Solve the inequality |2x-1|<5.

Remove the modulus using the double-inequality rule: -5\;<\;2x-1\;<\;5 Isolate the term in x in all three parts: -4\;<\;2x\;<\;6 Divide all three parts by 2: -2\;<\;x\;<\;3 State the complete solution set: -2<x<3 Answer: -2<x<3

Which of the following is the complete solution set of (x-3)(x-1)(x+2)(x+4)>0?

Write down the inequality to be solved: (x-3)(x-1)(x+2)(x+4)>0 State the critical values: x=-4, x=-2, x=1, x=3 Sketch the graph of the left-hand side: y=(x-3)(x-1)(x+2)(x+4) Read the required region from the sketch: (x-3)(x-1)(x+2)(x+4)\;>\;0 Test the interval x 3: x=4:=144,=0\;\;true Decide whether x=-4 belongs to the solution set: x=-4:=RHS=0\;\;excluded Decide whether x=-2 belongs to the solution set: x=-2:=RHS=0\;\;excluded Decide whether x=1 belongs to the solution set: x=1:=RHS=0\;\;excluded Decide whether x=3 belongs to the solution set: x=3:=RHS=0\;\;excluded Check a value inside the solution set: x=-20: 139104\;>\;0\;is true Check a value outside the solution set: x=-4: 0\;>\;0\;is false Select the option describing this solution set: x 3 Answer: x 3

Further Maths Inequalities Practice Questions

Free Further Maths Inequalities practice questions with full step-by-step worked solutions. Covers modulus, modulus-inequality, critical-values, quadratic-inequality. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions

Question 1

2 markseasy

Solve the inequality

\left|{2x-1}\right|<5

Show worked solution

Worked solution

Remove the modulus using the double-inequality rule
$-5\;<\;2x-1\;<\;5$
For a positive constant $k$ , $\left|u\right|<k$ is equivalent to $-k<u<k$ .
Isolate the term in $x$ in all three parts
$-4\;<\;2x\;<\;6$
The constant $-1$ is removed from all three parts at once.
Divide all three parts by $2$
$-2\;<\;x\;<\;3$
Dividing by a positive number does not reverse the inequality.
State the complete solution set
$-2<x<3$
This is the complete set of values of $x$ satisfying the inequality.

Answer

-2<x<3

Question 2

2 markseasy

Which of the following is the complete solution set of

\left|{2x-7}\right|\ge 1

Show worked solution

Worked solution

Split the modulus into two separate cases
$2x-7\;\le\;-1\quad\text{or}\quad 2x-7\;\ge\;1$
For a positive constant $k$ , $\left|u\right|>k$ is equivalent to $u<-k$ or $u>k$ .
Solve the first case
$2x-7\le -1\;\Rightarrow\;x\le 3$
Rearranging the first branch gives one part of the solution.
Solve the second case
$2x-7\ge 1\;\Rightarrow\;x\ge 4$
Rearranging the second branch gives the other part.
Select the option describing this solution set
$x\le 3\quad\text{or}\quad x\ge 4$
This is the complete set of values of $x$ satisfying the inequality.

Answer

x\le 3\quad\text{or}\quad x\ge 4

Question 3

4 marksintermediate

Which of the following is the complete solution set of

\frac{8}{2x-1}>4

Show worked solution

Worked solution

Write down the inequality to be solved
$\frac{8}{2x-1}>4$
Identify the two sides and the direction of the inequality.
Do not multiply through by the denominator
$\text{sign of }2x-1\text{ is unknown}$
The denominator may be positive or negative, so multiplying by it could reverse the inequality.
Move every term to one side
$\frac{8}{2x-1}-\left(4\right)\;>\;0$
A single expression compared with zero can be analysed by sign.
Write the left-hand side as a single fraction
$-\frac{4\left(2x-3\right)}{2x-1}>0$
Putting everything over a common denominator is the standard first move.
Multiply by the square of the denominator, which is positive
$-4\left(2x-3\right)\left(2x-1\right)>0$
Multiplying by $\left(2x-1\right)^{2}>0$ is safe and clears the fraction without reversing the inequality.
State the critical values
$x=\frac{1}{2},\quad x=\frac{3}{2}$
These are the values at which the two sides are equal or the expression is undefined.
Select the option describing this solution set
$\frac{1}{2}<x<\frac{3}{2}$
This is the complete set of values of $x$ satisfying the inequality.

