GRADE 12 MATHEMATICS JUNE EXAM 2026: MATHEMATICS PAPER 1 GR12 JUNE MATHS P1MEMO: GAUTENG JUNE

Añadido: 2026-06-07

[00:00:03]Hello great welcome to this session where we shall be discussing the proposed memorandum okay for the grade 12 mathematics paper one for the health province okay so this paper was written today and the memo is yet to come out but in the meantime we can together go through it and compare and see what exactly was expected of the learners Okay. So, time is 3 hours. The max basically 150. We have 30 pages plus an additional page. Okay. The information one. Okay. Instructions are we have nine questions that we need to work with. And from the nine, we need to answer all of them. Okay. In the spaces that have been provided clearly show the calculations, diagrams, graphs, etc. that you have used in determining your answers. Answers only will not necessarily be awarded marks. You may use the unapproved calculator that's nonprogrammable unless stated otherwise.

[00:01:02]Okay, we're going to round off to decimals unless if we told not to.

[00:01:07]Diagrams are not always to scale.

[00:01:09]There's an information page. No pages may be torn out of this. Okay.

[00:01:14]Candidates are supposed to retain the question paper or remove it. The examination group may not.

[00:01:21]Okay. Because this is your answer book as well. Okay. Question papers must be returned to me later. Okay. Answers only written in black and blue. Do not write in margins. Okay. Indicate the questions that you've answered. Write in it line the work that you don't want to be marked. The event that you have you need an additional one. You write the number of questions. Okay. All right. So those are the instructions. There are quite a lot. But question number one, let's see how did we find this paper and were examiners really fair to the learners or they were cruel. Okay. So based on this let's start with the discussion. Let's dive into question number one. Solve for x which is the algebra. So with this 3x^2 - x =0 it's already a standard equation. The only thing we have to do is factor out. So take the three. No we're going to take out the x. Not the 3x. Okay take out the x. We're going to have 3x - 1 is = 0. And then it's either x is a zero or the 3x - 1 is a zero. So our first solution is there. Then from this we're going to move the one to that side. 3x is a 1. Divide both sides by 3.

[00:02:36]x becomes 1 / 3. Okay. Then 1.1.2 2x^2 + 6x = 6. Correct two decimal places. So remember this condition is a hint to say that we're going to use the quadratic formula. But before we start with the quadratic formula, we need to create the standard equation. Meaning that's going to be 2x^2 + 6 x - 6 = 0. So our a is a 2, the b is a 6 and then the c is -6.

[00:03:09]from the quadratic formula b + -<unk> b^ 2 - 4 a c / a okay so our b we need to do the substitution it's going to be 6 + -<unk> 6 2 - 4 the a is a 2 the c is a -6 we are dividing this by 2 * 2 now someone will ask is it okay if I start by dividing through by the two it's very much allowed. Okay, start by dividing by two you get x² + 3x - 3 that is okay.

[00:03:51]Okay. So from uh what we have okay if we substitute this into the calculator let's see calculator where are you? My calculator is misbehaving now.

[00:04:06]Calculator is misbehaving.

[00:04:18]Let's try it out again.

[00:04:25]Okay, we saying fraction so it's a -6 plus square root of um saying 6 2 - 4 * 2 and then times -6 we close the bracket we divide by 2 * 2 so with a plus I'm getting 0 7 9.

[00:04:52]Okay.

[00:04:54]And then let's see with a minus. Okay.

[00:04:57]We said we can use this. Delete that.

[00:05:01]Put a minus. We get a negative answer which is -3, 79. So -3, 7 9 becomes our answer. Okay. Then 1.1.3. This is a quadratic inequality. Okay. So we need to move everything to one side. So x^2 - 2x when the 8 goes that side it's going to be + 8 is less or equal to z. Let me divide through by negative. So it's x^2 + 2x - 8 is greater or = 0. Let's use the factors or the quadratic formula.

[00:05:40]The two numbers I multiply to give me -8. But when I plus them I get a positive2. It's going to be x + 4 x - 2 is greater or = 0. So our critical values x is going to be -4 from this bracket or x is going to be a positive2 from that bracket. But we need to represent the solutions.

[00:06:04]So if I start with um a parabola. Okay.

[00:06:08]So this a -4 that is a two. If I use my simplified version, I would have a parabola of that nature. Okay, we look for values where the graph is above the x. So this is the region of interest.

[00:06:24]That is the region of interest. Some people use arrows, some people put a plus. Whichever way if I'm to use the original version still -4 positive2 and then I'll have a frown but the graph should be less or equal to zero. So I have this and then I have this.

