MEMO 2026 MATHEMATIC P.1 GRADE 12 JUNE GAUTENG (THUNDEREDUC) GRADE 12 MATHEMATICS PAPER 1

Added: 2026-06-06

[00:00:03][music] [music] Good day everyone. Welcome to yet another episode where we shall be discussing the mathematics paper one for the grade 12s June 20 26. Okay. So according to the paper, okay, it was a standard paper with uh 3 hours of time and we had to get all the 150 marks.

[00:00:35]They're telling us we have 30 pages and an information page. Okay, so they decided to give us the grade one style of question paper whereby we write all our answers within this paper. Okay, so the paper is made up of nine questions and we need to answer all the questions within the spaces that have been provided. Mure the calculations diagrams are not necessarily to scale answers only will not guarantee full marks. A calculator approved may be used if necessary. We must round off answers two decimal places. Okay. Information page is indicated or included. No pages may be torn from the paper.

[00:01:15]All right. draw a neat line through any work that should not be marked and all of those instructions.

[00:01:23]Okay, question number one which is the algebra.

[00:01:27]So 1.1.1 they want us to solve for x. So in this case we have uh the standard form of the equation. We can just factor out what is common and in this case going to be x which means we're going to have 3x - 1 is = 0. It's either x is a zero or the bracket of 3x - 1 should be a zero. And if that's the case, then x becomes a 1 / 3. Okay. So that is 1.1.1.

[00:01:57]1.1.2 grade 12s. We have this quadratic equation and we have a condition of correct to two decimal places. So what we do? Remember we said we must always start by creating a standard equation.

[00:02:11]So that's going to be 2x^2 + 6 x. We're going to move the 6 to that side. So - 6 is a 0. If you want, you can divide through by the 2. So I'll have x^2 + 3x - 3 is a zero. Meaning that our a is a 1. Our b is a 3. Our c is a -3. So even if you don't divide through, but you go ahead and use the quadratic formula, you should arrive at the same answers. So b + -<unk> b ^ 2 - 4 a c / a. Let's substitute. So -3 + -<unk> 3^ 2 - 4. The a is 1 and the c is -3.

[00:02:56]And then we are dividing this by 2 * 1.

[00:03:00]So the first x value that I'm getting is a -3a 7 9. My second x value is a 0a 7 9. That is if I round off to two decimal places.

[00:03:15]All right. Then we move on to the next question. I hope we haven't made any mistake in this case. So we move on. 1.3 [snorts] which is the quadratic inequality. So still we need to form the standard equation. So -x^2 - 2x. Let's move the 8 to the left. It's going to be a + 8 is less or equal to 0.

[00:03:40]I'll choose to divide through by the negative. So I'll have x^2 + 2x. The 8 is going to become negative. And then the inequality symbol will also have to change to greater or equal to. Now in this quadratic equation or inequality, I have to factor out. Okay? If I use my factors for me to get a 8.

[00:04:03]Okay, I'll have a four and maybe a two.

[00:04:07]Since the 8 is negative and the two is positive, which means the four must be a negative and this should be a positive greater or equal to zero.

[00:04:18]So, does this make sense? It doesn't. I think the four must be positive and then the two must be negative. Okay. So meaning that our critical values x is going to be a two or x is going to be a -4. And if that is the case with that kind of scenario for us to represent our solutions I'm going to use this simplified version which means x needs to be greater than 2 or x needs to be less than -4. If I'm to use um a parabola which means -4 a positive2 we can sketch this parabola to be smiling.

[00:05:03]Okay, if I'm using the simplified version so where do we find it to be greater or above? So it's this area or that area. Alternatively, if people use the original format, okay, of what was given, which means your parabola is going to frown with a negative four as a root and a positive2 and then you shade off the bottom part.

[00:05:27]All right. So that is how the solutions for 1.1.3 were to be represented grid curves.

[00:05:36]All right. Then we move on to 1.1.4.

[00:05:39]we have 2x - 2x = 12. So with this kind of question, remember from the papers that we worked on the liopo and the northwest, we said you can let 2 to the x be a k. Okay, meaning that this 2 to the 2x is the same as 2 to the x but squared. So that will be like a k^ 2 - a k - 12 is equal to z. Okay, pull the 12 to that side. We get a standard equation. Let's factor this or you can use your quadratic formula. Okay, so that's going to be a k - 4 a k + 3.

[00:06:19]Okay, so the first bracket k is going to be a four. The second bracket k is going to be a -3. But the question is asking for the values of x. We are solving for x, not k. So replace k with a 2 to the x. So 2 to the x is going to be a 4 or 2 to the x is going to be a -3. And then we know that 4 can also be written as 2^ 2. So 2 to the x is going to be the same as 2 squared. Okay? Drop the bases meaning that x is a 2. Now, we said yesterday and the days before that during our exponential equations, it's very rare to raise a number to another number and end up with a negative answer. So, in this case, we're just going to say not applicable because the solution does not exist. And then 1.1.5, they gave you guys the simplest of the equation with SDS. Okay. So step number one is already sorted for us. The root has been isolated. Then step number two, we need to square both sides. So I'll square that. I'll square that. When I square the root sign 4, I'll have a 4x - 3 is = x^2. Okay? And then I'll create the standard equation. Move everything to one side. So x^2 - 4x + 3 is = 0. And then in the event of this I'll use my factors again. Someone will say why do you love your factors? Again factors I find them easier than the quadratic.

