Equations and inequalities with parameters

9000104402

Level: 
A
Find a set of the values of the real parameter \(a\) which ensure that the following equation has no solution. \[ 2a^{2}x - ax - 2a = -1 \]
\(\left \{0\right \}\)
\(\left \{\frac{1} {2}\right \}\)
\(\left \{-\frac{1} {2}\right \}\)
\(\left \{-\frac{1} {2}; \frac{1} {2}\right \}\)

9000104403

Level: 
A
Find a set of the values of the real parameter \(a\) which ensure that the following equation has infinitely many solutions. \[ 3a^{2}x - 2ax + 4 = 6a \]
\(\left \{\frac{2} {3}\right \}\)
\(\left \{-\frac{2} {3}\right \}\)
\(\left \{0\right \}\)
\(\left \{0; \frac{2} {3}\right \}\)

9000104405

Level: 
A
Find a set of the values of the real parameter \(a\) which ensure that the following equation has a unique solution. \[ a^{3}x + 3 = 3a^{2}x + a \]
\(\mathbb{R}\setminus \left \{0;3\right \}\)
\(\left \{0\right \}\)
\(\left \{0;3\right \}\)
\(\mathbb{R}\setminus \left \{3\right \}\)

9000104501

Level: 
A
Consider equation \[ \frac{x - 3} {a} = \frac{a - x} {3} + 2 \] with an unknown \(x\in \mathbb{R}\) and a real parameter \(a\in \mathbb{R}\setminus \{0\}\). Identify a statement which is not true.
For \(a\mathrel{\in }\{ - 3;0\}\) we have \(x = \frac{1} {a+3}\).
For \(a\mathrel{\notin }\{ - 3;0\}\) we have \(x = a + 3\).
If \(a = -3\), then the equation has infinitely many solutions.

9000104502

Level: 
A
Solve the following equation with unknown \(x\) and a real parameter \(a\in\mathbb{R}\setminus\{-1\}\). \[\frac{x} {a+1} = x - a\]
\(\begin{array}{cc} \hline \text{Parameter} & \text{Solution set}\\ \hline a=0 & \mathbb{R} \\ a\notin\{-1;0\} & \{a+1\} \\\hline \end{array}\)
\(\begin{array}{cc} \hline \text{Parameter} & \text{Solution set}\\ \hline a=0 & \mathbb{R} \\ a\notin\{-1;0\} & \emptyset \\\hline \end{array}\)
\(\begin{array}{cc} \hline \text{Parameter} & \text{Solution set}\\ \hline a=0 & \emptyset \\ a\notin\{-1;0\} & \{a+1\} \\\hline \end{array}\)

9000104505

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