[00:06:44]Alternatively what I'll do I can say that x needs to be less or= to -4 or x must be greater or equal to 2. All right. So that is how I would have answered 1.1.3. Then 1.1.4 2x - 2x = 12. Let's create a standard equation here. 2x - 2 x - 12 is = 0. What can we do? K method. So let 2 to the x be a k. We saw a scenario like this in the lopo as well as the northwest papers. meaning that the 2x is the same as 2 to the x but squared. Meaning that we're going to have a k^ 2 - k - 12 is = 0. Okay. Then we need to factor out two numbers. You multiply you get -12. You plus you get a -1. It's going to be k - 4 and then a k + 3. Okay. Is equal to z. And then uh with our dot product simply means that our k is a 4 or the k is a -3. But the question was asking us to solve for x and not k. So replace k with a 2 to the x. So 2 to the x is a 4 or 2 to the x is a -3.

[00:08:11]So we know that 4 can also be written as 2^ 2 with a base of 2. So 2 to the x is the same as 2^ 2. Drop the bases. It means that x is a two. If we look at this version of ours, we do not have a scenario where a number raised to the number gives us a negative answer. So, we're going to say this is not applicable. We cannot have a solution.

[00:08:35]Meaning that the only solution is x = 2.

[00:08:40]All right. And then 1.1.5 we have the equation with ss. This one is a very simplified version. Okay. So step number one has already been done for us.

[00:08:51]They've isolated the side. Step number two, we need to square both sides.

[00:08:58]Okay? So it's going to be 4x - 3 = x^2.

[00:09:03]I'm going to move everything to the right. So it's going to be x^2 - 4x + 3 = 0. I can factor out. Okay.

[00:09:16]So I need a negative sum. So I can use -3 and -1.

[00:09:22]Okay. So x - 3 is a zero or x -1 becomes a zero. So if x - 3 is a zero, it means that x is a positive 3 or x is a positive 1. And if that is true, we do not stop there. Remember we need to test our answers. So if we go to the original version of the 4x - 3 = x. If we use 3, let's just punch that in the calculator.

[00:09:57]We are saying that uh square root of 4 * 3 - 3. What do we get? We get a 3. They told us it should be equivalent to x which is a three. So meaning that this is a solution. If our x is a one, let's just move this back. Delete. Put a one.

[00:10:19]Left hand side gives us a one. Right hand side should give us x which is a one. Which means the two solutions do stand and satisfy the equation.

[00:10:29]All right. Then 1.2 which is the simultaneous we have 4 2x + 1 = 16 y / 2.

[00:10:39]Okay. Okay. So if I call this equation one, I call this equation number two.

[00:10:44]The only tricky part is playing around with this. So if this is a base of four, we can also write 16 as a base of 4.

[00:10:53]Okay? So it's going to be 4 to the 2x + 1 is equivalent to 4^ 2. Okay? But then there's an exponent of y / 2. So I'll say y / 2. And then since the bases are the same, I'll have 4. 2x + 1 is equivalent to 4. This cancels that. I'll have y. I'll drop the bases. Meaning 2x + 1 gives me a y. I'll call this equation number three. And then we're going to say that we substitute equation three into equation number two. Where we see y, we replace the whole of that. So it's going to be x^2 - 2x into 2x + 1.

[00:11:36]Let's take the one that side is going to be a + 1 is equivalent to zero. So that is the step of our substitution. Then the next step should be the opening up.

[00:11:45]So it's going to be x^ 2 -2x * 2x is going to give us -4x^2 -2x * 1 is a -2x. Going to say + 1 is equivalent to zero. Let's uh play around with the like terms. That's going to be a - 3x^2 - 2x + 1 is a 0. I'll divide through by a1 which is going to be 3x^2 + a 2x - 1 is equivalent to zero. And then factors again. Okay, I love factors cuz quadratic formula is something else for me. So for me to get the product of one has to be one and one. And now since I want a sum of positive, I'll make this a positive and then I'll make this a negative. Okay? Equals to zero. And then uh the first bracket 3x - 1 gives us a zero or x + 1 gives us a zero. Move the one that side 3x becomes a 1.ide divide by 3, x becomes 1 / 3 or take the one that side, x becomes -1. So those are the x values. But then we also need the y values. y was 2x + 1. Okay, so I just need to substitute this into the calculator. Okay, so we are saying 2 * 1 / 3.

[00:13:15]Okay. And then we are adding a 1, we get a 5 over 3. Okay. Alternatively, if our x is a one, which is negative.

