9000028301
Level:
B
The following equation has a solution
\(x = 1\). Find
the sum of the remaining real solutions.
\[
x^{3} - 7x + 6 = 0
\]
\(- 1\)
\(1\)
\(0\)
\(2\)
B
9000026403
Level:
B
To solve the equation with absolute value using interval method we have to divide
the domain of the equation by the zero points of the subexpressions in the absolute
value. Find all these points.
\[
|x + 1| + |2x - 1| = 3
\]
\(-1,\ \frac{1}
{2}\)
\(- 3\)
\(1,\ -\frac{1}
{2}\)
\(0\)
9000026106
Level:
B
Find the solution set of the following inequality.
\[
\frac{x + 3}
{x - 1} > 1
\]
\(\left (1;\infty \right )\)
\(\left (-\infty ;1\right )\)
\(\left (-\infty ;3\right ] \)
\(\left [ 3;\infty \right )\)
9000026404
Level:
B
To solve the equation with absolute value using interval method we have to divide
the domain of the equation by the zero points of the subexpressions in the absolute
value. Find all these points.
\[
2|x - 2| + |2 - x| = 1 + |x|
\]
\(2,\ 0\)
\(-2,\ 2,\ 0\)
\(-1,\ 2\)
\(-1,\ 2,\ 0\)
9000026108
Level:
B
An inequality is solved graphically as shown in the picture. The solution is marked in red. Choose the corresponding inequality.
\(2\leq \frac{x+3}
{x} \)
\(2\geq \frac{3}
{x}\)
\(2\geq \frac{x+3}
{x} \)
\(2\leq \frac{3}
{x}\)
9000028102
Level:
B
Given graph of the linear function \(f\),
find the solution set of the inequality \(f(x) > 0\).
\((-4;\infty )\)
\(\emptyset \)
\((-4;2)\)
\((-2;\infty )\)
9000028101
Level:
B
Given graphs of the linear functions \(f\)
and \(g\), find the solution
set of the inequality \(f(x) > g(x)\).
\((1;\infty )\)
\(\mathbb{R}\)
\((0;1)\)
\((-\infty ;1)\)
9000026405
Level:
B
Assuming \(x\in (-\infty ;1)\),
write the following equation in the form which does not contain an absolute value.
\[
3 = |x - 1|
\]
\(3 = -x + 1\)
\(3 = x - 1\)
\(3 = -x - 1\)
\(3 = x + 1\)
9000026406
Level:
B
Assuming \(x\in (-3;2)\),
write the following equation in the form which does not contain an absolute value.
\[
|x + 3| = |x - 2|
\]
\(x + 3 = -x + 2\)
\(x + 3 = x - 2\)
\(- x - 3 = -x + 2\)
\(- x - 3 = x + 2\)
9000026407
Level:
B
Assuming \(x\in \left [ \frac{1}
{2};\infty \right )\),
write the following equation in the form which does not contain an absolute value.
\[
|x + 1| + |2x - 1| = 3
\]
\(x + 1 + 2x - 1 = 3\)
\(x + 1 - 2x - 1 = 3\)
\(x + 1 - 2x + 1 = 3\)
\(- x - 1 + 2x - 1 = 3\)