[00:07:57]Okay only use quadratic where I see I cannot get the factors. So if that is x and x we can have a three and a one since the sum is negative. So both brackets should be negative. The product is positive. Okay. So meaning that x is a three. Why am I writing four? So x is a three or x is a 1. Remember we do not stop there. We need to test our values. Okay.

[00:08:27]So if I'm to test the answers that I got to see whether the left hand side gives me the right hand side. If I use three, we know that 4 * 3 is 12. 12 - 3 is a 9.

[00:08:39]Square root of 9 is a 3. And our x was a 3. So 3 is a solution. If x is a 1, 4 * 1 is a 4 - 3 is a 1. 1 gives us a 1, which is the same value as x. So, which means both values do stand for this case. We do not cancel anything out.

[00:09:00]Then 1.2 1.2 simultaneous.

[00:09:04]Okay.

[00:09:12]Okay. So we have the quadratic equation which is uh we can say that is equation two and then we need to create equation one and simplify it. Okay. So from equation one we can see that 4 to the 2x + 1 is the same as 16. We can write 16 as a base of 4 just to make life easy for ourselves cuz 16 is the same as 4^ 2. Alternatively someone can change the 4 to a 2 and so on. So I'll just say 4^ 2. We are multiplying by y / 2. Okay, drop the bases. So 2x + 1 is equivalent to y. Call this equation 3. So we substituting equation three into equation number two. Okay. So where there's y, we put a 2x + 1. So it's going to be x^2 + a 2x into 2x + 1.

[00:10:04]We are saying let me say + one is equal to zero. Okay, I take this one that side.

[00:10:10]So x^2 + that's going to be 4 x^2 + a 2x + a 1 is equal to zero.

[00:10:21]Okay, we open up this and what do we end up with?

[00:10:28]There's something something I messed up.

[00:10:30]I'm changing signs for who? For what? I don't know.

[00:10:36]This is incorrect.

[00:10:38]Okay, our original sign was a minus, so it's a negative. So, it's going to be -4x^2 - 2x.

[00:10:47]Okay, does that make sense?

[00:10:55]I think it makes sense.

[00:10:57]It makes a lot of sense.

[00:11:03]Okay, so x^2 - 4x^2 gives us a -3x^2. We are subtracting a 2x. We are adding a 1 is equal to 0. Let's divide through by so it's 3x^2 + a 2x - 1 is = 0. I'll again go to the factors. Okay. Open my two brackets. That's a 3x and x. And this going to be a one and a one. Now since I want my sum to be positive, this will be a minus. This should be a plus is equal to z. So this gives me 3x - 1 is equal to 0 or x + 1 is a zero. So from the first equation x becomes 1 / 3 or x becomes -1. Great. We need to substitute the x back into the equation of y and see that y is going to be 2 * 1 / 3 + 1.

[00:11:59]Okay. 1 3 + 1. And then we get our answer for y which should be a 5 over 3.

[00:12:09][snorts] Okay. And then for this case my y is going to be 2 * -1 + 1. We're going to get a 1. Okay. So y is a 5 / 3 or y is a 1. x is a 1 / 3 or x is a1.

[00:12:25]And then 1.1.3.

[00:12:27]We last saw this questions in grade 11 for these exponents.

[00:12:34]Okay, let's see. So base 2 base 2 this is a base of four which means we need to create a base two there. So it's going to be the fourth root which is going to be 2^ 2 1 3 1 3 when you multiply it you're going to get this which is 20 26.

[00:12:54]We are subtracting 2 2025 over 2 to the 207.

[00:13:01]Okay. So please do not be tempted to subtract that and you get a one or just to know these are exponents for as long as there's a sign of a plus or minus.

[00:13:10]You have to do the factorization. So the quickest way to do is let the denominator lead you. Okay. The our denominator is a 2 to the 2017 which means we need to factoriize the 2017 from the top part. Okay, so if I remove 17, I'm only going to be left with two to the exponent of 9. Okay, because I know that 9 + 17 gives me a 26. And then I'm subtracting 25 - 17 is going to give me 8, which is 2 to the 8. Okay, but I'm dividing by 2 to the 2017.

[00:13:48]Everything is under the fourth root.

[00:13:52]Okay. Then what I notice that this 2017 cancels the 2017. I'm left with um [clears throat] the fourth root of 2 to the 9 - 2 8. Let's factoriize out again.

[00:14:04]I'll take out 2 to the 8. I'll be left with a 2 - 1. Everything is to the 4th root. 2 - 1 gives me just a one. So we are left with just two to the exponent of 8. And we know that the fourth root is the same as 1 / 4 exponent. Then we multiply these two which will give us just a 2 squared and 2 gives us just a four as our answer. All right. Then 1.4 determine the value of p for which the roots of this equation are real. Okay.

[00:14:42]So we know that when roots are real, the discriminant needs to be greater or equal to zero.

[00:14:50]Okay.

[00:14:51]So what we going to do? We're going to say that our x² let's move the px that side minus px. Let's move the 1 + 1 is equivalent to zero. Our discriminant is given as the b ^ 2 - 4 a c. Our b is a p. Our a is a 1. Our c is a 1. Let's substitute. Okay, we are saying that p^ 2 - 4. Okay, we say the a is 1, the c is 1 should be greater than zero. So p² gives us just a p ^ 2. Okay, this going to be min -4 is greater than zero.

[00:15:31]That's a difference of two squares.

