Đáp án:
$12.C\quad T= \left[\dfrac{1}{4};1\right]$
$13. B \quad y \leq 1, \, \forall x \in \Bbb R$
Giải thích các bước giải:
Câu 12:
$y = sin^6x + cos^6x$
$= (sin^2x + cosx^2)(sin^4x - sinx^2xcos^2x + cos^4x)$
$= (sin^2x + cos^2x)^2 - 3sin^2xcos^2x$
$= 1 - \dfrac{3}{4}(2sinxcosx)^2$
$= 1 - \dfrac{3}{4}sin^22x$
$= 1 - \dfrac{3}{8}(1 - cos4x)$
$= \dfrac{5 + 3cos4x}{8}$
Ta có:
$-1\leq cos4x \leq 1$
$\Leftrightarrow -3 \leq 3cos4x \leq 3$
$\Leftrightarrow 2 \leq 5 + 3cos4x\leq 8$
$\Leftrightarrow \dfrac{1}{4} \leq \dfrac{5 + 3cos4x}{8} \leq 1$
Hay $\dfrac{1}{4} \leq y \leq 1$
Vậy $T = \left[\dfrac{1}{4};1\right]$
Câu 13:
$y = sin^4x + cos^4x$
$= (sin^2x + cos^2x)^2 - 2sin^2xcos^2x$
$= 1 - \dfrac{1}{2}(2sinxcosx)^2$
$= 1 - \dfrac{1}{2}sin^22x$
$= 1 - \dfrac{1}{4}(1 - cos4x)$
$= \dfrac{3 + cos4x}{4}$
Ta có: $-1 \leq cos4x \leq 1$
$\Leftrightarrow 2 \leq 3 + cos4x \leq 4$
$\Leftrightarrow \dfrac{1}{2} \leq \dfrac{3 + cos4x}{4} \leq 1$
Hay $\dfrac{1}{2} \leq y \leq 1$
Vậy $y \leq 1, \, \forall x \in \Bbb R$