[00:13:29]Okay. Delete. Delete. Delete. That's a1.

[00:13:34]We end up with a1. So -1 3 and 5 over 3. So that is the simultaneous part for 6 months. All right. 1.3 we last saw such questions in grade 11.

[00:13:49]Now they are haunting us here. Okay. But the fourth root of 4 to the 103 - 2 the 25 / 2 to the 207.

[00:14:01]So the hint is very simple here. Since there's a plus sign in between exponents, okay, we need to factoriize.

[00:14:09]But before we factoriize, let's make sure that the bases are the same. So 4 can also be written as 2^ 2. Okay. 103 - 2 to the 2025 over 2 to the 207. We can write this to the exponent of 1 / 4. Okay. So 2 * 10 13 gives us 2 20 26 - 2 2025 / 2 to the 207.

[00:14:42]Please and please do not be tempted to multiply this a quarter into the terms inside. Okay. So for us to factoriize we cannot take out 2025.

[00:14:56]I don't know we can do that but I prefer to take out the 2017 first because I want to kill the denominator.

[00:15:04]Okay. But you can also go ahead and take out what is on top 20 25 and then you're good to go. I don't know whichever.

[00:15:13]Let's see if we take out 20 25. Okay, 2017 was going to be fine, but it's fine. Let's take out 20 25. 2 to the 20 25.

[00:15:26]Okay, we're going to be left with a 2 to the 1 minus 1 because 2 20 25 is coming out. Okay, I'm dividing this by the 20. 27 everything is to the exponent 1 / 4. Now I know that 2 - 1 is a one. So I'll only be left with 2 to the 20 25 over 270.

[00:15:55]Now we are dividing the same bases or we dividing exponents with the same bases which means we need to subtract. So what is 2017 or 202us 2017?

[00:16:09]This gives us 8. Meaning that this entire thing will reduce to something like 2 to the 8. Everything is to the exponent of a quarter. And then the 8 and the quarter are going to multiply.

[00:16:21]You'll be left with 2^ 2. 2^ 2 becomes a fourth. So you're still correct if you factoriize the 2017.

[00:16:30]meaning that you'll have something like a 2 to the 9us a 2 to the 8 over 2 to the 27. This cancels you take out the 2 to the 8 you'll be left with the 2 - one and then everything is the fourth root and then you'll arrive at the same answer. All right then 1.4 determine the values of p for which the roots of this equation are real. So roots can only be real if the discriminant is greater or equal to 0. Okay, so we know that the discriminant is the b ^ 2 - 4 a c greater equal to 0. So let's create the standard form from here. It's going to be x^ 2 - px + 1 is equivalent to zero.

[00:17:17]So our a is a 1. Our b is ap.

[00:17:25]We substitute. So p^ 2 - 4 * 1 * 1 is greater or = 0. So negative number squared we get the same thing. So it's a p ^ 2 - 4 is greater or = 0. Factoriize p - 2 or p + 2 is greater or = 0. So our critical values will be P being a two or P being a -2. If we solve this. So representing the solution for the because this is a parabola.

[00:18:03]Okay. If that's a -2, this is a positive2. Where do we find the graph to be above?

[00:18:10]Okay. Meaning that P needs to be less than -2 or P needs to be greater than two. we are saying equal to.

[00:18:21]All right. So those are the solutions that would have obtained in that case.

[00:18:26]So question one was fair enough. Okay.

[00:18:31]Then we move on to question number two.

[00:18:34]The first three terms of a quadratic are given below. -125 -16. They want us to write the next two terms. Okay. So what we do? We need to start by seeing the pattern. How is it going? So you need to plus no we subtract a three and then we subtract a one. This we are adding a two. So if you add a two there you're going to get a 1. And then -16 + 1 we're getting a -15.

[00:19:04]You plus a 2 you're going to get a 3.

[00:19:06]-15 + a 3 you're going to get -12. So -12 and5 those are the next three terms. So ne no the next two terms. So -15 and the -12.

[00:19:20]All right. So do not rearrange them and leave them like that. Then they want us to determine the nth term. So since this is a quadratic, we know that the second difference should be equated to 2 a. The first term of the first difference is a 3 a + b. The first term is a + b + c. So mean that 2 a is a is a positive two.

[00:19:43]Divide both sides by 2. A becomes a one.

[00:19:47]The 3 a + b must be equated to the -3.

[00:19:52]And if our a is a 1, we substitute 3 * 1 + b is a -3. Then we have to move the three that side. So b is going to be -3 - 3 which is a -6. Then lastly a + b + c we say it should be equivalent to 12. Okay. But our a was a 1 + -6. Our c is a 12. This is a -5 + c is a -12.

