\(y=2sin^2x+3sinx.cosx+cos^2x\)

\(=-\left(1-2sin^2x\right)+\dfrac{3}{2}sin2x+\dfrac{1}{2}\left(2cos^2x-1\right)+\dfrac{1}{2}\)

\(=-cos2x+\dfrac{3}{2}sin2x+\dfrac{1}{2}cos2x+\dfrac{1}{2}\)

\(=\dfrac{3}{2}sin2x-\dfrac{1}{2}cos2x+\dfrac{1}{2}\)

\(=\dfrac{\sqrt{10}}{2}\left(\dfrac{3}{\sqrt{10}}sin2x-\dfrac{1}{\sqrt{10}}cos2x\right)+\dfrac{1}{2}\)

\(=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\)

Vì \(sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)\in\left[-1;1\right]\)

\(\Rightarrow y=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\in\left[-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2};\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\right]\)

\(\Rightarrow y_{min}=-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=-1\Leftrightarrow...\)

\(y_{max}=\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=1\Leftrightarrow...\)

Lời giải:

Ta có:

\(B=\cos 2x+\sqrt{1+2\sin ^2x}=\cos ^2x-\sin ^2x+\sqrt{1+2\sin ^2x}\)

\(=1-2\sin ^2x+\sqrt{1+2\sin ^2x}\)

Đặt \(\sin ^2x=t(t\in [0;1])\). Khi đó:
\(B=1-2t+\sqrt{1+2t}\)

\(B'=\frac{1}{\sqrt{1+2t}}-2=0\Leftrightarrow t=-\frac{3}{8}\) (loại)

Lập bảng biến thiên suy ra:

\(B_{\max}=B(0)=2\)

\(B_{\min}=B(1)=\sqrt{3}-1\)

\(y=\dfrac{3}{4}sinx-\dfrac{1}{4}sin3x+cos2x-1\)

Hàm \(\dfrac{3}{4}sinx\) có chu kì \(T_1=2\pi\)

Hàm \(\dfrac{1}{4}sin3x\) có chu kì \(T_2=\dfrac{2\pi}{3}\)

Hàm \(cos2x\) có chu kì \(T_3=\dfrac{2\pi}{2}=\pi\)

\(\Rightarrow y\) có chu kì \(T=BCNN\left(2\pi;\dfrac{2\pi}{3};\pi\right)=2\pi\)

\(y=1-cos2x+2sin2x+6=2sin2x-cos2x+7\)

\(y=\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sin2x-\dfrac{1}{\sqrt{5}}cos2x\right)+7\)

Đặt \(\dfrac{2}{\sqrt{5}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)

\(y=\sqrt{5}sin\left(2x-a\right)+7\)

\(\Rightarrow-\sqrt{5}+7\le y\le\sqrt{5}+7\)

Đề thiếu. Bạn xem lại đề.

a/ \(0\le cos^2x\le1\Rightarrow2\le y\le\sqrt{7}\)

\(y_{min}=2\) khi \(cos^2x=1\)

\(y_{max}=\sqrt{7}\) khi \(cos^2x=0\)

b/ \(y=\frac{2}{1+tan^2x}=\frac{2}{\frac{1}{cos^2x}}=2cos^2x\le2\)

\(\Rightarrow y_{max}=2\) khi \(cos^2x=1\)

\(y_{min}\) ko tồn tại

c/ \(y=1-cos2x+\sqrt{3}sin2x=2\left(\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x\right)+1\)

\(y=2sin\left(2x-\frac{\pi}{6}\right)+1\)

Do \(-1\le sin\left(2x-\frac{\pi}{6}\right)\le1\Rightarrow-1\le y\le3\)

\(B=-2\left(x^2-\dfrac{3}{2}x\right)+5=-2\left(x^2-2x.\dfrac{3}{4}+\dfrac{9}{16}\right)+5+\dfrac{9}{16}=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{25}{16}\le\dfrac{25}{16}\)

dấu = xảy ra khi x=3/4

vậy Bmax=....

tik mik nha

B=-2x²-3x+5

 =-2(x²-\(\dfrac{3}{2}\)x)+5

 =-2(x²-2.\(\dfrac{3}{4}\)x+\(\dfrac{9}{16}\))+\(\dfrac{71}{16}\)

=-2(x-\(\dfrac{3}{4}\))²+\(\dfrac{71}{16}\)≤\(\dfrac{71}{16}\)

Dấu "=" xảy ra⇔x-\(\dfrac{3}{4}\)=0⇔x=\(\dfrac{3}{4}\)

MaxB=\(\dfrac{71}{16}\)⇔×=\(\dfrac{3}{4}\)

Ta có: \(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)