[00:15:33]Meaning that we need to factoriize p - 2 p + 2.

[00:15:40]Okay. So meaning that we have a p minus 2 and the p + 2 is greater than zero. We can put the equal sign because we say greater or equal to. Okay. No stress.

[00:15:50]Then um we need to get the critical values here.

[00:15:54]Okay. Our critical values p should be a 2 or p must be a -2. Now since our inequality is greater meaning that p needs to be less than two or p needs to be greater than two. So I don't know why the question of the question said values of P. So which means it makes sense with our solutions. All right then 1 four is done and that is the end of the 34 marks. Okay. These were all attainable for this question one. Question number two they telling us that the first three terms of a quadratic are given below.

[00:16:32]They want us to get the next two terms.

[00:16:35]Okay. So, -12, -15, -16. What are they doing here? They are subtracting a three. Okay. What are they doing here?

[00:16:45]They are subtracting a one. Okay. And then when I look at the second difference from -3 to this, they adding a two. Which means if I add a two, I'm going to get um a 1. And then 16 + 1 gives me a -15.

[00:17:04]Okay? If I plus a 2, that's going to be a 3. And then -15 + 3 gives me -12. So -12 and -15 are the next two terms that I am obtaining.

[00:17:19]All right. Then uh we look at the next question. 2.2.

[00:17:25]2.2.

[00:17:28]They're asking us to determine the nth term of the pattern and write the answer in the form of tn is n squ and all of that. So what we have to do is we know that this is equated to the 2 a. The -3 is the 3 a + b [snorts] and then our -12 will be equated to the a + b + c. Okay, let's work it out and see 2 a is 2, which means a is a 1. 3 a + b should be equated to -3. But if our a is a 1.

[00:18:02]Okay, meaning that we're going to move the three to that side and b is going to be a -6. Then a + b + c should give us a -12. a is a 1. B is a -6 + c is a -12.

[00:18:18]This becomes a -5 + c is a -12. Move the five that side going to become a a positive. Okay? and then we end up with a -7. So our tn okay is going to be n^ 2 - 6 n - 7.

[00:18:34]Okay, that is what I'm getting. Remember this is not the official memorandum.

[00:18:40]It's just the proposed memorandum. We are discussing to see how the learners found the paper. All right. 2.3 determine which terms of the pattern are positive. Okay. So for terms to be positive it simply means that our tn must be greater than zero. Okay that's according to my understanding. Okay. So if I get the n^ 2 - 6 n - 7 and I include the inequality of greater than zero. If I'm to solve this with my factors again I'll have n - 7 n + 1 is greater than zero. our critical values our critical values will be n being a seven or n being a negative one. So when I represent the solutions, we know since we're talking about the terms, okay, and we are looking for the position of the terms, terms should never be negative, okay? So this -1 part falls away, okay?

[00:19:47]But the solutions were supposed to be n is less than -1 or n is greater than seven. So this is not applicable because the num the positions can never be negative. So which terms are positive?

[00:20:02]So terms that are greater than seven. So from the eighth term, the 9th, the 10th and so on and so forth. All right. Then 2.4. Which two consecutive terms of the quadratic have a first difference of 1 13th? Now we know that in a quadratic pattern the [clears throat] first difference always forms an arithmetic.

[00:20:26]Okay. So if I have -3, -1 and 2, I think that's enough for me to get the rule.

[00:20:31]Okay, we are saying -3, -1, this gives us a 2. So -3 is our a and then 2 is going to be our d. So from the tn or the general term of an arithmetic a nus1 into d is going to be -3 + n -1 * 2.

[00:20:53]When I open this up, I'll have 2 n - 2, but then there's a -3, which is basically 2 n - 5. So, this is the general term. Now, we are looking for the position, meaning that I'll replace tn with 113 that's a 2 n - 5. I'll move the -5 to that side. Will be 118 is equivalent to n. Divide both sides by two. Okay? Meaning that our n is going to be what is 1 a / 2? I'm getting a 59.

[00:21:28]Okay. So meaning that our solution should be between t59 and t60.

[00:21:36]Okay. So between T59 and T60 of the quadratic because remember our first term of the of the the first difference the first term of the first difference is between T1 and T2 meaning that if we look at the 59th term is going to be between T59 and T60.

[00:22:00]All right. So that is that is that is with question number two which I believe was also a fair question. Okay, 14 marks in the bag. We go to question number three. We have an arithmetic. Okay, we have 5 9 13 and so on up to 1,01.

[00:22:20]Okay, determine the number of terms in the sequence. So we know that this 1,1 is the last term which can also be represented as tn. So that we look for n. So t n is a + n -1 into d. Okay, we can do the substitution where there's tn I'll put 1 0 1. Our first term is a five plus the n minus one. Our d which is the common difference is basically a four.

[00:22:49]So 5 + 4 is a 9. So I'll put a four there and then I'll open up. So 1 0 1 is equivalent 5 + 4 * n is 4 n. And then 4 * 1 is a 4. Okay, let's play around with this now. So 1 0 1 is equivalent to this and that is a 1 + 4 n. I'll move the one that side which means going to be just 1,000 is equivalent to 4 n. Divide both sides by 4. So 1,00ide by 4 I'm getting a 250.

[00:23:23]So we have 250 terms in this sequence. All right. Then 3.2 they want us to prove that for any arithmetic sequence where the first term is a and the common difference is d that the sum is given by this formula. Wow.

