Tìm GTLN-GTNN của hàm số: a) 4/(1 +2sin ²x) b) 3/(1+ √2+sin ²x ) 03/08/2021 Bởi Autumn 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 $\sin x=\pm1$ Bình luận
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 $\sin x=\pm1$