Question

Tìm min, max

a) \(y=\sqrt{7-3cos^2x}\)

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

c) \(y=2sin^2x+\sqrt{3}sin2x\)

Answer 1

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