[00:23:43]Okay. So I know majority of us might have forgotten on how to work on this. But what we do or what we say is that um if our sn okay if we are given the sum of the sequence. So the first term is a our second term will be a + d. The third term will be a + 2d.

[00:24:05]Okay. And so on and so forth. I can look at the last term. My last term is going to be l. Okay. Which l is basically a 2 a plus no no it's not a 2 a. My last term should be the nth term. Okay. A + n minus one into d. Okay. So I'll say that this is l - 2d. The second last will be l + d. And then the last is l. This is a minus.

[00:24:35]Why am I putting a plus? Okay. Then uh if this is equation one, I just need to flip this around. Okay. start with what was last and then I'll end with what was first. So that's going to be L plus this is going to be L - D + L - 2D D plus whatever is there. And then this going to be A + 2D a + D and then lastly it's going to be A. That is our equation number two. Okay. Then I need to plus the two equations. Meaning that SN and SN I'll get twice Sn. And then A + L I'm going to get A + L. A + L. a + l and a + l a number of times. Okay. Now for n terms, okay, I'll multiply this by n.

[00:25:25]Meaning that twice sn is the same as n into a + l. Divide both sides by 2.

[00:25:32]Meaning that it's n / 2 into a + l. But remember, we say that our l is a + n -1 into d. So it's n / 2 into a + a + n -1 into d then a + a gives us the 2 a.

[00:25:52]Okay. So that is how we would have claimed the formax.

[00:25:58]All right. Then 3.3 which is the sigma notation evaluate the following. Okay. 3 P min - 2 since the P is just a product of three or P is a product with three what am I saying okay whatever it is but this looks to be linear meaning that it's more of an arithmetic okay but just to you know not take chances we can start by substituting so if our P is -1 so 3 * -1 - 2 is going to give us um a5 Okay. And then what if P is a zero?

[00:26:37]Because from -1 we go to zero. That's going to be -2. So -5, -2. Let's try one.

[00:26:45]Okay, that's going to be a one. Okay, let's see if this makes sense with what we are looking at. So meaning that you're plusing a three. You plus a three. Okay, so our common difference is a three, which is this three here. All right. and the first term is -5.

[00:27:04]Okay. So what do they want us to do?

[00:27:06]They want us to find the sum okay between -1 and 19. So meaning for us to get the number of terms the last value minus the first one and then you plus one. So that's going to be equivalent to 21 terms.

[00:27:23]Okay. So we have the number of terms, we have our a and then we have our d. Then go to the sum of a the sum of an arithmetic. So n / 2 into should we use the last term? I think we can whereby p is 19. So it's 3 * 19 - 2.

[00:27:45]Let's just confirm that 3 * 19 - 2 gives us 55.

[00:27:52]Okay. So we can just use what we just derived. So it's going to be 21 / 2 into -5 + 55.

[00:28:01]So 21 / 2 * 50 which will be 21 * 25 which gives me 525.

[00:28:10]Okay. But even if you use the long format of the 2 a + n - 1 into d you should still get the same answer. All right. So that is how we would have claimed the five marks 3.4. form.

[00:28:26]What a question. Okay, let's see. A metal road of 15 9375 m is cut into 80 pieces. Okay. Whose lengths form a geometric sequence and then the pieces are then used to support the structure.

[00:28:39]The pieces are welded vertically starting with the longest piece to the two support metal beams in the diagram below. The length of the longest is 128 times the length of the shortest.

[00:28:54]Okay. So if the length of the shortest piece which is the eighth piece is x.

[00:29:00]Okay. So if that is x meaning that the first one should be8 128 * x 3.4.1 they're asking us to write the length of the longest. So it's going to be 128 x m.

[00:29:14]Okay. then determine the common ratio of the sequence that is formed by the pieces of the road. Now since this is a geometric okay if we take this to be our a and we know that the t8 is x we can use the general term of an arithmetic so mean that tn is a r n minus one okay great welves let's see so our tn which is the eighth term okay so I'll say that the eighth term is given as a r to the exponent of 7. Let's substitute our eighth term is x. Our a, we said it's 128x * r to the exponent of 7. Let's divide both sides by 128x so that we isolate the r. Meaning that r 7 is going to be x over 128x.

[00:30:09]Okay, cancel out the x and the x shall be left with a one. So r to the 7 is 1 / 128. and let's find the 7th root.

[00:30:21]Okay, meaning that r is going to be 1 / 2. Alternatively, what somebody would have also done is for you to look at um we said r 7 is equivalent to 1 / 128.

[00:30:39]And then uh you can express this 128 as a fraction sorry as an exponential. So r 7 should be a 2 to the 7.

[00:30:51]Okay. And then when you take out the r to the 7 is the same as 2 to the -1 with a 7 which means that r is 2 to the -1 which is a half. Okay. Which is also 0 5. So our common ratio is 0a 5. And then hence meaning that from where we stop they want us to get the length of the longest piece which is the 128x. For us to get that length we need to find the value of x. So that's what the question is indirectly asking us to do first.

[00:31:26]Okay. So what I will do [clears throat] since I have the common ratio I'll try to look at the 15 7 what what and that was like the sum okay meaning that if we add up the length of all these pieces we're going to get a 15a 9375 okay so I'll say if our sn okay okay s8 is 15a 9375 the n is 8. So the formula is given as a into 1 - r to the exponent of n 1 - r.

