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\(y=\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)+\cos4x-1\)
\(\sin^6x+\cos^6x=\left(\sin^2x+\cos^2x\right)\left(\sin^4x-\sin^2x\cdot\cos^2x+\cos^4x\right)\\ =\left(\sin^2x+\cos^2x\right)^2-3\sin^2x\cdot\cos^2x=1-\dfrac{3}{4}\sin^22x\)
Do \(0\le\sin^22x\le1\Leftrightarrow\dfrac{3}{4}\cdot0\ge-\dfrac{3}{4}\sin^22x\ge-\dfrac{3}{4}\)
\(\Leftrightarrow1\ge1-\dfrac{3}{4}\sin^22x\ge1-\dfrac{3}{4}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{4}{3}\ge\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)\ge\dfrac{1}{4}\cdot\dfrac{4}{3}=\dfrac{1}{3}\)
Ta có \(-1\le\cos4x\le1\)
\(\Leftrightarrow\dfrac{1}{3}-1-1\le\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)+\cos4x-1\le\dfrac{4}{3}+1-1\\ \Leftrightarrow-\dfrac{5}{3}\le y\le\dfrac{4}{3}\)
Vậy \(y_{min}=-\dfrac{5}{3};y_{max}=\dfrac{4}{3}\)
\(y=\dfrac{4}{3}\left(sin^6x+cos^6x\right)+cos4x-1\)
\(y=\dfrac{4}{3}\left(\dfrac{5}{8}+\dfrac{3}{8}cos4x\right)+cos4x-1\)
\(y=\dfrac{3}{2}cos4x-\dfrac{1}{6}\)
\(-1\le cos4x\le1\Rightarrow-\dfrac{5}{3}\le y\le\dfrac{4}{3}\)
\(y_{min}=-\dfrac{5}{3}\) khi \(cos4x=-1\)
\(y_{max}=\dfrac{4}{3}\) khi \(cos4x=1\)
1.
\(-1\le sin7x\le1\Rightarrow-7\le y\le3\)
\(y_{min}=-7\) khi \(sin7x=-1\)
\(y_{max}=3\) khi \(sin7x=1\)
2.
Miền xác định của hàm \(D=R\backslash\left\{0\right\}\) là miền đối xứng
\(y\left(-x\right)=\frac{cos\left(-4x\right)}{2\left(-x\right)}=-\frac{cos4x}{2x}=-y\left(x\right)\)
Hàm lẻ
\(y=\sqrt{3}cos2x+2sinxcosx-2\)
\(=\sqrt{3}cos2x+sin2x-2\)
Ta có: \(\left|\sqrt{3}cos2x+sin2x\right|\le\sqrt{\left(\sqrt{3}\right)^2+1^2}=2\)
Do đó \(-2\le\sqrt{3}cos2x+sin2x\le2\)
\(\Leftrightarrow-4\le\sqrt{3}cos2x+sin2x-2\le2\).
Ta có: \(\left|\sqrt{3}cosx-sinx\right|\le\sqrt{\left(\sqrt{3}\right)^2+\left(-1\right)^2}=2\)
Do đó \(-2\le\sqrt{3}cosx-sinx\le2\)
\(y=4\left(1-sin^2x\right)+2sinx+2=-4sin^2x+2sinx+6\)
Đặt \(sinx=t\in\left[-1;1\right]\Rightarrow y=f\left(t\right)=-4t^2+2t+6\)
\(-\dfrac{b}{2a}=\dfrac{1}{4}\in\left[-1;1\right]\)
\(f\left(-1\right)=0\) ; \(f\left(\dfrac{1}{4}\right)=\dfrac{25}{4}\); \(f\left(1\right)=4\)
\(\Rightarrow y_{max}=\dfrac{25}{4}\) khi \(sinx=\dfrac{1}{4}\)
\(y_{min}=0\) khi \(sinx=-1\)
Ta có: \(y=4cos^2x+2sinx+2=4-4sin^2x+2sinx+2=-4sin^2x+2sinx+6=-\left(4sin^2x-2sinx+\dfrac{1}{16}-\dfrac{1}{16}-6\right)=-\left(2sin^2x-\dfrac{1}{4}\right)^2+\dfrac{97}{16}\)
Ta có: \(-\left(2sin^2x-\dfrac{1}{4}\right)^2\le0\Rightarrow y\le\dfrac{97}{16}\)
Vậy \(y_{max}=\dfrac{97}{16}\)
Bạn chú ý viết đề bài bằng công thức toán.
Phần a là \(\sqrt{\frac{\sin x+3}{2}}\) hay\(\sqrt{\sin x+\frac{3}{2}}\)?
\(y=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right).sinx.cosx\)
\(=\left(cos^2x-sin^2x\right).\dfrac{1}{2}\left(2sinx.cosx\right)=\dfrac{1}{2}cos2x.sin2x\)
\(=\dfrac{1}{4}sin4x\)
Do \(-1\le sin4x\le1\Rightarrow-\dfrac{1}{4}\le y\le\dfrac{1}{4}\)
\(y_{min}=-\dfrac{1}{4}\) khi \(sin4x=1\)
\(y_{max}=\dfrac{1}{4}\) khi \(sin4x=1\)
\(y=\sqrt{1+cos4x}-2\)
+) \(y=\sqrt{1+cos4x}-2\ge-2\)
\(\Rightarrow min=-2\Leftrightarrow cos4x=-1\Leftrightarrow4x=\pi+k2\pi\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
+) \(cos4x\in\left[-1;1\right]\Rightarrow1+cos4x\le2\Rightarrow y=\sqrt{1+cos4x}-2\le\sqrt{2}-2\)
\(\Rightarrow max=\sqrt{2}-2\Leftrightarrow cos4x=1\Leftrightarrow4x=k2\pi\Leftrightarrow x=\dfrac{k\pi}{2}\)