Đặt \(sinx=t\left(t\in\left[-1;1\right]\right)\).

\(\Rightarrow y=f\left(t\right)=-2t^2+3t-1\)

\(\Rightarrow y_{min}=min\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{3}{4}\right)\right\}=f\left(-1\right)=-6\)

\(y_{max}=max\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{3}{4}\right)\right\}=f\left(\dfrac{3}{4}\right)=\dfrac{1}{8}\)

Chọn A

1.

\(y=\frac{1}{2}sin2x-1\)

Do \(-1\le sin2x\le1\Rightarrow-\frac{3}{2}\le y\le-\frac{1}{2}\)

\(y_{min}=-\frac{3}{2}\) ; \(y_{max}=-\frac{1}{2}\)

2.

\(y=5+5\left(\frac{4}{5}cosx-\frac{3}{5}sinx\right)=5+5cos\left(x+a\right)\) với \(cosa=\frac{4}{5}\)

Do \(-1\le cos\left(x+a\right)\le1\Rightarrow0\le y\le10\)

\(y_{min}=0\) ; \(y_{max}=10\)

Sửa: \(y=3\sin x+4\cos x+2\)

Áp dụng BĐT Bunhiacopski được:

\(\left(3\sin x+4\cos x\right)^2\le\left(3^2+4^2\right)\left(\sin x^2+\cos x^2\right)=25\)

\(\Leftrightarrow-5\le3\sin x+4\cos x\le5\\ \Leftrightarrow-3\le3\sin x+4\cos x+2\le7\\ \Leftrightarrow y_{min}=-3\\ y_{max}=7\)

\(M^2=\left(3sinx+4cosx\right)^2\le\left(3^2+4^2\right)\left(sin^2x+cos^2x\right)=25\)

\(\Rightarrow-5\le M\le5\)

\(\Rightarrow M_{max}=5\) ; \(M_{min}=-5\)

\(t=\sin x;t\in\left[-1;1\right]\)

Xét hàm f(t) trên [-1;1]

\(f\left(-1\right)=2+3+1=6\)

\(f\left(1\right)=2-3+1=0\)

\(f\left(\frac{3}{4}\right)=2.\frac{9}{16}-3.\frac{3}{4}+1=-\frac{1}{8}\)

\(\Rightarrow\left\{{}\begin{matrix}y_{max}=6;"="\Leftrightarrow\sin x=-1\\y_{min}=-\frac{1}{8};"="\Leftrightarrow\sin x=\frac{3}{4}\end{matrix}\right.\)

a, \(y=3-4sin^2x.cos^2x=3-sin^22x\)

Đặt \(sin2x=t\left(t\in\left[-1;1\right]\right)\).

\(\Rightarrow y=f\left(t\right)=3-t^2\)

\(\Rightarrow y_{min}=minf\left(t\right)=2\)

\(y_{max}=maxf\left(t\right)=3\)

b, \(y=f\left(t\right)=\dfrac{-2}{3t-5}\left(t\in\left[0;1\right]\right)\)

\(\Rightarrow y_{min}=minf\left(t\right)=\dfrac{2}{5}\)

\(y_{max}=maxf\left(t\right)=1\)