[00:32:04]When we do the substitutions 15, 9375 is equivalent to our first term. We said it's 1.8x into 1us. Our common ratio is a half or we can just say 0a 5 to the exponent of 8.

[00:32:21]Okay. All of this I'm dividing by 1 minus 0a 5. [snorts] Okay. Then um I just need to simplify this. So I'm going to multiply both sides by this because that's going to be a 0a 5. So that's going to be a 225 over 32 is equivalent 128x into when I simplify this in the calculator I'm getting 225 over 256.

[00:32:51]Okay. And then I need to divide this by this. So 225 over 32. We are saying divided by 225 / 256 should give us 128x.

[00:33:05]Simplifying this I'm ending up with just 8 is equivalent 128x. So divide by 128ide by 128. This cancels that.

[00:33:16]meaning that our x okay our x is 1 / 16.

[00:33:22]What was the question asking for?

[00:33:24]Determine the length of the longest piece. So the longest piece which is the first term. Okay, it's going to be 1.8 * 1 / 16 and this gives us 8 m. All right, so that is how I would have done it. I mean it looks lengthy but if you have a short time of doing it you are welcome to share with us. So that is question number three 20 marks was also achievable not bad. Okay in as much as yes this it could have been challenging somewhere somehow but still doable.

[00:33:58]Question number five which is the functions we are given f of x to the 1 / 3 to the exponent of x. They want us to get the inverse. So the inverse will just be the log with the same base.

[00:34:10]Okay, of 1 / 3 exponent x. So that is the inverse. Alternatively, someone will use okay because if they change this to a 3 to thex, it would be a negative log x base 3. Okay, in that format, but whichever format you use, we still appreciate it. Then they want us to sketch both f and f inverse on the same set of axis, we must at least indicate one point. Now with exponentials we know that the form is a to the x. If the a [clears throat] okay if a is greater than one then the graph will be increasing in that sort of direction. But our a is a fraction which means the graph should be decreasing.

[00:34:54]So if x is a zero from this point the y is going to be a one. Okay. So if I make this a one, it simply means that my graph should be moving in this sort of direction towards the x-axis.

[00:35:10]Okay, so this is f. And then if I choose a point, let me choose a point like um one.

[00:35:21]If I use one, that's going to be 1 / 3.

[00:35:25]No, let me choose a point of negative.

[00:35:27]Let me choose a negative point. maybe -1.

[00:35:30]So 1 / 3 to the -1 gives us a 3. So this point here is going to be -1 and a three. Okay, that is my point of choice.

[00:35:41]It could have been any other point.

[00:35:42]Okay, someone would have used one and a third. No stress. Then for us to draw the inverse, remember it's all about switching points. Okay, what was a one on the y must become a one on the x.

[00:35:56]Okay. And then this point of -13 should be a 3 and a -1 which is somewhere there. And then you just have to sketch this towards the yaxis because that is going to be our new asytote.

[00:36:12]Okay. So this is going to be our inverse of f. They said indicate this point is going to be a 3 and1. So depending on the points that you used all you have to do is switch them around. So that's the inverse together with the original graph 6 marks. All right. Then 4.3 determine the values of x for which okay the inverse is greater than -3.

[00:36:40]So meaning that if there's a -3 somewhere here and this graph is just continuing.

[00:36:49]Okay. If this is a -3, they want us to get the value of x.

[00:36:54]Okay, values of x where the graph is above. Meaning that it will only be above. Okay, from this point up to this point, that's where the graph is above the -3 line. So for me to get a -3, okay, you're saying that our y is a -3.

[00:37:10]Let's substitute this -3 into the original f ofx. So 1 / 3 to the -3 gives me a 27. Okay, meaning that if x is a -3, y is going to be a 27 for the original graph. Okay, so -3 and the 27 is somewhere then meaning that if I switch these two points, then it's going to be a no, it's going to be a positive 27 and a -3. So for which values of x is the graph above? So between 0 and 27.

[00:37:47]Okay. So I'll say 0 is less x is less 27.

[00:37:53]Okay. So that is how I would have approached this question as well. So 10 marks pretty much achievable. Okay. We can't complain. Question number five. We have three graphs. We have a parabola.

[00:38:09]We have a straight line. And we also have another straight line. The graph cuts Q and T. And then the graphs G and H meet at point R just as we can see here. And then um the X intercepts of G and H and S and Q respectively. Okay, we can see this. We can see that. And then P is the turning point. Thank you. First things first, they want us to show with necessary calculations that the graphs J H are perpendicular to each other. Now from our paper two we know that lines or graphs are perpendicular when the gradients or the product of the gradients gives us 1. But from this equation of g we know that a straight line equation is always mx + c. So the first number or the coefficient of x is always our gradient. So the gradient of g is a half and then the gradient of h is -2. If I multiply these two a half * -2, I'm getting -1. There are four g of x is perpendicular to h of x. That's all we have to do. All right. And then can we determine the coordinates of r? Where is r? r is a point where the two graphs are meeting. So whenever graphs are meeting at that intersection, the two graphs have to be equal to each other.

[00:39:35]So I'll say that um our g of x separating the two. So simplifying the left hand side I'm going to get a 5 / 2x to the right I'm going to get a -10.

[00:39:48]Multiply both sides by 2 5x is a -20ide by 5 it means that our x is a4.

[00:39:57]So we can see even from the graph that they're meeting on the negative side. So if that's a -4, we need to find the y value by substituting this -4 into any of the equations. Okay? So y is going to be -2, -4 -4.

