Question

Tìm GTLN, GTNN:

a, \(y=4\sin^2x-4\sin x+3\).

b, \(y=\cos^2x+2\sin x+2\).

c, \(y=\sin^4x-2\cos^2x+1\).

Answer 1

a.

Tìm min:

$y=(4\sin ^2x-4\sin x+1)+2=(2\sin x-1)^2+2$
Vì $(2\sin x-1)^2\geq 0$ với mọi $x$ nên $y=(2\sin x-1)^2+2\geq 0+2=2$

Vậy $y_{\min}=2$

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Mặt khác: 

$y=4\sin x(\sin x+1)-8(\sin x+1)+11$

$=(\sin x+1)(4\sin x-8)+11$

$=4(\sin x+1)(\sin x-2)+11$

Vì $\sin x\in [-1;1]\Rightarrow \sin x+1\geq 0; \sin x-2<0$

$\Rightarrow 4(\sin x+1)(\sin x-2)\leq 0$

$\Rightarrow y=4(\sin x+1)(\sin x-2)+11\leq 11$

Vậy $y_{\max}=11$

 

Answer 2

b.

$y=\cos ^2x+2\sin x+2=1-\sin ^2x+2\sin x+2$

$=3-\sin ^2x+2\sin x$
$=4-(\sin ^2x-2\sin x+1)=4-(\sin x-1)^2\leq 4-0=4$

Vậy $y_{\max}=4$.

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Mặt khác:

$y=3-\sin ^2x+2\sin x = (1-\sin ^2x)+(2+2\sin x)$

$=(1-\sin x)(1+\sin x)+2(1+\sin x)=(1+\sin x)(1-\sin x+2)$

$=(1+\sin x)(3-\sin x)$

Vì $\sin x\in [-1;1]$ nên $1+\sin x\geq 0; 3-\sin x>0$

$\Rightarrow y=(1+\sin x)(3-\sin x)\geq 0$

Vậy $y_{\min}=0$

Answer 3

c.

$y=\sin ^4x-2\cos ^2x+1=\sin ^4x-2(1-\sin ^2x)+1$

$=\sin ^4x+2\sin ^2x-1$

$=(\sin ^4x-1)+(2\sin ^2x-2)+2$

$=(\sin ^2x-1)(\sin ^2x+1)+2(\sin ^2x-1)+2$

$=(\sin ^2x-1)(\sin ^2x+3)+2$

Vì $\sin x\in [-1;1]$ nên $\sin ^2x\leq 1$

$\Rightarrow (\sin ^2x-1)(\sin ^2x+3)\leq 0$

$\Rightarrow y=(\sin ^2x-1)(\sin ^2x+3)+2\leq 2$

Vậy $y_{\max}=2$

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$y=\sin ^4x+2\sin ^2x-1=\sin ^2x(\sin ^2x+2)-1$

Vì $\sin ^2x\geq 0$ nên $\sin ^2x(\sin ^2x+2)\geq 0$

$\Rightarrow y=\sin ^2x(\sin ^2x+2)-1\geq 0-1=-1$
Vậy $y_{\min}=-1$