1.02l, 1.02s, 1.02t - Modulus graphs, equations and inequalities | OCR A-Level Mathematics A

Turn modulus notation into distances, straight-line branches and clear solution sets. You will sketch a folded line, solve equations at its intersections and decide which intervals satisfy an inequality.

Modulus as distance and bounds

How can the same two boundary points describe either a middle interval or two outside intervals? Imagine a number-line alarm that accepts positions at most 2 units from 3. It accepts 1 and 5, but also everything between them; an alarm accepting positions more than 2 units away chooses the outside instead.

Modulus

The modulus of a real number is its distance from zero, written $|x|$ . Thus $|5|=|-5|=5$ and $|0|=0$ . It is non-negative, rather than always positive. The distance between $x$ and $a$ is $|x-a|$ .

For a negative input, changing its sign makes the distance positive. In particular, $|-5|=-(-5)$ , not $-5$ .

Modulus and distance bounds

For real $x$ ,

|x|=\begin{cases}x&x\ge0\\-x&x<0.\end{cases}

For real $a$ and $b>0$ ,

|x-a|<b\quad\Longleftrightarrow\quad a-b<x<a+b.

Replace both strict signs by inclusive signs for $|x-a|\le b$ . An outside condition uses “or”: $|x-a|>b$ means $x<a-b$ or $x>a+b$ .

Not in the formula sheet: learn it.

Try the distance picture before the algebra. Both tell you which side of each boundary to keep.

Worked example 1: an inclusive distance

Question. Solve $|x-3|\le2$ .

Plan. Find the positions 2 units either side of 3, then keep the interval between them.

\begin{aligned} -2&\le x-3\le2 &&\text{distance bound}\\ 1&\le x\le5 &&\text{add 3 throughout} \end{aligned}

Why this step? A distance at most 2 allows movement in either direction from 3, including both endpoints.

Answer. $1\le x\le5$ .

Check. At $x=1$ and $x=5$ the distance is 2; at $x=3$ it is 0. All three are accepted.

Change just the comparison to “more than”. The boundary positions stay put, but the accepted region changes.

Worked example 2: an outside distance

Question. Solve $|x-3|>2$ .

Plan. Use the same boundaries and keep the two outside intervals, excluding equality.

\begin{aligned} x-3&<-2\quad\text{or}\quad x-3>2 &&\text{outside the bounds}\\ x&<1\quad\text{or}\quad x>5 &&\text{add 3} \end{aligned}

Why this step? Positions outside the interval are further than 2 from 3; a position cannot lie in both outside intervals at once.

Answer. $x<1$ or $x>5$ .

Check. $x=0$ gives distance 3 and works; $x=3$ gives 0 and fails. Neither endpoint works because its distance is exactly 2.

A distance cannot have a negative upper bound: $|x-a|\le-1$ has no solutions. The boundary case $|x-a|\le0$ gives only $x=a$ ; $|x-a|<0$ gives none.

Watch out

Wrong: $|x-3|>2$ gives $1<x<5$ .

Right: $|x-3|>2$ gives $x<1$ or $x>5$ . First decide whether the distance is inside or outside the bound.

The next check changes the centre and the distance, keeping the outside reasoning.

Folding a linear graph

A modulus graph keeps the part of a line above the horizontal axis and folds the part below it upwards.

You already know

A straight line $y=ax+b$ has gradient $a$ and vertical intercept $b$ . Its horizontal intercept is found by setting $y=0$ .

For $y=|2x-4|$ , the line underneath changes sign at $x=2$ . To the left, its output is negative, so that branch changes sign; to the right it stays as it is.

The line y equals 2x minus 4 folded upwards, with y equals 3 meeting the V at two labelled points

The V has straight arms, not the rounded bottom of a parabola.

A single linear modulus graph

For real $a\ne0$ ,

|ax+b|=|a|\left|x+\frac ba\right|.

Its vertex is $\left(-\frac ba,0\right)$ , its axis of symmetry is $x=-\frac ba$ , and the two branch gradients have magnitude $|a|$ . If $a=0$ , the graph is the horizontal line $y=|b|$ .

Not in the formula sheet: learn it.

Locate the zero before writing the branches: the sign of the expression inside the modulus decides which branch to use.

Worked example 3: folding a positive-gradient line

Question. Describe a sketch of $y=|2x-4|$ , giving its vertex, vertical intercept and branches.

Plan. Find the zero of the inside expression, then change its sign exactly where it is negative.

\begin{aligned} 2x-4=0&\quad\Longrightarrow\quad x=2 &&\text{vertex input}\\ |2(0)-4|&=4 &&\text{vertical intercept}\\ y&=\begin{cases}4-2x&x<2\\2x-4&x\ge2.\end{cases} &&\text{use the sign of }2x-4 \end{aligned}

Why this step? Only negative outputs are reflected; positive outputs already equal their modulus.

Answer. Draw straight branches meeting at $(2,0)$ , through $(0,4)$ , with left gradient $-2$ and right gradient 2.

