# Modulus Functions, Graphs & Inequalities | Class #06 | A2 Mathematics 9709

Lesson on Functions for CAIE A2 Level mathematics, taught by Arham of Math vs Arham.

- Page: https://mathvsarham.com/lectures/5-rziOlksuY/
- YouTube: https://www.youtube.com/watch?v=5-rziOlksuY
- Channel: https://www.youtube.com/@MathVsArham
- Details: CAIE | A2 Level | Lesson
- Topic: Functions
- Published: 2026-08-05

## Transcript

Auto-generated from YouTube captions; mathematical notation may be inaccurate.

modulus functions. Modulus functions means the functions inside this two vertical lines. This is called the modulus. Whatever that you have inside it, the two vertical lines, these this is called your modulus. And the this modulus function what it does is it will convert anything which is negative into positive and if it is something that is positive it will remain positive. So positive it will remain positive and a negative value after passing through the modulus function will become positive. For example, if you find the modulus, the absolute value function. The modus can modulus function is also known as

absolute value function. It will get you the absolute value. Absolute value of four will be equal to 4 and the absolute value of minus4 will also be equal to 4. You have this function in your calculator. If you will see this over here uh modulus is somewhere over here. So modulus and minus4 is also four and + 4 is also four. So if you put in negative it will convert into positive. This will give you the absolute value. Absolute value is the distance from the zero. Okay, that that's something uh later on the basic or the more logical part or if we go deeper we'll go on deeper. So right now the more relevant part is now

you know that anything positive still remains positive after the modus and anything negative will become positive. So now if you have modulus equation solve the following for example it is x - 4 which = to x something like this minus x they are asking you to solve this. So how do you do this whenever you have this is the modus example of a modus function. So when you open up any when when try to solve this what you do you square on both sides >> when you square on both sides you will get x - 4 square is equ= to - x whole square and if you open it up x² - 8 x + 16 is equ= to x² and the x² x² will cancel out so you will have - 8 x +

16 = 0 and therefore your x value will come out - 16 upon - 8 for 2. >> And if you input this complete thing again back into the function to into the question to check if you have got the right answer, you can confirm as well. I forgot about this example. You will the method is the same. I formed the question myself. So there are some issues over So okay let's suppose it is 3 - x modulus is equ= to 4. So how do you expand this? You will expand it the usual way. So whole square squaring on both sides = to 4 square. So you will if after expansion you will get 9 - 6x + x² = to 16. And when you move everything on one side,

you will get x² - 6x and this would be minus 7. This will be equal to 0. And then you can solve it quadratically and you will get the answer 1 - 6 - 7. So your answer for x is -1 and 7. Just like that. So you square on both side and then uh simplify it you will get the answer. And if you use if you if you check this answer back into the question if you put them back in the question you will see that it is satisfying. Look if I put in x is minus one. So 3 minus min -1 modus is it equals to four. So yes you can see four modulus will be equal to 4. So 4 is equ= to 4. And if you have 3 - 7 is equal to

4. So you will see -4 modulus is equals to 4 and minus 4 modulus is four. So both are satisfied. They can confirm their answer like this as well. Whatever the solution that you get, you can put it back into the question and see if it is satisfying or not. For let's suppose this one you have modulus 2x + 1 is equ= to modulus x + 5. So what would you do? You would square on both sides. 2x + 1 whole square and x + 5 whole square. And then you expand it. So 4x² + 4 x + 1 x² + 10 x + 25. Do everything on one side. So 3x² and this will become - 6x. This will become -4. Will be equal to zero. And again use your calculator. Now if if

you're in A2 then this would work. Okay. Using your calculator to get the final answer it is okay. You can do that. So you can see x = -2 and x = to 4. So this is one of the ways of solving uh these modulus function. Another way of solving these modus function uh is using sometimes by sketching as well. You should also learn how to sketch modulus functions. Sketching modulus function. So this is also included in a syllus. How do they work? How do they work? For example, you have y= to x. If it is only simple y= to x, what will be the graph? It would be a straight line passing through the origin.

