Toán Tìm giá trị min max của Y=sin6x-cos6x +2020 05/10/2021 By Remi Tìm giá trị min max của Y=sin6x-cos6x +2020
Đáp án: $\min y=-\dfrac1{\sqrt2}+2020$ dấu “=” xảy ra $\Leftrightarrow x=-\dfrac{\pi}{24}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$ $\max y=\dfrac1{\sqrt2}+2020$ dấu “=” xảy ra $\Leftrightarrow x=\dfrac{\pi}{8}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$. Giải thích các bước giải: $y=\sin 6x-\cos 6x+2020$ $=\dfrac1{\sqrt2}\sin\left({6x-\dfrac{\pi}4}\right)+2020$ Do $-1\le\sin x\le1$ $\forall x$ $\Rightarrow-1\le\sin\left({6x-\dfrac{\pi}4}\right)\le1$ $\Rightarrow -\dfrac1{\sqrt2}+2020\le\dfrac1{\sqrt2}\sin\left({6x-\dfrac{\pi}4}\right)+2020\le\dfrac1{\sqrt2}+2020$ Vậy $\min y=-\dfrac1{\sqrt2}+2020$ dấu “=” xảy ra $\Leftrightarrow\sin\left({6x-\dfrac{\pi}4}\right)=-1$ $\Leftrightarrow 6x-\dfrac{\pi}4=-\dfrac{\pi}2+k2\pi$ $\Leftrightarrow x=-\dfrac{\pi}{24}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$ $\max y=\dfrac1{\sqrt2}+2020$ dấu “=” xảy ra $\Leftrightarrow\sin\left({6x-\dfrac{\pi}4}\right)=1$ $\Leftrightarrow 6x-\dfrac{\pi}4=\dfrac{\pi}2+k2\pi$ $\Leftrightarrow x=\dfrac{\pi}{8}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$. Trả lời
Đáp án:
$\min y=-\dfrac1{\sqrt2}+2020$ dấu “=” xảy ra $\Leftrightarrow x=-\dfrac{\pi}{24}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$
$\max y=\dfrac1{\sqrt2}+2020$ dấu “=” xảy ra $\Leftrightarrow x=\dfrac{\pi}{8}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$.
Giải thích các bước giải:
$y=\sin 6x-\cos 6x+2020$
$=\dfrac1{\sqrt2}\sin\left({6x-\dfrac{\pi}4}\right)+2020$
Do $-1\le\sin x\le1$ $\forall x$
$\Rightarrow-1\le\sin\left({6x-\dfrac{\pi}4}\right)\le1$
$\Rightarrow -\dfrac1{\sqrt2}+2020\le\dfrac1{\sqrt2}\sin\left({6x-\dfrac{\pi}4}\right)+2020\le\dfrac1{\sqrt2}+2020$
Vậy $\min y=-\dfrac1{\sqrt2}+2020$ dấu “=” xảy ra $\Leftrightarrow\sin\left({6x-\dfrac{\pi}4}\right)=-1$
$\Leftrightarrow 6x-\dfrac{\pi}4=-\dfrac{\pi}2+k2\pi$
$\Leftrightarrow x=-\dfrac{\pi}{24}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$
$\max y=\dfrac1{\sqrt2}+2020$ dấu “=” xảy ra $\Leftrightarrow\sin\left({6x-\dfrac{\pi}4}\right)=1$
$\Leftrightarrow 6x-\dfrac{\pi}4=\dfrac{\pi}2+k2\pi$
$\Leftrightarrow x=\dfrac{\pi}{8}+k\dfrac{\pi}{3}$ $(k\in\mathbb Z)$.