Given $\left\{ \begin{array} { l } x + y \geq 2 \\ y \geq 0 \\ x + 2 y - 3 \leq 0 \end{array} \right.$, find the maximum value of $z = y - 2 x$ as $\_\_\_\_$
If $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { l } 2 x + y - 2 \leqslant 0 , \\ x - y - 1 \geqslant 0 , \\ y + 1 \geqslant 0 , \end{array} \right.$ then the maximum value of $z = x + 7 y$ is $\_\_\_\_$
If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x + y \geqslant 0 , \\ 2 x - y \geqslant 0 , \\ x \leqslant 1 , \end{array} \right.$ then the maximum value of $z = 3 x + 2 y$ is $\_\_\_\_$ .
Which of the following inequalities always holds? ( ) A. $a ^ { 2 } + b ^ { 2 } \leq 2 a b$ B. $a ^ { 2 } + b ^ { 2 } \geq - 2 a b$ C. $a + b \geq - 2 \sqrt { | a b | }$ D. $a + b \leq 2 \sqrt { | a b | }$
1. If $M = \{ x \mid \sqrt { x } < 4 \}$ and $N = \{ x \mid 3 x \geqslant 1 \}$, then $M \cap N =$ A. $\{ x \mid 0 \leqslant x < 2 \}$ B. $\left\{ x \left\lvert \, \frac { 1 } { 3 } \leqslant x < 2 \right. \right\}$ C. $\{ x \mid 3 \leqslant x < 16 \}$ D. $\left\{ x \left\lvert \, \frac { 1 } { 3 } \leqslant x < 16 \right. \right\}$
If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x + y \geqslant 2 , \\ x + 2 y \leqslant 4 , \end{array} \right.$ then the maximum value of $z = 2 x - y$ is A. $- 2$ B. 4 C. 8 D. 12
Let $x , y$ satisfy the constraints $\left\{ \begin{array}{l} -2x + 3y \leqslant 3 \\ 3x - 2y \leqslant 3 \\ x + y = 1 \end{array} \right.$ . Let $z = 3x + 2y$ . The maximum value of $z$ is $\_\_\_\_$ .
[Elective 4-5: Inequalities] Given $f(x) = 2|x - a| - a , \ a > 0$ . (1) Solve the inequality $f(x) < x$ ; (2) If the area enclosed by $y = f(x)$ and the coordinate axes is 2 , find $a$ .
The solution set of the inequality $\frac{x-4}{x-1} < 2$ is ( ) A. $\{x \mid -2 \leq x \leq 1\}$ B. $\{x \mid x < -2\}$ C. $\{x \mid -2 \leq x < 1\}$ D. $\{x \mid x > 1\}$
104- Suppose the solution set of the inequality $\dfrac{((m^2-1)x^2 - 4mx + 4)(x - 3\sqrt{x} + 2)}{3x - 3} > 0$, for $x > \dfrac{3}{2}$, is $[2, 4]$. What is the value of $m$? (1) $-2$ (2) zero (3) $1$ (4) $2$
14. Function $f$ is strictly decreasing and its range is the set of all negative values. If $f(m^2 - m - 5) < f(-3 + 2m - m^2)$, how many correct integer values does $m$ have? \[
\text{(1) } 1 \qquad \text{(2) } 2 \qquad \text{(3) } 3 \qquad \text{(4) zero}
\]
In a triangle with angles $P$, $Q$, $R$, let $\alpha$, $\beta$, $\gamma$ be the angles $\angle QCR = 2P$, $\angle QIR = Q + R$, $\angle QOR = P + Q/2 + R/2$ respectively. Show that $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} > \frac{1}{45}$.
The set of all real numbers $x$ satisfying the inequality $x ^ { 3 } ( x + 1 ) ( x - 2 ) \geq 0$ is (A) the interval $[ 2 , \infty )$ (B) the interval $[ 0 , \infty )$ (C) the interval $[ - 1 , \infty )$ (D) none of the above
The set of values of $m$ for which $mx^2 - 6mx + 5m + 1 > 0$ for all real $x$ is (A) $m < \frac{1}{4}$ (B) $m \geq 0$ (C) $0 \leq m \leq \frac{1}{4}$ (D) $0 \leq m < \frac{1}{4}$
The set of all real numbers $x$ such that $x ^ { 3 } ( x + 1 ) ( x - 2 ) \geq 0$ is: (a) the interval $2 \leq x < \infty$ (b) the interval $0 \leq x < \infty$ (c) the interval $- 1 \leq x < \infty$ (d) none of the above.
The set of all real numbers $x$ such that $x ^ { 3 } ( x + 1 ) ( x - 2 ) \geq 0$ is: (a) the interval $2 \leq x < \infty$ (b) the interval $0 \leq x < \infty$ (c) the interval $- 1 \leq x < \infty$ (d) none of the above.
The set of values of $m$ for which $m x ^ { 2 } - 6 m x + 5 m + 1 > 0$ for all real $x$ is (a) $m < \frac { 1 } { 4 }$ (b) $m \geq 0$ (c) $0 \leq m \leq \frac { 1 } { 4 }$ (d) $0 \leq m < \frac { 1 } { 4 }$.
The set of all real numbers $x$ satisfying the inequality $x ^ { 3 } ( x + 1 ) ( x - 2 ) \geq 0$ is (A) the interval $[ 2 , \infty )$ (B) the interval $[ 0 , \infty )$ (C) the interval $[ - 1 , \infty )$ (D) none of the above