[00:20:23]Don't leave the negative. Then let's take the five that side. So -12 + 5 we get the -7. So the t n will be n 2 - 6 n. Okay. We cannot use a plus. Okay. So minus 6 n then - 7. So this is our general term and then uh determine which terms of the pattern are positive. So in terms of positive meaning that the tn must be greater than zero. So since we have n squ - 6 n do we say - 7 or + 7? It's - 7. Okay. So we're going to say this needs to be greater than zero. And since I love the factors a lot, I'll factor this out. So this is n and n numbers I multiply to give me -7. I plus them to give me -6. That will be a -7 + 1.

[00:21:26]Okay. So my values or my critical values n will be a no a positive 7 or n must be a1.

[00:21:36]Since we focusing on greater, so it means that a needs to be less than -1 or a needs to be greater than seven. But now since we're talking about the position which terms, we can never have a negative position. So we can as well cross this out and say this is not applicable or in other words, we don't have solutions where n is less than -1.

[00:22:04]Okay. So the main focus for positivity you're going to have something of that nature. Okay. So I need to be liberated if I'm wrong in that instance. 2.4.

[00:22:15]Which two consecutive terms in the quadratic will have the first difference to be 113? One thing we know about the first differences is that those first differences they form what we call um an arithmetic. And if that is true, okay, if that is true, which means we need to use the general term of an arithmetic. So this is the quadratic. We say this is a three. This was -16 - one and then this is a two. So our focus is this. So if this is your a, this is your common difference. We can say that tn is going to be a nus1 into d. If a is -3 n -1 * 2 when you open up you're going to get 2 n - 2 but we are subtracting a three meaning that our tn okay is a 2 n minus 5. So we need to replace the tn with 113. We want to find out what is the position of 113. So 113 is 2 n - 5. Move the five that side.

[00:23:24]we're going to get 18 is n / 2 which means n is a 59.

[00:23:33]Let me just confirm here 118 oops divided by two. Okay, we get a 59.

[00:23:42]So meaning that our final answer should be between t59 and t60. Why am I choosing that? Because our first difference here, the first term of the first difference is between t_1 and t2 of the quadratic. Which means if this was to continue, we have 113 somewhere here. It will be between t59 and t16.

[00:24:08]All right. So that is how question number two is going to be answered which wasn't bad again. Okay. And then question number three given the arithmetic. Okay. 5 9 13 up to 101. They want us to determine the number of terms. So since we are given the last term which is tn we need to know that five is our first term and what is the difference? 9 - 5 is a four. So our d is a four. Let's get the general term first. Tn is a + n minus one into d. So we're looking for the position of n. So where there is tn let's replace 1 a is a 5 n -1 * 4 you can open up can do whatever magic you want to do so 5 + 4 n - 4 gives us1 so 5 - 4 is a 1.

[00:25:08]Okay I'm going to go I'll try to at least show all the steps as much as it's going to take time. Okay so let's move the one to that side. So, 1 - 1 gives us a,000 is equivalent to 4 n. Divide both sides by 4 meaning that n is a 250. So, how many terms are there? We have 250 terms. All right. So, 3.2 can we prove that for any arithmetic sequence with the first term a and a common difference d that the sum is given as that. So step number one, we know that the last term is given as error which is the same as a + n minus one into d which is like the tn. Okay.

[00:25:54]So if I'm to look at the first or sum of the first terms, the first term is a the second term will be a + d. The third term will be a + 2d since we keep on adding the difference plus can just skip whatever is in the middle there. We know that the last term is going to be L. The second last term must be L minus D. And then the third last term should be L - D. I'll call this equation one. Okay.

[00:26:25]Then I'll rewrite SN. I'll rewrite this equation but starting with the last term first. So I'll have L plus we have L minus D plus the L - 2D plus keep what is in the middle and then plus we have the A + 2D plus the A + B and then plus A. You don't need to use three terms. You can even use the two terms there. Okay. And then uh we need to plus the two equations. Okay. Plus we are saying this is equation number two.

[00:26:59]So equation one plus equation 2. So sn and sn will give us sn. A + l we're going to get a + l. The same story there. A + l a + l up to the end. Okay.