Check. At $x=1$ and $x=3$ , the output is 2 on both sides of the vertex.

Now reverse the sign of the entire expression inside the modulus. This is a useful surprise: the folded graph does not change at all.

Worked example 4: a negative-gradient line

Question. Describe the graph and branches of $y=|-2x+4|$ .

Plan. Find where $-2x+4$ changes sign and compare the resulting branches with the previous graph.

\begin{aligned} -2x+4&\ge0\quad\Longleftrightarrow\quad x\le2 &&\text{positive branch}\\ y&=\begin{cases}-2x+4&x\le2\\2x-4&x>2\end{cases} &&\text{fold negative outputs}\\ |-2x+4|&=|-(2x-4)|=|2x-4| &&\text{opposite numbers have equal distances} \end{aligned}

Why this step? Modulus loses the sign of the whole input. A descending line and its negative produce the same V.

Answer. The vertex is $(2,0)$ , the vertical intercept is $(0,4)$ , and the graph is unchanged.

Check. At $x=4$ , $|-8+4|=4$ , the same output as $|8-4|$ .

This folding rule is for one modulus sign applied to a linear expression. Do not use a smooth curve as a substitute for its straight branches.

Watch out

Wrong: reflecting the entire line $y=2x-4$ to obtain $y=4-2x$ .

Right: keep $y=2x-4$ where $x\ge2$ , and reflect only the branch where $x<2$ .

Apply the same branch method with a new zero.

Equations and admissible intersections

A solution of a modulus equation is an input where the modulus graph meets the graph on the other side.

On the diagram above, $y=3$ meets $y=|2x-4|$ twice. Graphs show how many intersections to expect; the branch equations give exact values.

Removing a modulus in an equation

For real $u$ and $v$ ,

|u|=v\quad\Longleftrightarrow\quad v\ge0\ \text{and}\ (u=v\ \text{or}\ u=-v).

Equivalently, require $u^2=v^2$ and $v\ge0$ . Without the sign condition, squaring can introduce false solutions.

Not in the formula sheet: learn it.

A horizontal line above the vertex cuts both arms; one below the horizontal axis cannot meet this V at all.

Worked example 5: two horizontal intersections

Question. Solve $|2x-4|=3$ .

Plan. Set the inside expression equal to 3 and to $-3$ .

\begin{aligned} 2x-4&=3\quad\text{or}\quad2x-4=-3 &&\text{two ways to have modulus 3}\\ x&=\frac72\quad\text{or}\quad x=\frac12 &&\text{solve each line} \end{aligned}

Why this step? The modulus is 3 for the two signed inputs 3 and $-3$ .

Answer. $x=\frac12$ or $x=\frac72$ , the horizontal coordinates of the two intersections with $y=3$ .

Check. The inside outputs are $-3$ and 3, whose moduli are both 3.

Change one feature: the other graph is now a sloping line. A candidate from extending an arm backwards may fail its branch condition.

Worked example 6: rejecting a false intersection

Question. Solve $|2x-4|=3x+1$ .

Plan. Split at $x=2$ , solve each linear equation and check its branch.

\begin{aligned} x\ge2:&\quad 2x-4=3x+1\ \Longrightarrow\ x=-5 &&\text{fails }x\ge2\\ x<2:&\quad4-2x=3x+1\ \Longrightarrow\ x=\frac35 &&\text{lies on this branch} \end{aligned}

Why this step? Each straight-line equation represents only one arm of the V, so its solution must lie on that arm.

Answer. The only solution is $x=\frac35$ .

Check. At $x=\frac35$ , both sides are $\frac{14}{5}$ . At the rejected $x=-5$ , they are 14 and $-14$ , so the unsquared equation is false.

Squaring would preserve the equality of the two squares at $-5$ . It would not repair the negative right-hand side of the actual equation.

Watch out

Wrong: $|2x-4|=3x+1$ has both roots of $(2x-4)^2=(3x+1)^2$ .

Right: check the branch or substitute into the unsquared equation; $x=-5$ is not a solution.

Your next sloping line meets both arms. Establish that both candidates are admissible.

Choosing intervals from modulus inequalities

Intersections locate the boundaries of a graphical inequality; vertical order decides which intervals to keep.

For $|2x-4|<3$ , look for the V below the horizontal line. Between its two intersections the V is lower; outside, it is higher. With a sloping line, the acceptable set need not be a middle interval.

Graphical comparison

For graphs $y=\mathrm f(x)$ and $y=\mathrm g(x)$ ,

\mathrm f(x)<\mathrm g(x)

selects the inputs where the first graph is below the second. Include their intersections for $\le$ , and exclude them for $<$ . Use only inputs where both functions are defined.

Not in the formula sheet: learn it.

Use the picture to choose the intervals, then algebra to locate their exact endpoints.

Worked example 7: below a horizontal line

Question. Solve $|2x-4|<3$ graphically, giving exact boundaries.

