B

9000029301

Level: 
B
Find the solution set of the following inequality. \[ \left (x - 1\right )\left (x - 2\right )\left (x - 3\right )\geq 0 \]
\(\left [ 1;2\right ] \cup \left [ 3;\infty \right )\)
\(\left (-\infty ;\infty \right )\)
\(\left (-\infty ;1\right )\cup \left (2;3\right )\)
\(\emptyset \)
\(\{0\}\)

9000026409

Level: 
B
Consider the following equation. \[ |2x - 4| = 5x - 7 \] Solving the equation on the intervals where it is possible to evaluate the absolute value we get equations on partial subintervals as follows. \[\begin{aligned} \text{for }x &\in (-\infty ;2)\colon &\text{for }x &\in [ 2;\infty )\colon & & & & \\ - 2x + 4 & = 5x - 7 &2x - 4 & = 5x - 7 & & & & \\ - 7x & = -11 & - 3x & = -3 & & & & \\x & = \frac{11} {7} &x & = 1 & & & & \end{aligned}\] Find the solution set of the original equation.
\(\left \{\frac{11} {7} \right \}\)
\(\left \{\frac{11} {7} ;1\right \}\)
\(\left \{1\right \}\)
\(\emptyset \)