - \sin{\dfrac{\pi}{4}}\sin{\left (4\pi - 8x \right )} = \dfrac{\sqrt{2}}{2}\sin{8x} \leq \dfrac{\sqrt{2}}{2}$
Dấu "=" xảy ra khi $\sin{8x} = 1$
Bài 3.
a) $y = \sin^6{x} + \cos^6{x} = 1 - \dfrac{3}{4}\sin^2{2x} \geq \dfrac{1}{4}$
Dấu "=" xảy ra khi $\sin{2x} = \pm 1$
b) $y = 2(1 + \sin{2x}\sin{3x}) - \dfrac{1}{2}(\cos{4x} + \cos{6x})$
$= 2 + \cos{x} - \cos{5x} - \cos{x}\cos{5x} = (\cos{x} + 1)(1 - \cos{5x}) + 1 \geq 1$
Dấu "=" xảy ra khi $\cos{x} = -1$ hoặc $\cos{5x} = 1$
c) $y = \dfrac{\cot{x} - \tan{x}}{1 + \cos{4x}}$
$= \dfrac{\dfrac{\cos{x}}{\sin{x}} - \dfrac{\sin{x}}{\cos{x}}}{2\cos^2{2x}} = \dfrac{\cos{2x}}{2\sin{x}\cos{x}\cos^2{2x}}$
$= \dfrac{1}{\sin{2x}\cos{2x}} = \dfrac{2}{\sin{4x}}$
Do $x \in \left (0; \dfrac{\pi}{4} \right )$ nên $4x \in (0; \pi)$
Vậy $0 < \sin{4x} \leq 1$. Vậy: $y \geq 2$. Dấu "=" xảy ra khi $\sin{4x} = 1$
Bài viết đã được chỉnh sửa nội dung bởi Phạm Hữu Bảo Chung: 15-08-2013 - 21:00
