Equations and Inequalities with Parameters

9000104505

Level: 
A
Solve the following equation with unknown \(x\) and a real parameter \(a\in\mathbb{R}\setminus\{-3,3\}\). \[\frac{a-x} {a-3} - \frac{6a} {a^{2}-9} = \frac{x-3} {a+3} \]
\(\begin{array}{cc} \hline \text{Parameter} & \text{Solution set}\\ \hline a=0 & \emptyset \\ a\notin\{-3,0,3\} & \left\lbrace\frac{a^2-9}{2a}\right\rbrace \\\hline \end{array}\)
\(\begin{array}{cc} \hline \text{Parameter} & \text{Solution set}\\ \hline a=0 & \mathbb{R} \\ a\notin\{-3,0,3\} & \left\lbrace\frac{a^2-9}{2a}\right\rbrace \\\hline \end{array}\)
\(\begin{array}{cc} \hline \text{Parameter} & \text{Solution set}\\ \hline a=0 & \mathbb{R}\setminus\{0\} \\ a\notin\{-3,0,3\} & \left\lbrace\frac{a^2-9}{2a}\right\rbrace \\\hline \end{array}\)

9000104307

Level: 
B
Assuming \(a\in \left (0,2\right )\), solve the following inequality. \[ a\left (a - 2\right )x > 1 \]
\(\left (-\infty , \frac{1} {a\left (a-2\right )}\right )\)
\(\left ( \frac{1} {a\left (a-2\right )},\infty \right )\)
\(\emptyset \)
\(\left \{ \frac{1} {a\left (a-2\right )}\right \}\)

9000034701

Level: 
B
Identify a set of the values of the real parameter \(m\) which ensure that the equation \[ \frac{m} {x} - 8 = \frac{1} {x} -\frac{m + 3} {2} \] has solution \(x = 2\).
\(\left \{7\right \}\)
\(\left \{10\right \}\)
\(\left \{6\right \}\)
\(\left \{\frac{5} {2}\right \}\)