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

Which of the following is the complete solution set of x^2-1x^2-5x+6≥ 0?

Write down the inequality to be solved: x^2-1x^2-5x+6≥ 0 Do not multiply through by the denominator: sign of x^2-5x+6 is unknown Write the left-hand side as a single fraction: ((x-1)(x+1))/((x-3)(x-2))≥ 0 Multiply by the square of the denominator, which is positive: (x-3)(x-2)(x-1)(x+1)≥ 0 State the critical values: x=-1, x=1, x=2, x=3 Sketch the polynomial obtained: y=(x-3)(x-2)(x-1)(x+1) Note where the expression is undefined: x≠ 2, x≠ 3 Test the interval x 3: x=4:=(15)/(2),=0\;\;true Decide whether x=-1 belongs to the solution set: x=-1:=RHS=0\;\;included Decide whether x=1 belongs to the solution set: x=1:=RHS=0\;\;included Select the option describing this solution set: x≤ -1 1≤ x 3 Answer: x≤ -1 1≤ x 3

Hard Further Maths Inequalities Questions

Challenging, exam-style Further Maths Inequalities questions with worked solutions. Stretch yourself on the hardest modulus, modulus-vs-linear, case-analysis, graphical-method problems.

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Further Maths34 questionsStep-by-step solutions

Question 1

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

Question 2

9 markschallenging

Which of the following is the complete solution set of

\frac{x^{2}-1}{x^{2}-5x+6}\ge 0

Show worked solution

Worked solution

Write down the inequality to be solved
$\frac{x^{2}-1}{x^{2}-5x+6}\ge 0$
Identify the two sides and the direction of the inequality.
Do not multiply through by the denominator
$\text{sign of }x^{2}-5x+6\text{ is unknown}$
The denominator may be positive or negative, so multiplying by it could reverse the inequality.
Write the left-hand side as a single fraction
$\frac{\left(x-1\right)\left(x+1\right)}{\left(x-3\right)\left(x-2\right)}\ge 0$
Putting everything over a common denominator is the standard first move.
Multiply by the square of the denominator, which is positive
$\left(x-3\right)\left(x-2\right)\left(x-1\right)\left(x+1\right)\ge 0$
Multiplying by $\left(x^{2}-5x+6\right)^{2}>0$ is safe and clears the fraction without reversing the inequality.
State the critical values
$x=-1,\quad x=1,\quad x=2,\quad x=3$
These are the values at which the two sides are equal or the expression is undefined.
Sketch the polynomial obtained
$y=\left(x-3\right)\left(x-2\right)\left(x-1\right)\left(x+1\right)$
The polynomial changes sign only at its roots, which are the critical values.
Note where the expression is undefined
$x\neq 2,\quad x\neq 3$
These values must be excluded from the final answer.
Test the interval $x<-1$
$x=-2:\quad\text{LHS}=\frac{3}{20},\quad\text{RHS}=0\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $-1<x<1$
$x=0:\quad\text{LHS}=-\frac{1}{6},\quad\text{RHS}=0\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $1<x<2$
$x=\frac{3}{2}:\quad\text{LHS}=\frac{5}{3},\quad\text{RHS}=0\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $2<x<3$
$x=\frac{5}{2}:\quad\text{LHS}=-21,\quad\text{RHS}=0\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $x>3$
$x=4:\quad\text{LHS}=\frac{15}{2},\quad\text{RHS}=0\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Decide whether $x=-1$ belongs to the solution set
$x=-1:\quad\text{LHS}=\text{RHS}=0\;\Rightarrow\;\text{included}$
The two sides are equal here, so the value is accepted by a weak inequality.
Decide whether $x=1$ belongs to the solution set
$x=1:\quad\text{LHS}=\text{RHS}=0\;\Rightarrow\;\text{included}$
The two sides are equal here, so the value is accepted by a weak inequality.
Select the option describing this solution set
$x\le -1\quad\text{or}\quad 1\le x<2\quad\text{or}\quad x>3$
This is the complete set of values of $x$ satisfying the inequality.

