Đặt \(sinx=t\in\left[-1;1\right]\)

\(\Rightarrow y=f\left(t\right)=4t^2-3t-1\)

Xét hàm \(f\left(t\right)\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=\dfrac{3}{8}\in\left[-1;1\right]\)

\(f\left(-1\right)=6\) ; \(f\left(\dfrac{3}{8}\right)=-\dfrac{25}{16}\) ; \(f\left(1\right)=0\)

\(\Rightarrow y_{min}=-\dfrac{25}{16}\) khi \(sinx=\dfrac{3}{8}\)

\(y_{max}=6\) khi \(sinx=-1\)

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

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

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

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

Sửa: \(y=3\sin x+4\cos x+2\)

Áp dụng BĐT Bunhiacopski được:

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

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

Chọn A

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

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

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

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

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

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

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

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