Question

tìm gtnn gtln ạ

Answer 1

`y=sin 2x`   `, AA x in [0;\pi/2]`

`+,D=RR`

  `=>` H/s xác định trên `K=[0;\pi/2]`

`+,` Do `y=sin 2x` đồng biến trên `[0;\pi/2]`

    `=>sin 0 <= sin 2x <= sin \pi/2`

 `<=>0 <= sin 2x <= 1`

`<=>0 <= y <= 1`

Vậy `Max _y=1<=>sin 2x=1<=>x=\pi/4 in K`

       `Mi n _y=0<=>sin 2x=0<=>x=0 in K`

Answer 2

\(0< =x< =\dfrac{\Pi}{2}\)

nên \(0< =2x< =\Pi\)

\(\Leftrightarrow sin2x\in\left[0;1\right]\)

\(y_{Min}=0\) khi \(2x\in\left\{0;\Pi\right\}\)

hay \(x\in\left\{0;\dfrac{\Pi}{2}\right\}\)

\(y_{max}=1\) khi \(2x=\dfrac{\Pi}{2}\)

hay \(x=\dfrac{\Pi}{4}\)