Plan. Find the intersections with $y=3$ , then select where the V lies below it.

\begin{aligned} 2x-4&=\pm3 &&\text{intersection boundaries}\\ x&=\frac12,\ \frac72 &&\text{solve the two equations}\\ \frac12&<x<\frac72 &&\text{V below the line} \end{aligned}

Why this step? The inequality asks for vertical order, so the crossings alone are not the complete answer.

Answer. $\frac12<x<\frac72$ .

Check. At $x=2$ , $0<3$ ; at $x=0$ , $4<3$ is false. Both equality endpoints are excluded.

Change only the other graph to a sloping line, keeping the same modulus and strict comparison. Compare each arm with the line and restrict the answer to that arm.

Worked example 8: below a sloping line

Question. Solve $|2x-4|<3x+3$ .

Plan. Compare the two straight arms with $y=3x+3$ , then join the permitted inputs.

\begin{aligned} x\ge2:&\quad 2x-4<3x+3\ \Longrightarrow\ x>-7 &&\text{all of this arm works}\\ x<2:&\quad4-2x<3x+3\ \Longrightarrow\ x>\frac15 &&\text{keep }\frac15<x<2 \end{aligned}

Why this step? The first inequality is solved inside $x\ge2$ , not on the whole number line. Its answer contributes $[2,\infty)$ , not $(-7,\infty)$ .

Answer. Joining both branches gives $x>\frac15$ .

Check. At $\frac15$ , both sides are $\frac{18}{5}$ , so that endpoint is excluded. At $x=-1$ , the inequality becomes $6<0$ , so that input fails.

You can also obtain the answer by sketching the V and line: they meet once, after which the line stays strictly above both arms.

Watch out

Wrong: square an inequality with a potentially negative right-hand side and keep every resulting interval.

Right: use branches or vertical order. A negative number cannot be an upper bound for a non-negative modulus.

Keep the graph interpretation as you solve the next two-branch comparison.

Comparing two distances

When both sides are moduli, both are non-negative, so squaring preserves their order.

This is a useful distinction from the previous part: there is no hidden negative right-hand side to reject. Algebra can compare the distances without drawing two V-shaped graphs.

Comparing moduli by squares

For real $u$ and $v$ ,

|u|\le|v|\quad\Longleftrightarrow\quad u^2\le v^2.

The same equivalence holds with $<$ , $>$ , $\ge$ or $=$ . It works because squaring preserves order for non-negative numbers.

Not in the formula sheet: learn it.

The difference of the two squares gives a familiar quadratic sign problem.

Worked example 9: two inclusive distances

Question. Solve $|x-1|\le|2x+1|$ .

Plan. Square both non-negative sides, factor their difference and select its sign intervals.

\begin{aligned} (x-1)^2&\le(2x+1)^2 &&\text{both sides non-negative}\\ 0&\le3x^2+6x=3x(x+2) &&\text{expand and collect}\\ x&\le-2\quad\text{or}\quad x\ge0 &&\text{product non-negative outside} \end{aligned}

Why this step? It is the two moduli that are squared, not two arbitrary signed expressions whose order might reverse.

Answer. $x\le-2$ or $x\ge0$ .

Check. At $x=-1$ , the distances are 2 and 1, so the middle interval fails. At $x=-2$ , both are 3 and equality is accepted.

Change only the sign to a strict comparison. The quadratic boundaries stay the same and equality is now removed.

Worked example 10: the same comparison made strict

Question. Solve $|x-1|<|2x+1|$ .

Plan. Repeat the squared comparison, retaining only positive values of the difference.

\begin{aligned} (x-1)^2&<(2x+1)^2 &&\text{order preserved}\\ 3x(x+2)&>0 &&\text{collect the difference}\\ x&<-2\quad\text{or}\quad x>0 &&\text{exclude zero product} \end{aligned}

Why this step? Strictness changes endpoint membership, not the sign of the product between its roots.

Answer. $x<-2$ or $x>0$ .

Check. At $x=1$ , $0<3$ works. At $x=0$ , $1<1$ fails.

The opening alarm problem and this comparison have the same final decision: locate the boundaries, then decide which intervals satisfy the actual sign.

Watch out

Wrong: $|u|\le|v|$ gives $u\le v$ .

Right: compare $u^2$ and $v^2$ , or compare all valid straight-line branches. For example, $|-3|\le|2|$ is false although $-3\le2$ is true.

Finish by changing the two distances rather than the comparison sign.

Quick Recap

Modulus is a non-negative distance; zero is allowed.
Fold only the negative-output part of a line to sketch its modulus.
Equations locate intersections; inequalities also require a choice of intervals and endpoints.
Check branch conditions when the other side can be negative. Squaring two moduli preserves their comparison.

The alarm's boundaries are 1 and 5: distance at most 2 selects $[1,5]$ , while distance more than 2 selects $x<1$ or $x>5$ .

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1.02l, 1.02s, 1.02t - Modulus graphs, equations and inequalities

Modulus as distance and bounds

Folding a linear graph

Equations and admissible intersections

Choosing intervals from modulus inequalities

Comparing two distances

Quick Recap

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