[00:40:14]That's going to be 8 -4, which gives us a -6. So it simply means that r is going to be a -4 and a -6.

[00:40:26]All right. Then 5.3, they want us to get the length of qt. So from here up to here and we are seeing that the graph that is passing through both Q and T is the parabola and it is passing through along the X-axis. So meaning that these are X intercepts where the Y must be a zero. So I'm going to get the equation of the parabola which is -x^2 - 6x + 7 and then I'll equate it to zero.

[00:40:57]Okay. -6 + 7. Okay. So divide through by that's going to be x^2 + 6 x - 7 is 0. I need to factor again or quadratic formula.

[00:41:11]That's going to be x.

[00:41:13]That's x. This is going to be a 7 - 7 which gives us 8 unit. And then uh 5.4.

[00:41:23]They're telling us that wx is an equation.

[00:41:27]Okay, the equation of a line that is obtained by shifting the graph g perpendicularly along h. What does this mean? If this a 90 even after shifting the 90 must be maintained, okay, up until it reaches point q. So if we have another graph, okay, the one that we are calling w, this w should be a 90. What do we know about 90 and 90? They are corresponding. And what does that [clears throat] tell us about the line WQ and RS? They are parallel. And when the lines are parallel, it simply means that the gradients are the same. So this was a half of X - 4.

[00:42:11]Okay, if I'm not mistaken, which means the gradient here is also going to be a half. But we said that the coordinates of Q are -7 and 0. What is the question asking for? They want the y intercept of W. So where is the graph meeting W? Okay. So what I'll do for two marks I think there's a quicker way but I don't know if I'm trying to get the equation of W, it will be a half of X + C. They want me to get C. Let's use the point -7 and 0. So 0 is going to be a half into -7 meaning that our c is going to be a positive 7 / 2. So that is the y intercept.

[00:43:00]Then um 5.4.2 if w meets the line of symmetry of f at a point m that is not shown they want us to get the distance m for four marks.

[00:43:13]Okay, they're meeting at M. Then they want us to get the length of P M. So first things first, we need to first of all determine what X values here at the turning point. And if that is possible, then we can substitute into the difference between the two graphs or you can get the value of Y at P the value of Y at M and then we subtract. So from the equation of -x^2 - 6 + 7, you can get your turning point. normally b / 2 a. So it's a negative with a -6 / 2 * -1 which gives us um a -3. Alternatively get the -7 you plus the one you divide by 2 you're still going to get a3.

[00:44:02]Now for me to get my p okay no I don't want to do that it's going to take time. Let's say what is f of -3. So if x is -3, it's going to be with a -3^ 2 - 6 into -3 + 7. What does this give me? It gives me a good 16 cuz that's 18. 25 - 9 is a 16. And then let's look at w x is a -3. Remember we said it's a half into -3. And then we said it's + 7 / 2.

[00:44:42]Okay, minus that and that it's giving me just a two. So meaning that the y value here is a 2. The y value here is a 16.

[00:44:52]So meaning that our PM will be 16 - 2 which gives us 14 units. Alternatively, subtract the two equations. -x^2 - 6 x + 7. You're subtracting that straight line which is a half x you said + 7 / 2 and then replace x with a -3. You should be able to get the 14.

[00:45:16]Okay. I bet to be corrected in that regard.

[00:45:21]All right. So then we move on to 5 4.2.

[00:45:27]No, that is what we answered. Okay. We say that is 14 units. All right. Then um that brings us the end of question five.

[00:45:35]Not bad. Okay. I prophesy that there are learners that are going to get the total in this paper the way things are going just with the five questions that we've done. [clears throat] Okay. I can smell 150 out of 150. All right. Question number six. F of x is x - 2x - 15. Jo of x is the hyperola with those unknowns.

[00:45:57]The symptoms are meeting at this meaning that this one will somehow relate to the P. The -2 relate to the Q. Okay, AB intersects passing through. But we only know the graph of F. So let's use F and we know that on the X-axis Y is a zero.

[00:46:14]So we're going to have the X² is the turning point.

[00:46:19]Okay. And the turning point, okay? And uh the x coordinate of the turning point they're telling us it's a one cuz the question told us that one and two graph f x ais blah blah blah blah. And then um did they me they mentioned something about the turning point. Okay. Turning point of g of a graph f lies in the vertical simp lies in the vertical simp.

[00:46:50]Okay. So what we can do we can substitute and say what is our f of 1.

[00:46:55]Okay so remember we have a 1^ 2 - 2 * 1 - 15. Okay that's going to be -7 + 1 which is a -6. So meaning that this is a 1 and -16. Now when you talk of range you're dealing with the y values. So this graph can only be valid from -16 upwards. So we're going to say that y should be greater or equal to 16. All right. And then 6.4 consider the graph of the hyperola. They want us to write down the values of P and Q. So we know that the Q is already -2. Okay.

[00:47:38]From the asympto here. And then the P remember the sign must always change. So this P I'm going to make it a1.

[00:47:47]Okay. So I'm going to say P is a1, Q is a -2. And then can we show that A is 8. So we are saying Y is A over X -1 - 2. We need one point where the graph of G is passing and that is the 5 and Z.

[00:48:09]Okay, I think that's the best point we can ever use. Five and Z.

[00:48:13]Okay, I think that's the best point we can ever use. So 5 and 0 I'll say 0 is equal to a over 5 - 1 - 2. I'll move the two that side meaning that 2 is a / 4.