And this would be the line y = x. But if I put in this equation modulus y= to modulus x, this is telling me that even if the value of x is negative, your value will y for y would always be positive. Isn't it? Because because of the modus. So your values of x can be negative, positive both. But after you input the value of x inside the modus function everything will come out as positive. So whatever the function for y= to x was you reflect the part which is below which is negative in the opposite direction. So now this would be your graph for y is equals to modus x because anything which is negative it

has reflected it in the x-axis. It is now in the positive y value right because of the mod model modus y cannot be negative because of the modus whatever the value of x is y will never be positive. So if your x is positive your y will remain positive but if your x is negative your y still remains positive because the modus because negative value modus is still positive value. So everything reflected in xx. You can say like this. For example, for example, you have y = x - x + 2. So you will first of all forget about the modulus part. First you sketch it the usual way. Find the x intercept. Find the y intercept.

Just like the usual way. So y intercept forget about the modulus. Let's suppose it is - x + 2. So your y intercept easily you know it is the value of c which is 2. So let's suppose this is your two. Your graph is passing through this point. And then your x intercept your x intercept will be when your y is 0. 0= to - x + 2. So x= to 2. x = to 2. So your line right now is passing like this right because you have the x intercept you have the y intercept and without the modus this will keep on going downward as well in the negative y as well but because now you have a modulus your this graph would be reflected.

So this is the the purple one is your modulus with modulus. Okay, do you get it? The modulus will just reflect everything which is below the axis which is negative into the positive region. >> Same happens with quadratic graphs as well. For example, you have y = to x² - 3. So if I sketch this thing directly without the model, this would be something like this without the model. The y intercept is minus 3. Let's suppose this is minus 3 and the turning point is also minus 3. So quadratic graph the normal one would be like this easy. But now the problem is that the modus part. So what will the modus do? It will reflect everything

that is below the quadratic uh sorry below the x-axis it will convert it into positive. Now it will pass through positive 3. The y intercept it was previously minus 3 but after the modulus it got reflected. So now this is the final answer. Okay. So look look at this reflected >> that is below the x-axis has been reflected upwards. And now to solve these kind of equations modulus 3 - x modulus - 4 there is another way graphically as well you can plot this part separately y= to modulus 3 - x modulus and you can graph this part separately y= to 4 on the same sketch and the point of intersection would be a solution look so

modulus 3 - x is this graph and uh is equ= to 4. y = to 4. So look it is intersecting at the points the values of x are -1 and four. The same two x values that we have found where x values the points of intersection are minus1 and four. So this is another method graphically solving equation. If you want to solve any of the equation over here, you can separate the graphs. Sorry, you can separate the equation into two graphs and you plot them and then the point of intersection will get you the solution. For example, let me do the we solved over here. Right? Where was it? Yes, this one. You will get the same answer

using the graph as well. 2x + 1. So, this is your graph for this 2x + 1. And then you have modus x + 5. So you can see they both are intersecting at x is 4 and x is minus2 interesting. So you can solve them simultaneously like this as well. >> If you are confused it's just logical part of it otherwise you can do it algebraically as well. Okay. Now the problematic part is you know how to modulus graph you know how to solve simple modus equations but now you have modulus inequalities the questions are so these are the modulus inequalities x + 3 modulus less than x - 4 modulus so this is something that's different so

how do you solve this >> it's very similar But only one or few two few things are different. You will add it to uh okay. Now this is modulus inequality. So whatever the inequality is I have brought in a modus equation. Where is the inequality? X modulus less than 5 + 2x solve or find the values of x set of values of x for which is the for which this is the question. So first of all first step is forget about the inequality forget about inequality. The modulus x 5 + 2x then solve it the usual way. x² is = to 5 + 2x whole square. x² is = to 25 + 20x + 4 x². Move everything on one side. So 25 + 20x and then you will have

uh + 3x 4. and then solve this simultaneously. >> Wait, hold on. >> 3 20 25. So your x is coming out as -5 and your other value of x is - 5 upon 3. Now what do you do? You create a you make a number line a basic number line and you you make you draw these two points. So minus5 is more is lesser. So - 5 to the left side and minus 5 upon 3. So and now you make region on your number line. So this is your region one. The region to the left of 5, the region which is between these two roots and the region which is to the right side of 5 upon 3. Then what do you do? You take in any random value from the three regions. So for this one

you can take let's suppose minus 10 you can take minus two you can take zero from here and what do you do you put these values you take any random value from within the regions and then you input these values in the in your question back so if your x is let's suppose -10 so this becomes -10 modus less than 5 + 2 - 10 modulus and now simplify 10 modulus would be and 5 - 20. So 10 less than -5. So 10 less than 15. So is 10 less than 15? Yes. >> Yes. >> This means that this part is this region is part of the solution. This is satisfying the question. Then if if your x is -2. So this would be -2 less than 5

