Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(5-2\cdot cos^2x\cdot\sin^2x\)
\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)
\(=5-2\cdot\left(\frac12\cdot\sin2x\right)^2=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)
Ta có: \(0\le\sin^22x\le1\)
=>\(-\frac12\le-\frac12\cdot\sin^22x\le0\)
=>\(-\frac12+5\le-\frac12\cdot\sin^22x+5\le0+5\)
=>\(\frac92\le-\frac12\cdot\sin^22x+5\le5\)
=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)
=>\(4:\frac{3\sqrt2}{2}\ge\frac{4}{\sqrt{-\frac12\cdot sin^22x+5}}\ge\frac{4}{\sqrt5}\)
=>\(\frac{2\sqrt2}{3}\ge y\ge\frac{4\sqrt5}{5}\)
Do đó: \(y_{\max}=\frac{2\sqrt2}{3}\) khi \(\sin^22x=1\)
=>\(cos^22x=0\)
=>cos2x=0
=>\(2x=\frac{\pi}{2}+k\pi\)
=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)
\(y_{\min}=\frac{4\sqrt5}{5}\) khi \(\sin^22x=0\)
=>sin 2x=0
=>\(2x=k\pi\)
=>\(x=\frac{k\pi}{2}\)
b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)
\(=3\cdot\sin^2x+5\cdot cos^2x-4\left(cos^2x-\sin^2x\right)-2\)
\(=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos^2x+4\cdot\sin^2x-2\)
\(=7\cdot\sin^2x+cos^2x-2=7\cdot\sin^2x+1-\sin^2x-2=6\cdot\sin^2x-1\)
Ta có: \(0\le\sin^2x\le1\)
=>\(0\le6\sin^2x\le6\)
=>\(0-1\le6\sin^2x-1\le6-1\)
=>-1<=f(x)<=5
f(x) min=-1 khi \(\sin^2x=0\)
=>sin x=0
=>\(x=k\pi\)
f(x) max=5 khi \(\sin^2x=1\)
=>\(cos^2x=0\)
=>cosx=0
=>\(x=\frac{\pi}{2}+k\pi\)
Đề là:
\(y=\sqrt{4-3cos^23x}+1\) đúng không nhỉ?
Ta có:
\(0\le cos^23x\le1\Rightarrow1\le\sqrt{4-3cos^23x}\le2\)
\(\Rightarrow2\le y\le3\)
\(y_{min}=2\) khi \(cos^23x=1\)
\(y_{max}=3\) khi \(cos3x=0\)
\(-1\le cos\left(\sqrt{x}+\dfrac{\pi}{4}\right)\le1\Rightarrow-5\le y\le5\)
\(y_{max}=5\) khi \(cos\left(\sqrt{x}+\dfrac{\pi}{4}\right)=1\)
\(y_{min}=-5\) khi \(cos\left(\sqrt{x}+\dfrac{\pi}{4}\right)=-1\)
a: \(5-2\cdot cos^2x\cdot\sin^2x\)
\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)
\(=5-2\cdot\left\lbrack\frac12\cdot2\cdot\sin x\cdot cosx\right\rbrack^2=5-2\cdot\left\lbrack\frac12\cdot\sin2x\right\rbrack^2\)
\(=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)
\(0\le\sin^22x\le1\)
=>\(0\ge-\frac12\sin^22x\ge-\frac12\)
=>\(0+5\ge-\frac12\sin^22x+5\ge-\frac12+5\)
=>\(5\ge-\frac12\sin^22x+5\ge\frac92\)
=>\(\frac92\le-\frac12\sin^22x+5\le5\)
=>\(\sqrt{\frac92}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)
=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)
=>\(\frac{2}{3\sqrt2}\ge\frac{1}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1}{\sqrt5}\)
=>\(\frac{2\cdot4}{3\sqrt2}\ge\frac{1\cdot4}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1\cdot4}{\sqrt5}\)
=>\(\frac{4\sqrt2}{3}\ge y\ge\frac{4}{\sqrt5}\)
=>\(y_{\max}=\frac{4\sqrt2}{3}\) khi \(-\frac12\cdot\sin^22x+5=\frac92\)
=>\(-\frac12\cdot\sin^22x=-\frac12\)
=>\(\sin^22x=1\)
=>\(cos^22x=0\)
=>cos2x=0
=>\(2x=\frac{\pi}{2}+k\pi\)
=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)
\(y_{\min}=\frac{4}{\sqrt5}\) khi \(-\frac12\cdot\sin^22x+5=5\)
=>\(\sin^22x=0\)
=>sin 2x=0
=>\(2x=k\pi\)
=>\(x=\frac{k\pi}{2}\)
b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)
\(=3\left(1-cos^2x\right)+5\cdot cos^2x-4\left(2\cdot cos^2x-1\right)-2\)
\(=3-3\cdot cos^2x+5\cdot cos^2x-8\cdot cos^2x+4-2=-6\cdot cos^2x+5\)
Ta có: \(0<=cos^2x\le1\)
=>\(0\ge-6\cdot cos^2x\ge-6\)
=>\(0+5\ge-6\cdot cos^2x+5\ge-6+5\)
=>5>=y>=-1
Do đó: \(y_{\min}=-1\) khi \(-6\cdot cos^2x+5=-1\)
=>\(-6\cdot cos^2x=-6\)
=>\(cos^2x=1\)
=>\(\sin^2x=0\)
=>sin x=0
=>\(x=k\pi\)
y max=5 khi \(-6\cdot cos^2x+5=5\)
=>\(-6\cdot cos^2x=0\)
=>cosx=0
=>\(x=\frac{\pi}{2}+k\pi\)
\(x\in\left[\dfrac{1}{4};\dfrac{3}{2}\right]\Rightarrow\pi x\in\left[\dfrac{\pi}{4};\dfrac{3\pi}{2}\right]\)
\(\Rightarrow cos\left(\pi x\right)\in\left[-1;\dfrac{\sqrt{2}}{2}\right]\)
\(y_{max}=\dfrac{\sqrt{2}}{2}\) khi \(x=\dfrac{1}{4}\)
\(y_{min}=-1\) khi \(x=1\)
Đặt \(sin^24x=t\left(t\in\left[0;1\right]\right)\)
\(y=1-8sin^22x.cos^22x+2sin^42x\)
\(=1-2sin^24x+2sin^42x\)
\(\Rightarrow y=f\left(t\right)=1-2t+2t^2\)
\(y_{min}=min\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=\dfrac{1}{2}\)
\(y_{max}=max\left\{f\left(0\right);f\left(1\right);f\left(\dfrac{1}{2}\right)\right\}=1\)
\(y=-1-cos^2\left(2x+\dfrac{\pi}{3}\right)\)
\(=-\dfrac{3}{2}+\dfrac{1}{2}-cos^2\left(2x+\dfrac{\pi}{3}\right)\)
\(=-\dfrac{3}{2}-\dfrac{1}{2}\left[2cos^2\left(2x+\dfrac{\pi}{3}\right)-1\right]\)
\(=-\dfrac{3}{2}-\dfrac{1}{2}cos\left(4x+\dfrac{2\pi}{3}\right)\)
Vì \(cos\left(4x+\dfrac{2\pi}{3}\right)\in\left[-1;1\right]\)
\(\Rightarrow min=-\dfrac{3}{2}-\dfrac{1}{2}=-2\Leftrightarrow cos\left(4x+\dfrac{2\pi}{3}\right)=1\)
\(\Rightarrow max=-\dfrac{3}{2}+\dfrac{1}{2}=-1\Leftrightarrow cos\left(4x+\dfrac{2\pi}{3}\right)=-1\)