Đáp án:
12) Không có đáp án
13) $A.\,\dfrac{33}{8};\, -2$
Giải thích các bước giải:
12) $y = f(x)= \sqrt{x + 3} - 2\sqrt{5 - 2x}$
$TXD: D = \left[-3;\dfrac{5}{2}\right]$
$y' = \dfrac{1}{2\sqrt{x +3}} + \dfrac{2}{\sqrt{5 - 2x}} > 0,\,\forall x \in D$
$\Rightarrow$ Hàm số đồng biến trên $\left[-3;\dfrac{5}{2}\right]$
$\Rightarrow \begin{cases}M = \mathop{\max}\limits_{x \in \left[-3;\tfrac{5}{2}\right]}y = f\left(\dfrac 52\right) = \dfrac{\sqrt{22}}{2}\\m = \mathop{\min}\limits_{x \in \left[-3;\tfrac{5}{2}\right]}y = f\left(-3\right) = -2\sqrt{11}\end{cases}$
$\Rightarrow M + m =\dfrac{\sqrt{22} - 4\sqrt{11}}{2}$
13) $y = 2\cos^2x - 3\sin x + 1$
$\to y = 2(1-\sin^2x) - 3\sin x + 1$
$\to y = -2\sin^2x - 3\sin x + 3$
$\to y = -2\left(\sin x +\dfrac 34\right)^2 +\dfrac{33}{8}$
Ta có:
$-1 \leq \sin x \leq 1$
$\to -\dfrac14 \leq \sin x +\dfrac34 \leq \dfrac74$
$\to 0 \leq \left(\sin x +\dfrac 34\right)^2 \leq \dfrac{49}{16}$
$\to -\dfrac{49}{8} \leq -2\left(\sin x +\dfrac 34\right)^2 \leq 0$
$\to - 2 \leq -2\left(\sin x +\dfrac 34\right)^2 +\dfrac{33}{8} \leq \dfrac{33}{8}$
Vậy $\min y = -2;\, \max y = \dfrac{33}{8}$