+ 2 - 2. So 2 is less than 5 - uh 4 is 1 1. So is 2 less than 1? No. So this part this is not part of the solution. And then if x is zero. So if x is 0. So 0 modulus is less than 5 - 0 modulus. So 0 is less than five. That means this is part of solution. So as your final answer you will say that your x should be less than minus 5 and your x should be more than - 5 upon 3. Okay. Do you get it? G s the roots the usual way squaring on both good sign find the roots and then plotting them on a number line finding the different regions they're getting random values from the different regions and then inputting back in the question

the parts which are satisfying your question are part of the solution others are not this is your answer. Forget about inequality squaring on both sides. So x² + 2x + 1 x² - 8 x + 16 x² x² cancel out. So you will have 10 x is equ= to 15 x is coming out as 1.5. So only one rule. So what would you do? 1.5. So any value which is above 1.5 and any value which is below one. Let's suppose below it is zero. You can check any one of the region. Don't need to check one of the both of them because if if you find one to be the solution, the other one won't be the answer. If you find the one which you are solving for

as the wrong part, then the other part will be solution. So if your x is zero, if your x is 0, this would become 0 + 1 modulus greater than 0 - 4 modulus. So 1 is greater than 4. No. So this part is not the part of solution. So this is the answer. So this means that x should be more than 1.5. Okay. >> Yes. Okay, let me do this one. So 3x - 2 is greater than equal to x + forget about inequality expand using the identity. 9 x² - 12 x + 4 x² + 8 x + 16. Move everything on one side. So 8 x² this would become - 20x this would become - 12 - 20 0. Now solve it quadratically. 8 - 20 - plug. So x is coming out as.5

and x is coming out as three. Now number line.5 over here three over here. So the region value from between them the easiest value that can help you to solve it is zero. Over here it can be let's suppose five. over here just take it minus one. So if x = -1 3 * -1 - 2 greater than equal to -1 + 4. So - 3 - 2 - 5 - 5 modulus is 5 and 5 is greater than equal to 3. This is part of the solution. Then zero if x is zero. If x is 0 0 - 2 modulus greater than equal to 0 + 4. So 2 modulus is not equal to four. No, this is not part of the solution. And then lastly if x is 5. So 3 * 5 - 2 greater than equal to 5

5 + 4 is uh 9. Right? Yes. So 15 - 2 is 13 is greater than equal to 9. Yes. This part of the solution. So x should be less than equal to.5 and x should be greater than equal to 3. This is your answer. The three can now be multiplied inside the bracket as well. That is also one of the ways. Or you can keep it outside and square everything. That is another way. But the method everything is the same. It's still the same. So squaring on both sides, forget about the inequality. You will get 3 2x + 2x - 1 square greater is equals to x + 4. So this would be 9 and 4 x² - 4x + 1 = to x² + 8 x + 16 and then you

can work it out completely. The rest of it is the same. Let's see something name that we can everything is the same. Now we complete this one which I have given you right now and then try these two as well in front of me. Complete the one which I have started and then the next these two. Okay. 36 x² 36 x + 9 x² + 8 x + 16 x² - x 35 x² - x - Open pres. I think the answer at the back is is wrong. 36 x² 35 x² - 44x and then okay whatever the solution that you have what I think that is right x 5 and - 1 by 7 and 7 by 5 >> and then what would you do completed x is less than - 7 and greater and 7 by 5 to

inequality 3 modulus 2x -1 is greater than uh modulus x + 4 >> and then make the regions from the using the number line x less than 7 or greater than by five giving you the final answer after putting everything the values from the number line everything 7 by 5 what is the next step number line and number line is as a and now you have three regions region one two and three and then you will put in values from this region let's suppose This is minus one. You can take minus one from here. From here you can take zero. From here you can take two. And then you will input minus 1 0 and two in the question and satis and

find if which one satisfies. >> Okay. Okay. you will get 9 is greater than three. Yes, 9 is greater than three. This is part of the solution. >> H >> and then put in zero. Put in two. >> Oh, okay. Okay. Okay. >> 3 0 - 1 >> 0 + 4. So three is greater than four. No is not part of that means >> uh 4 - 1 >> 1 3 9 9 >> 9 is greater than 7. >> Yes. So your x is greater than 7 upon 5 and xative 1 upon 7 less than. >> Okay. M the next I'm stopping the recording.