Tìm min max cua y=sinx /sinx-2cosx+3 Tìm min max cua hàm số y=2+sin2x-2√2 cos2x 03/07/2021 Bởi Mary Tìm min max cua y=sinx /sinx-2cosx+3 Tìm min max cua hàm số y=2+sin2x-2√2 cos2x
a) $y = \dfrac{\sin x}{\sin x – 2\cos x + 3}$ $\Leftrightarrow y\sin x – 2y\cos x + 3y = \sin x$ $\Leftrightarrow (y – 1)\sin x – 2y\cos x + 3y = 0$ Phương trình có nghiệm $\Leftrightarrow (y – 1)^2 + (2y)^2 \geq (3y)^2$ $\Leftrightarrow 5y^2 – 2y + 1 \geq 9y^2$ $\Leftrightarrow 4y^2 + 2y – 1 \leq 0$ $\Leftrightarrow \dfrac{-1 – \sqrt5}{4} \leq y \leq \dfrac{-1 + \sqrt5}{4}$ Vậy $\min y = \dfrac{-1 -\sqrt5}{4};\, \max y = \dfrac{-1 +\sqrt5}{4}$ b) $y = 2 + \sin2x – 2\sqrt2\cos2x$ $\Leftrightarrow \sin2x – 2\sqrt2\cos2x = y – 2$ Phương trình có nghiệm $\Leftrightarrow 1^2 + (2\sqrt2)^2 \geq (y – 2)^2$ $\Leftrightarrow (y-2)^2 \leq 9$ $\Leftrightarrow -3 \leq y – 2 \leq 3$ $\Leftrightarrow – 1 \leq y \leq 5$ Vậy $\min y = -1;\, \max y – 5$ Bình luận
`a) y = (sin x)/(sin x – 2cos x + 3)` Dễ dàng thấy `sin x – 2cos x + 3 ne 0` `=> ysin x – 2ycos x + 3y = sin x` `<=> ysin x – 2ycos x + 3y – sin x = 0` `<=> (y – 1)sin x – 2ycos x = -3y` Để phương trình có nghiệm `<=> (y – 1)^2 + (2y)^2 >= (-3y)^2` `<=> y^2 – 2y + 1 + 4y^2 >= 9y²` `<=> 4y² ≤ 1 – 2y` `<=> 4y² + 2y – 1 ≤ 0` `<=> (-1 – \sqrt{5})/4 ≤ y ≤ (\sqrt{5} – 1)/4` `b) y = 2 + sin 2x – 2sqrt{2}cos 2x` Ta có: `-sqrt{1^2 + (2\sqrt{2})^2} ≤ sin 2x – 2sqrt{2}cos 2x ≤ sqrt{1^2 + (2\sqrt{2})^2}` `<=> -3 ≤ sin 2x – 2sqrt{2}cos 2x ≤ 3` `<=> -1 ≤ y ≤ 5` Bình luận
a) $y = \dfrac{\sin x}{\sin x – 2\cos x + 3}$
$\Leftrightarrow y\sin x – 2y\cos x + 3y = \sin x$
$\Leftrightarrow (y – 1)\sin x – 2y\cos x + 3y = 0$
Phương trình có nghiệm
$\Leftrightarrow (y – 1)^2 + (2y)^2 \geq (3y)^2$
$\Leftrightarrow 5y^2 – 2y + 1 \geq 9y^2$
$\Leftrightarrow 4y^2 + 2y – 1 \leq 0$
$\Leftrightarrow \dfrac{-1 – \sqrt5}{4} \leq y \leq \dfrac{-1 + \sqrt5}{4}$
Vậy $\min y = \dfrac{-1 -\sqrt5}{4};\, \max y = \dfrac{-1 +\sqrt5}{4}$
b) $y = 2 + \sin2x – 2\sqrt2\cos2x$
$\Leftrightarrow \sin2x – 2\sqrt2\cos2x = y – 2$
Phương trình có nghiệm
$\Leftrightarrow 1^2 + (2\sqrt2)^2 \geq (y – 2)^2$
$\Leftrightarrow (y-2)^2 \leq 9$
$\Leftrightarrow -3 \leq y – 2 \leq 3$
$\Leftrightarrow – 1 \leq y \leq 5$
Vậy $\min y = -1;\, \max y – 5$
`a) y = (sin x)/(sin x – 2cos x + 3)`
Dễ dàng thấy `sin x – 2cos x + 3 ne 0`
`=> ysin x – 2ycos x + 3y = sin x`
`<=> ysin x – 2ycos x + 3y – sin x = 0`
`<=> (y – 1)sin x – 2ycos x = -3y`
Để phương trình có nghiệm
`<=> (y – 1)^2 + (2y)^2 >= (-3y)^2`
`<=> y^2 – 2y + 1 + 4y^2 >= 9y²`
`<=> 4y² ≤ 1 – 2y`
`<=> 4y² + 2y – 1 ≤ 0`
`<=> (-1 – \sqrt{5})/4 ≤ y ≤ (\sqrt{5} – 1)/4`
`b) y = 2 + sin 2x – 2sqrt{2}cos 2x`
Ta có:
`-sqrt{1^2 + (2\sqrt{2})^2} ≤ sin 2x – 2sqrt{2}cos 2x ≤ sqrt{1^2 + (2\sqrt{2})^2}`
`<=> -3 ≤ sin 2x – 2sqrt{2}cos 2x ≤ 3`
`<=> -1 ≤ y ≤ 5`