Question

Đã trả lời bởi chuyên gia

Answer 1

Câu 1:

`(\log_{2}^{2}x-\log_{2}x^3+2)\sqrt{32-2^x} \ge 0`

ĐK: `0 < x \le 5\ (1)`

Giải `\log_{2}^{2}x-\log_{2}x^3+2=0`

`⇔ \log_{2}^{2}x-3\log_{2}x+2=0`

`⇔ (\log_{2} x-1)(\log_{2}x-2)=0`

`⇔` $\left[ \begin{array}{l} \log_{2} x-1=0\\ \log_{2} x-2=0\end{array} \right.$

`⇔` $\left[ \begin{array}{l} \log_{2} x=1\\ \log_{2} x=2\end{array} \right.$

`⇔` $\left[ \begin{array}{l} x=2\\x=4\end{array} \right.$

Giải `\sqrt{32-2^x}=0`

`⇔ 32-2^x=0`

`⇔ 2^x=32`

`⇔ x=5`

Ta có BXD:

$\begin{array}{|c|cc|}\hline \text{$x$}&\text{$0$}&\text{}&\text{}2&\text{}&\text{}4&\text{}&\text{}5\\\hline \text{$(\log_{2}^{2}x-\log_{2}x^3+2)\sqrt{32-2^x}$}&\text{}&\text{}+&\text{0}&\text{}-&\text{}0&\text{}+&\text{}0\\\hline \end{array}$

Kết hợp `(1)` ta có:

`⇒ x \in (0;2] ∪ [4;5]`

`⇒ x \in {1;2;4;5}`

Vậy có 4 số nguyên `x` thỏa mãn đề bài

Câu 2:

`f(x)=\int f'(x)`

`f(x)=\int\ (3-5cos\ x) dx`

`f(x)=3x-5sin\ x+C`

Mà `f(0)=5`

`⇔ 3.0-5sin\ 0+C=5`

`⇒ C=5`

`⇒ f(x)=3x-5sin\ x+5`

Ta có:

`F(x)=\int f(x)`

$\int\limits_{0}^{\dfrac{\pi}{2}} f(x) dx=F\Bigg(\dfrac{\pi}{2}\Bigg)-F(0)$

$⇔ \int\limits_{0}^{\dfrac{\pi}{2}} (3x-5\sin\ x+5) dx=F\Bigg(\dfrac{\pi}{2}\Bigg)-5$

$⇔ F\Bigg(\dfrac{\pi}{2}\Bigg)=\int\limits_{0}^{\dfrac{\pi}{2}} (3x-5\sin\ x+5) dx+5$

$⇔ F\Bigg(\dfrac{\pi}{2}\Bigg) \approx 11,55$

`=>` Đáp án:

Câu 1: `\bb C`

Câu 2: `\bb B`

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Answer 2