Question

Đã trả lời bởi chuyên gia

Answer 1

Đáp án B

Hàm số $y = \cos^{4} x + \sin^{4} x$ có nghĩa $\forall x \in ℝ \Leftrightarrow D = ℝ$ .

Ta có $y = \cos^{4} x + \sin^{4} x = {(1 - \sin^{2} x)}^{2} + \sin^{4} x = 1 - 2 \sin^{2} x + \sin^{4} x + \sin^{4} x = 2 \sin^{4} x - 2 \sin^{2} x + 1$ .

$y = 2 {(\sin^{2} x)}^{2} - 2 \sin^{2} x + 1 = 2 ({(\sin^{2} x)}^{2} - \sin^{2} x + \frac{1}{2}) = 2 {(\sin^{2} x - \frac{1}{2})}^{2} + \frac{1}{2}$ .

$- 1 \leq \sin x \leq 1 \Leftrightarrow 0 \leq \sin^{2} x \leq 1 \Leftrightarrow \frac{- 1}{2} \leq \sin^{2} x - \frac{1}{2} \leq \frac{1}{2} \Leftrightarrow 0 \leq {(\sin^{2} x - \frac{1}{2})}^{2} \leq \frac{1}{4} \Leftrightarrow 0 \leq 2 {(\sin^{2} x - \frac{1}{2})}^{2} \leq \frac{1}{2}$ .

$\Leftrightarrow \frac{1}{2} \leq 2 {(\sin^{2} x - \frac{1}{2})}^{2} + \frac{1}{2} \leq 1 \Leftrightarrow \frac{1}{2} \leq y \leq 1$

Vậy $\min y = \frac{1}{2} \Leftrightarrow \sin^{2} x = \frac{1}{2} \Leftrightarrow [\begin{array}{l} \sin x = \frac{\sqrt{2}}{2} \Leftrightarrow x = \frac{π}{4} + k 2 π \\ \sin x = \frac{- \sqrt{2}}{2} \Leftrightarrow x = \frac{- π}{4} + k 2 π \end{array} (k \in ℤ)$ ;

$\max y = 1 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k π (k \in ℤ)$

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Answer 2