Answer

\frac{1}{2}<x<\frac{3}{2}

Question 4

6 markshard

Which of the following is the complete solution set of

\frac{5}{x+3}\le 1

Show worked solution

Worked solution

Write down the inequality to be solved
$\frac{5}{x+3}\le 1$
Identify the two sides and the direction of the inequality.
Do not multiply through by the denominator
$\text{sign of }x+3\text{ is unknown}$
The denominator may be positive or negative, so multiplying by it could reverse the inequality.
Move every term to one side
$\frac{5}{x+3}-\left(1\right)\;\le\;0$
A single expression compared with zero can be analysed by sign.
Write the left-hand side as a single fraction
$\frac{2-x}{x+3}\le 0$
Putting everything over a common denominator is the standard first move.
Multiply by the square of the denominator, which is positive
$-\left(x-2\right)\left(x+3\right)\le 0$
Multiplying by $\left(x+3\right)^{2}>0$ is safe and clears the fraction without reversing the inequality.
State the critical values
$x=-3,\quad x=2$
These are the values at which the two sides are equal or the expression is undefined.
Sketch the polynomial obtained
$y=-\left(x-2\right)\left(x+3\right)$
The polynomial changes sign only at its roots, which are the critical values.
Note where the expression is undefined
$x\neq -3$
These values must be excluded from the final answer.
Test the interval $x<-3$
$x=-4:\quad\text{LHS}=-5,\quad\text{RHS}=1\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $-3<x<2$
$x=-\frac{1}{2}:\quad\text{LHS}=2,\quad\text{RHS}=1\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Select the option describing this solution set
$x<-3\quad\text{or}\quad x\ge 2$
This is the complete set of values of $x$ satisfying the inequality.

Answer

x<-3\quad\text{or}\quad x\ge 2

Question 5

9 markschallenging

Which of the following is the complete solution set of

\left(x-3\right)\left(x-1\right)\left(x+2\right)\left(x+4\right)>0

Show worked solution

Worked solution

Write down the inequality to be solved
$\left(x-3\right)\left(x-1\right)\left(x+2\right)\left(x+4\right)>0$
Identify the two sides and the direction of the inequality.
State the critical values
$x=-4,\quad x=-2,\quad x=1,\quad x=3$
These are the values at which the two sides are equal or the expression is undefined.
Sketch the graph of the left-hand side
$y=\left(x-3\right)\left(x-1\right)\left(x+2\right)\left(x+4\right)$
The curve is a quartic crossing the $x$ -axis at the critical values.
Read the required region from the sketch
$\left(x-3\right)\left(x-1\right)\left(x+2\right)\left(x+4\right)\;>\;0$
The solution is the set of $x$ for which the curve lies on the required side of the $x$ -axis.
Test the interval $x<-4$
$x=-5:\quad\text{LHS}=144,\quad\text{RHS}=0\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $-4<x<-2$
$x=-3:\quad\text{LHS}=-24,\quad\text{RHS}=0\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $-2<x<1$
$x=-\frac{1}{2}:\quad\text{LHS}=\frac{441}{16},\quad\text{RHS}=0\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $1<x<3$
$x=2:\quad\text{LHS}=-24,\quad\text{RHS}=0\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $x>3$
$x=4:\quad\text{LHS}=144,\quad\text{RHS}=0\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Decide whether $x=-4$ belongs to the solution set
$x=-4:\quad\text{LHS}=\text{RHS}=0\;\Rightarrow\;\text{excluded}$
The two sides are equal here, so the value is rejected by a strict inequality.
Decide whether $x=-2$ belongs to the solution set
$x=-2:\quad\text{LHS}=\text{RHS}=0\;\Rightarrow\;\text{excluded}$
The two sides are equal here, so the value is rejected by a strict inequality.
Decide whether $x=1$ belongs to the solution set
$x=1:\quad\text{LHS}=\text{RHS}=0\;\Rightarrow\;\text{excluded}$
The two sides are equal here, so the value is rejected by a strict inequality.
Decide whether $x=3$ belongs to the solution set
$x=3:\quad\text{LHS}=\text{RHS}=0\;\Rightarrow\;\text{excluded}$
The two sides are equal here, so the value is rejected by a strict inequality.
Check a value inside the solution set
$x=-20:\quad 139104\;>\;0\;\text{is true}$
Substituting back into the ORIGINAL inequality confirms the interior of the solution set.
Check a value outside the solution set
$x=-4:\quad 0\;>\;0\;\text{is false}$
Substituting back into the ORIGINAL inequality confirms the boundary of the solution set.
Select the option describing this solution set
$x<-4\quad\text{or}\quad -2<x<1\quad\text{or}\quad x>3$
This is the complete set of values of $x$ satisfying the inequality.

Answer

x<-4\quad\text{or}\quad -2<x<1\quad\text{or}\quad x>3

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