Answer

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

Question 3

9 markschallenging

Which of the following is the complete solution set of

\frac{2}{x-2}\ge x-1

Show worked solution

Worked solution

Write down the inequality to be solved
$\frac{2}{x-2}\ge x-1$
Identify the two sides and the direction of the inequality.
Do not multiply through by the denominator
$\text{sign of }x-2\text{ is unknown}$
The denominator may be positive or negative, so multiplying by it could reverse the inequality.
Move every term to one side
$\frac{2}{x-2}-\left(x-1\right)\;\ge\;0$
A single expression compared with zero can be analysed by sign.
Write the left-hand side as a single fraction
$-\frac{x\left(x-3\right)}{x-2}\ge 0$
Putting everything over a common denominator is the standard first move.
Multiply by the square of the denominator, which is positive
$-x\left(x-3\right)\left(x-2\right)\ge 0$
Multiplying by $\left(x-2\right)^{2}>0$ is safe and clears the fraction without reversing the inequality.
State the critical values
$x=0,\quad x=2,\quad x=3$
These are the values at which the two sides are equal or the expression is undefined.
Sketch the polynomial obtained
$y=-x\left(x-3\right)\left(x-2\right)$
The polynomial changes sign only at its roots, which are the critical values.
Note where the expression is undefined
$x\neq 2$
These values must be excluded from the final answer.
Test the interval $x<0$
$x=-1:\quad\text{LHS}=-\frac{2}{3},\quad\text{RHS}=-2\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $0<x<2$
$x=1:\quad\text{LHS}=-2,\quad\text{RHS}=0\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $2<x<3$
$x=\frac{5}{2}:\quad\text{LHS}=4,\quad\text{RHS}=\frac{3}{2}\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $x>3$
$x=4:\quad\text{LHS}=1,\quad\text{RHS}=3\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Decide whether $x=0$ belongs to the solution set
$x=0:\quad\text{LHS}=\text{RHS}=-1\;\Rightarrow\;\text{included}$
The two sides are equal here, so the value is accepted by a weak inequality.
Decide whether $x=2$ belongs to the solution set
$x=2:\quad\text{denominator}=0\;\Rightarrow\;\text{excluded}$
The expression is undefined here, so this value is never a solution.
Decide whether $x=3$ belongs to the solution set
$x=3:\quad\text{LHS}=\text{RHS}=2\;\Rightarrow\;\text{included}$
The two sides are equal here, so the value is accepted by a weak inequality.
Check a value inside the solution set
$x=-20:\quad -\frac{1}{11}\;\ge\;-21\;\text{is true}$
Substituting back into the ORIGINAL inequality confirms the interior of the solution set.
Select the option describing this solution set
$x\le 0\quad\text{or}\quad 2<x\le 3$
This is the complete set of values of $x$ satisfying the inequality.

