Tìm GTLN-GTNN của hàm số: a) 4/(1 +2sin ²x) b) 3/(1+ √2+sin ²x )

Đáp án: a.$2\le \dfrac{4}{1+2\sin^2x}\le 4$ 

             b.$ \dfrac{3}{1+\sqrt{3}}\le \dfrac{3}{1+\sqrt{2+\sin^2x}}\le \dfrac{3}{1+\sqrt{2}}$

Giải thích các bước giải:

a.Ta có:

$-1\le \sin x\le 1$

$\to 0\le \sin^2x\le 1$

$\to 1\le 1+2\sin^2x\le 2$

$\to 2\le \dfrac{4}{1+2\sin^2x}\le 4$

$\to GTLN(\dfrac{4}{1+2\sin^2x})=4$ khi $\sin x=0$

$\to GTNN(\dfrac{4}{1+2\sin^2x})=4$ khi $\sin x=\pm1$

b.Ta có $-1\le \sin x\le 1$

$\to 0\le \sin^2x\le 1$

$\to 1+\sqrt{2}\le 1+\sqrt{2+\sin^2x}\le 1+\sqrt{3}$

$\to \dfrac{3}{1+\sqrt{3}}\le \dfrac{3}{1+\sqrt{2+\sin^2x}}\le \dfrac{3}{1+\sqrt{2}}$

$\to GTLN (\dfrac{3}{1+\sqrt{2+\sin^2x}})=\dfrac{3}{1+\sqrt{2}}$ khi $\sin x=0$

      $GTNN (\dfrac{3}{1+\sqrt{2+\sin^2x}})=\dfrac{3}{1+\sqrt{3}}$ khi $\sin x=\pm1$