[00:48:29]Multiply both sides by 4 meaning that a is going to be 8.

[00:48:34]All right. Then 6.4.3. Hence determine the equation of axis of symmetry of g with a negative gradient. So that format is -x + c. We must use the coordinates of the asymptos. Let's do the substitution. So where there is y is a -2. Then with a 1 + c. You move that one that side is going to be a + 1 + c. So meaning that is a1. So the equation is -x -1. There are people that have that easier way of shifting based [clears throat] on the transformation that is happening with the coordinates because it is a y= x but then you're moving the y by one unit down so it's a -1. Okay. Then um next we have 6.5.

[00:49:29]Okay. They're telling us that if it's further given that the points C and F are -2A 41 and 0a 41 respectively.

[00:49:38]Determine the values of X for which G of X - F of X is greater than zero. So the point of intersection here, okay, they're telling us the X here is a 4A is it a four? No. 2A 4 1. Okay. And then this one is a 0 comma 0 or one positive. And if that is the case, where do we find okay? Let's restructure this.

[00:50:07]Let me move the f of x that side. So g of x is greater than f of x. So where do we find g to be above f. Okay. So g can only be above f in this region here.

[00:50:20]Okay. That's why we are seeing the hyperola to be above. And then also when you look at this area here okay the graph of g is just above it's above f. So what I'm going to do I'm going to start by writing down the values here. So -2a 41 is less x is less 0a 41 and then also when you look at between their one and then the value of b.

[00:50:52]So I'll say one is less x but it's also less five. So these are the two regions where g will be above. Remember g is the hyperola. All right. Then uh question 6.6 6.6 they're telling us that h of x okay h of x is -2x + k is a straight line that is drawn on the same set of axis as f and g. Determine the values of K for which the graphs G and H will not meet. Okay.

[00:51:29]So if we are to look at the two graphs, we're going to start by equating the two graphs and find or use the nature of roots to find a situation where the roots are nonreal. Meaning that our discriminant must be less than zero.

[00:51:45]Okay. So we are saying g of x should be equated to h of x. Our h is already -2x + k. What is our g?

[00:51:58]Okay, so from the previous sections, we say that g is the hyperola if I'm not mistaken. Yes. So it's going to be 8.

[00:52:09]Okay, it's going to be 8 over x -1 but minus a 2.

[00:52:18]So step number one, I can multiply through by the xus one. So I'll be left with 8 2 into x -1 is equivalent to -2x into x -1 + k into x -1. I'm trying to get rid of the denominator. Okay, let me try to open up and see. 8 - 2x that's going to be a + 2. This going to be a -2x^2 + a 2x.

[00:52:50]This is going to be a kx - k. Let me regroup. Move everything to the left. I'll have a positive 2x^2.

[00:53:01]When this one moves that side, I'll have a -4x.

[00:53:06]The k will be a - kx. This k will be a + k. And then I have 8 + 2 which gives me a 10. This should give me a zero. Okay.

[00:53:21]And then uh what do I see? I can combine these two since they have x and x. So it's going to be 2x^2.

[00:53:29]I'll say minus x into why am I doing that? Let me just put the bracket there first.

[00:53:41]Okay. This is going to be 4 + k into x and then a k + 10. So from this I know that my a is a two. The all of this will be our b and then the all of that will be our c. So let's substitute into the discriminant and say that b ^ 2 - 4 a c should be less than zero. So that's going to be negative with a 4 + k we are squaring this. This is four the a is a 2 and then this is a k + 10 is less than zero.

[00:54:19]What question? How many marks? Five. All right. But we know that anything negative is when you square it there will be a positive. So that's going to be 16 + 8 k + k 2 -4 and 2 is -8. So I'll say 8k and then this is -80 is less than zero. Let's simplify that.

[00:54:46]Let us simplify that from what I'm seeing.

[00:54:53]Okay. So what I see is that um the 8k and the 8k will cancel. We are only left with a k squ 2 16 and -8 gives us -64.

[00:55:06]Okay, this is a difference of two squares. It's going to be k minus 8 or k + 8 is less than zero. So our critical values will be k being 8 or k being -8.

[00:55:20]Meaning that our solution should be 8 is less k and is less 8. All right. So that gives me the answer for question number 6.6. I believe it was fair enough. Not so bad. Question number seven. Calculus. Calculus. Okay. They want us to find the frime of x which is the first derivative from the first principles. So remember the formula is limit as h ts to 0 f x + h minus f x everything is divided by h. So we have our f of x. What is missing is the f x + h meaning it's 4 into x + h 2 - 3 into x + h. Let's do the substitutions. So limit as h ts to 0 4 into x + h 2 - 3 into x + h you're dividing by 3 limit as h ts to zero we open up I'll keep the four open this bracket it's going to be x^ 2 xh h 2 open this - 3x - 3 h everything is over h so from our previous discussions remember we say that always the f of x must fall away.

[00:56:39]So it's 4 x^2 + 8 xh + 4 h 2 - 3x - 3 h.

[00:56:50]There's something I'm missing. There's something I'm missing. Okay. And that's a mistake that I thought I was going to make. I did not subtract the f of x.

