Đề thi cuối kì 2 Toán 10 Kết nối tri thức – Đề 08 (có đáp án)

Hàm số nghịch biến trên khoảng $$\left(0;3\right)$$;

Hàm số đồng biến trên khoảng $$\left(-\infty ;1\right)$$;

Hàm số nghịch biến trên khoảng $$\left(0;2\right)$$;

Hàm số đồng biến trên khoảng $$\left(-\infty ;3\right)$$.

$$x=-3$$;

$$x=\frac{3}{2}$$;

$$x=-\frac{3}{2}$$;

$$x=3$$.

$S=\left(-\infty ;2\right)\cup \left(2;+\infty \right)$;

$S=\mathbb{R}$;

$$S=\left(2;+\infty \right)$$;

$S=\left(-\infty ;-2\right)\cup \left(-2;+\infty \right)$.

$0$;

$1$;

$2$;

$4$.

$$\overrightarrow{n}\left(-2;-1\right)$$;

$$\overrightarrow{n}\left(2;-1\right)$$;

$$\overrightarrow{n}\left(-1;2\right)$$;

$$\overrightarrow{n}\left(1;2\right)$$.

$\{\begin{array}{l}x=1+3t\\ y=-4+2t\end{array}$;

$\{\begin{array}{l}x=3+2t\\ y=1-4t\end{array}$;

$\{\begin{array}{l}x=3+t\\ y=2-4t\end{array}$;

$\{\begin{array}{l}x=1-4t\\ y=3+2t\end{array}$.

$4x-5y-7=0$;

$4x+5y-17=0$;

$4x-5y-17=0$;

$4x+5y+17=0$.

$\{\begin{array}{l}x=-1+3t\\ y=2t\end{array}$;

$\{\begin{array}{l}x=3+3t\\ y=-1-t\end{array}$;

$\{\begin{array}{l}x=3+3t\\ y=-6-t\end{array}$;

$\{\begin{array}{l}x=3+3t\\ y=-1+t\end{array}$.

$$x+2y+1=0$$;

$$2x-y=0$$;

$$-x+2y+1=0$$;

$$-2x+4y-1=0$$;

$$d\left(O,\Delta \right)=\frac{\left|0.x+0.y+c\right|}{\sqrt{0^2+0^2}}$$;

$$d\left(O,\Delta \right)=\frac{a.0+b.0+c}{\sqrt{a^2+b^2}}$$;

$$d\left(O,\Delta \right)=\frac{\left|c\right|}{\sqrt{a^2+b^2}}$$;

$$d\left(M,\Delta \right)=\left|a.0+b.0+c\right|$$.

$$\left(-\frac{7}{2};-\frac{25}{2}\right)$$;

$$\left(\frac{11}{4};-\frac{25}{4}\right)$$;

$$\left(\frac{7}{2};\frac{25}{2}\right)$$;

$$\left(\frac{11}{4};\frac{25}{4}\right)$$.

nhọn;

vuông;

A và B đúng;

A và C sai.

$$I\left(-2;-3\right)$$;

$$I\left(2;3\right)$$;

$$I\left(4;6\right)$$;

$$I\left(-4;-6\right)$$.

${\left(x-1\right)}^2+{\left(y+2\right)}^2=9$;

${\left(x+1\right)}^2+{\left(y+2\right)}^2=9$;

${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$;

${\left(x+1\right)}^2+{\left(y-2\right)}^2=9$.

${\left(x-1\right)}^2+{\left(y-1\right)}^2=5$;

${\left(x-1\right)}^2+{\left(y-1\right)}^2=25$;

${\left(x-1\right)}^2+{\left(y-1\right)}^2=1$;

${\left(x-1\right)}^2+{\left(y-1\right)}^2=\frac{1}{5}$.

${\left(x+1\right)}^2+{\left(y+4\right)}^2=5$;

${\left(x-1\right)}^2+{\left(y-4\right)}^2=\sqrt{5}$;

${\left(x+1\right)}^2+{\left(y+4\right)}^2=\sqrt{5}$;

${\left(x-1\right)}^2+{\left(y-4\right)}^2=5$.

$$\frac{9}{5}$$;

$$\frac{12}{5}$$;

$$\frac{3}{5}$$;

$$\frac{6}{5}$$.

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\left(a>b>0\right)$$;

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b>0)$$;

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1\left(a>b>0\right)$$;

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=-1\left(a>b>0\right)$$.

$\frac{x^2}{9}-\frac{y^2}{16}=1$;

$\frac{x^2}{-9}-\frac{y^2}{16}=1$;

$\frac{x^2}{25}+\frac{{\left(-y\right)}^2}{9}=1$;

$\frac{x^2}{16}+\frac{y^2}{4}=1$.

$x=9$;

$x+\frac{9}{2}=0$;

$y+9=0$;

$x=-\frac{9}{4}$.

$$24$$;

$$48$$;

$$480$$;

$$60$$.

$$99$$;

$$50$$;

$$20$$;

$$10$$.

$5^5$;

$5!$;

$4!$;

$5$.

$C_6^2+C_9^4$;

$C_6^2{C}_{13}^4$;

$A_6^2{A}_9^4$;

$C_6^2{C}_9^4$.

$A_{11}^5$;

$C_{11}^5$;

$A_{11}^2.5!$;

$C_{10}^5$.

$$x^5-5x^{4}y+10x^3{y}^2-10x^2{y}^3+5x{y}^4-y^5$$;

$$x^5+5x^{4}y+10x^3{y}^2+10x^2{y}^3+5x{y}^4+y^5$$;

$$x^5-5x^{4}y-10x^3{y}^2-10x^2{y}^3-5x{y}^4+y^5$$;

$$x^5+5x^{4}y-10x^3{y}^2+10x^2{y}^3-5x{y}^4+y^5$$.

$$160$$;

$$80$$;

$$20$$;

$$40$$.

$$4$$;

$$8$$;

$$12$$;

$$16$$.

“ Tổng số chấm trong hai lần gieo là tám”.

“ Lần gieo đầu tiên được mặt hai chấm”.

“ Số chấm trong hai lần gieo là như nhau”.

“ Hai lần gieo đều xuất hiện mặt chấm chẵn”.

$A=\left\{\left(4,6\right),\left(6,4\right),\left(3,5\right),\left(5,3\right),\left(5,5\right)\right\}$;

$A=\left\{\left(4,6\right),\left(6,4\right),\left(5,5\right)\right\}$;

$A=\left\{\left(4,6\right),\left(5,5\right)\right\}$;

$A=\left\{\left(4,6\right),\left(5,3\right),\left(5,5\right)\right\}$.

$P\left(\overline{A}\right)=1-P\left(A\right)$;

$P\left(\overline{A}\right)>1$;

$0\le P\left(\overline{A}\right)\le 1$;

$P\left(A\right)+P\left(\overline{A}\right)=1$.

$\overline{A}$: “Trong $3$ quả bóng lấy ra có tối đa một quả bóng đỏ”;

$\overline{A}$: “Trong $3$ quả bóng lấy ra có nhiều nhất một quả bóng đỏ”;

$\overline{A}$: “Trong $3$ quả bóng lấy ra không có bóng đỏ”;

$\overline{A}$: “Trong $3$ quả bóng lấy ra có cả ba quả bóng đỏ”;

$\frac{2}{7}$;

$\frac{1}{21}$;

$\frac{37}{42}$;

$\frac{5}{42}$;

$$\frac{1}{2}$$;

$$\frac{7}{12}$$;

$$\frac{1}{6}$$;

$$\frac{1}{3}$$.

$3$;

$2$;

$5$;

$4$.