Answer

x\le 0\quad\text{or}\quad 2<x\le 3

Question 4

9 markschallenging

Which of the following is the complete solution set of

\left|{x^{2}-1}\right|\le 3

Show worked solution

Worked solution

Write down the inequality to be solved
$\left|{x^{2}-1}\right|\le 3$
Identify the two sides and the direction of the inequality.
Remove the modulus using the double-inequality rule
$-3\;\le\;x^{2}-1\;\le\;3$
For $k>0$ , $\left|q\right|<k$ means $-k<q<k$ .
Solve the upper inequality
$x^{2}-1\le 3\;\Rightarrow\;-2\le x\le 2$
This gives the first restriction on $x$ .
Solve the lower inequality
$x^{2}-1\ge -3\;\Rightarrow\;\text{true for all real }x$
This gives the second restriction on $x$ .
Both restrictions must hold at once
$-2\le x\le 2$
Take the intersection of the two solution sets.
State the critical values
$x=-2,\quad x=2$
These are the values at which the two sides are equal or the expression is undefined.
Test the interval $x<-2$
$x=-3:\quad\text{LHS}=8,\quad\text{RHS}=3\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $-2<x<2$
$x=0:\quad\text{LHS}=1,\quad\text{RHS}=3\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $x>2$
$x=3:\quad\text{LHS}=8,\quad\text{RHS}=3\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Decide whether $x=-2$ belongs to the solution set
$x=-2:\quad\text{LHS}=\text{RHS}=3\;\Rightarrow\;\text{included}$
The two sides are equal here, so the value is accepted by a weak inequality.
Decide whether $x=2$ belongs to the solution set
$x=2:\quad\text{LHS}=\text{RHS}=3\;\Rightarrow\;\text{included}$
The two sides are equal here, so the value is accepted by a weak inequality.
Check a value inside the solution set
$x=-2:\quad 3\;\le\;3\;\text{is true}$
Substituting back into the ORIGINAL inequality confirms the interior of the solution set.
Check a value outside the solution set
$x=-20:\quad 399\;\le\;3\;\text{is false}$
Substituting back into the ORIGINAL inequality confirms the boundary of the solution set.
Recall why you may not simply multiply by the denominator
$\text{the sign of }x-a\text{ is unknown}$
Multiplying an inequality by a negative quantity reverses it, so an unknown-sign denominator must never be cleared directly.
Recall the safe way to clear a denominator
$\left(x-a\right)^{2}>0\quad\text{for }x\neq a$
Multiplying by the SQUARE of the denominator is always safe because a square is positive.
Select the option describing this solution set
$-2\le x\le 2$
This is the complete set of values of $x$ satisfying the inequality.

Answer

-2\le x\le 2

Question 5

9 markschallenging

By sketching the graphs of both sides, or otherwise, solve

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

Show worked solution

Worked solution

Write down the inequality to be solved
$\frac{3}{x+1}\le x+3$
Identify the two sides and the direction of the inequality.
Do not multiply through by the denominator
$\text{sign of }x+1\text{ is unknown}$
The denominator may be positive or negative, so multiplying by it could reverse the inequality.
Move every term to one side
$\frac{3}{x+1}-\left(x+3\right)\;\le\;0$
A single expression compared with zero can be analysed by sign.
Write the left-hand side as a single fraction
$-\frac{x\left(x+4\right)}{x+1}\le 0$
Putting everything over a common denominator is the standard first move.
Multiply by the square of the denominator, which is positive
$-x\left(x+1\right)\left(x+4\right)\le 0$
Multiplying by $\left(x+1\right)^{2}>0$ is safe and clears the fraction without reversing the inequality.
State the critical values
$x=-4,\quad x=-1,\quad x=0$
These are the values at which the two sides are equal or the expression is undefined.
Sketch the polynomial obtained
$y=-x\left(x+1\right)\left(x+4\right)$
The polynomial changes sign only at its roots, which are the critical values.
Note where the expression is undefined
$x\neq -1$
These values must be excluded from the final answer.
Test the interval $x<-4$
$x=-5:\quad\text{LHS}=-\frac{3}{4},\quad\text{RHS}=-2\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $-4<x<-1$
$x=-\frac{5}{2}:\quad\text{LHS}=-2,\quad\text{RHS}=\frac{1}{2}\;\Rightarrow\;\text{true}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $-1<x<0$
$x=-\frac{1}{2}:\quad\text{LHS}=6,\quad\text{RHS}=\frac{5}{2}\;\Rightarrow\;\text{false}$
A single test value decides the sign on the whole interval, because the sign can only change at a critical value.
Test the interval $x>0$
$x=1:\quad\text{LHS}=\frac{3}{2},\quad\text{RHS}=4\;\Rightarrow\;\text{true}$
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
$x=-4:\quad\text{LHS}=\text{RHS}=-1\;\Rightarrow\;\text{included}$
The two sides are equal here, so the value is accepted by a weak inequality.
Decide whether $x=-1$ belongs to the solution set
$x=-1:\quad\text{denominator}=0\;\Rightarrow\;\text{excluded}$
The expression is undefined here, so this value is never a solution.
State the complete solution set
$-4\le x<-1\quad\text{or}\quad x\ge 0$
This is the complete set of values of $x$ satisfying the inequality.

Answer

-4\le x<-1\quad\text{or}\quad x\ge 0

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