[00:57:00]Okay. So our f of x is 4x^2 - 3x. You subtract 4x^2 + 3x - 4x^2 + 3x and then we are dividing by h. Okay, cancel what needs to be cancelled. So this um 4x^2 cancels the 4x 2 the - 3x and the positive 3x cancels. We are left with this. we need to factor out. Okay, so it's going to be h that's 8 x + 4 h - 3 / h and then you substitute h with a zero meaning that you have 8 x - 3 as our answer. All right, so apologies about the f of x that was left. I usually do that but I always make it a habit to go back. Then 7.2 determine the one the dx. So in this case factoriize first. Okay. So we know that x^2 - 9 is the same as x - 3 and then x + 3.

[00:58:01]Everything is over 3 - x. But we're talking of the dx. One thing we know is that x - 3 and 3 - x are not exactly the same, but we can create similarity if we introduce a negative. So if I have a negative by the top then I'll have 3 - x and then x + 3 over the bottom part is just 3 - x. Okay, we are still talking about dx. So this cancels that we left with a negative with that. So dx into -x - 3. So the derivative of that is just going to be a 1. The constant will die out. All right. So that is 7.2.

[00:58:46]Okay, we move on to 7.3. Given that y is this, can we show that dy dx is 4 roo<unk> of y. So before we get the the solution, let's try to make 2x - 3 the subject.

[00:59:07]Okay, meaning that that is going to be you take square roots on either sides.

[00:59:10]So this is going to be the root of y.

[00:59:13]Let's keep this in mind. Okay, let's open this because we need the dy dx. So, it's going to be 2x - 3 and then 2x - 3.

[00:59:22]So, y is going to be 2x and 2x. That's going to be 4x 2. 2x and 3 is going to be -6x - 6 x + 9. y is going to be 4x^2 - 12x + 9. And then the dy dx in this case is going to be 8x - 12. Okay. But the question wants us to express it in this form. So let's take out what is common.

[00:59:49]That is a four. We shall be left with a 2x - 3. But what did we say? We say 2x - 3 is root of y. So meaning that it is 4 * the root of y. Okay. So that is how we would have obtained that. Then now question number eight, we have a cubic function. Okay. Determine the coordinates of the intercepts with the axis with the axis. So we can start with the y intercept. Okay. Y intercept x should be a zero. And we know that the number the constant will always be the y intercept. So it's going to be 0 and four. And then for x intercept, okay, y is a zero. I'm going to equate this to zero. So x² x cub - okay minus 3x^2 + 4 is equivalent to zero. Now since we don't have any idea of what number we can substitute here cuz we always start with a try and error kind of format. So if we substitute x with a one, we're going to get 1 - 3 is a -2 + 4 doesn't give us a zero. Let's try two. 2 cub is 8 - 2 2 is 4. 4 * 3 is 12 + 4. So 4 + 4 is 12 - 12 which is a 0. Which means the correct solution is two. If x is a two, that's a root. Then I'm going to use this two and the synthetic. Okay? I always use what works best for me. Okay? People can still use the the long division. Okay? cuz they find pleasure and comfort in working with that. Coefficient of x cub is 1. x^2 is -3. We don't have x so it will be a zero. The constant is a four. Okay. Our first root is a two.

[01:01:43]Okay. And then we said we shade this off and then we write this down.

[01:01:50]Okay. And then from there x cubed.

[01:01:53]Okay.

[01:01:55]It was x cub let me just get the equation back. - 3x^2 + 4. So our first derivative is going to be 3x^2 - 6x. We must equate this to zero. Take out what is common which is 3x. That's going to be x - 2 is a zero. So the first part x is a zero or x is a 2. What was the question asking for? They want the coordinates, which means you need to find the y. Okay, so let's substitute zero into the original f. Okay, 0 0. I'm going to have a four. Okay, and then if I substitute x with a two, we say this thing gives us a zero. So the coordinates of the turning point will be 0 and four. And then we have 2 and 0.

[01:02:50]And then 8.3 this question was sweet guys I mean was systematic step by step and then they ask you to draw the graph these are sweet marks so let's indicate where the graph is cutting the intercepts okay but before that our a is positive now meaning that our graph should be in this sort of format okay so the intercepts we got was 2 and one positive one. Okay.

[01:03:25]Is that is it a positive one? Really can't be a positive one guys because this was a positive one which means a negative. Okay. So it's a negative one not a positive. So it's one.

[01:03:39]Okay. And then we have a positive2 which can be somewhere there. It's fine.

[01:03:45]Remember it's a sketch. Okay. So we got the y intercept as 0 and 4.

[01:03:52]Okay, 0 and four can be somewhere there.

[01:03:56]Okay, and then what did we find with the turning points? We found 0 and four again. So this is a turning point. Okay.

[01:04:04]And then 2 and 0 is also a turning point. So since we say the shape should be in that format, which means this is our maxima and this is our minima. Okay.

[01:04:13]So just connect those points and we get the neat graph.

[01:04:19]Okay, something like that.

[01:04:24]Okay, so that becomes our f. And then 8.4 for which values of x will the graph be concave up? So with concave up, we must find the second derivative of x to be greater than zero. But our first derivative remember was a 3x^2 - 6x 3x^2.

[01:04:48]Okay, we said 3x^2 - 6x. So our second derivative in this case is going to be 6x - 6 is greater than zero. So meaning that uh let's try to solve move the six divide by 6 meaning that x should be greater than one. All right then which is question number nine.

[01:05:12]A school fundraising committee is planning a metric dance for the grade four of learners. The weekly profit depends on the number of teachers involved in the activities. Relationship between the prophet and the number of teachers