[00:27:14]Now so for n terms okay for n terms we know that this is going to be the same as n * a + l. So if we have 10 of them which means we just multiply this by 10. Now since we're talking of n terms which we don't know we just multiply by n is equivalent twice s n. Divide both sides by two which means it's n / 2 into a + l. But our l was a + n -1 into d. We need to substitute back. So n / 2 is a + a + n -1 into d. So I'm replacing L with all of that. Then A + A will give us A + N -1 into D. And then that is what they wanted us to prove for max 3.3 sigma notation. Sigma notation. So looking at the contents within the brackets, this is an arithmetic because there's no exponent, there's no all of those gimmicks. So start by substituting. If p is a -1, so 3 * -1 - 2, what do we end up with?

[00:28:31]What do we end up with? You're going to end up with a -5 cuz -3 - 2 is a -5. If p 0 goes from 1, you go to 0 - 2, you're going to end up with a -2. And then if p is a 1, 3 - 2 is going to be a 1. So -5, -2 and 1 we are plusing a three. So our common difference is a three. Our first term is a five. So if we have a as a five, d as a three, the only thing we need to find is also the number of terms. Get the last minus the first, you plus one. So that's going to be 20 + 1 which gives us 21 terms. And then the sum that we looked at, we have two ways of doing it. You can use the formula that we just proved or we can use the A + L where L is the last term. Meaning that I have to substitute 19 into that. So let's use the original so that we don't confuse people.

[00:29:45]Okay. So that's going to be 21 / 2 into 2 * -5 + 21 - 1 that's going to be 20 * 3.

[00:29:58]Okay, let's see if we arrive at something proper. Okay, so we are saying we have 2 * -5.

[00:30:10]Mhm. We are plusing 20 * 3.

[00:30:17]What do we get? 525. Okay. 525 not 22.

[00:30:24]Okay. So that becomes our answer 525.

[00:30:28]So even if you look for t of 19 or the last term which will be 3 * 19 - 2.

[00:30:39]Okay. This must give you some funny answer which I don't know. Let's see.

[00:30:47]We are saying 3 * 19 - 2 you get 55. Okay.

[00:30:58]Meaning that it's going to be n / 2 which is 21 / 2 into -5 + 55. So 21 / 2 * 50 should give you also 525.

[00:31:12]All right. So that is 53.3 3.4.

[00:31:19]Okay. A metal road of 15 9375 m is cut into 80 pieces whose lengths form a geometric sequence. Okay. The pieces are then used to support a structure and the pieces are welded vertically. Starting with the longest piece to support the two metal beams in diagram shown below.

[00:31:38]The longest piece is 128 times the length of the shortest. Okay, this is key. And then if the length of the shortest piece which is the eighth piece is x m. Okay, meaning that if this is X and then this longest the first one is going to be 128 * X according to the information. All right, 3.4.1 write down the length of the longest of thank goodness. So that is it but m and then determine the common ratio. So for common ratio what we can do we know that this X is our T8.

[00:32:17]Okay. So let's use the general term of the geometric. So a r n minus one. So when you talk of t8 it's going to be a r 8 - 1 which is a 7. But the eighth term is x. Our first term is 128x.

[00:32:36]This is r to the 7. Let's divide both sides by 128x.

[00:32:42]So x over 12.8x 8 x is equivalent to r to the exponent of 7. I cross that. I cross that. I'll be left with the one on top. So r 7 is equivalent to 1 / 128.

[00:32:57]And then we can find the 7th root on either sides. Okay, let's see how we do that on the calculator. So I'll start by pressing shift and then the square root sign. And then I will move backwards backspace.

[00:33:12]Change that to a seven. and then get back inside. We said 1 over 128 what we get a half. Good. So r is a half or some people would prefer to write it as 0 5. Alternatively you could have used the exponents whereby your r 7 is equivalent 1 / 2 to the 7. Take that on top. r 7 is 27.

[00:33:41]Switch it around.

[00:33:44]Meaning out to the 7 is 2 to the1 with a seven. Drop the exponents. R 21 which is the same as a half or someone will say a half to the exponent of 7 is equal to r 7 whichever way you do it.

[00:34:03]Okay. So that is 3.4.2. Then hence determine the length of the longest piece. I mean, it's it's kind of unfair cuz if someone messes up that question, then there's no way they're going to answer this last question. There's no way. Maybe they should have said if r or if is a half, determine the length of the longest piece. So, for us to get the length of the longest piece, number one, we need to find the x first and then we can do the substitution. So, for us to get the x, I think we can use the 15, something. Okay? If we add up all the pieces, we're going to get that. So we saying our SA or S8 is a 15 comma it was 9 what was 93 75 m. Okay then uh we know that our n is 8. Our r is a half. What are we looking for? The value of x. So let's substitute. We know that Sn is going to be a into 1 - rn / 1 - r. Let's substitute. So 15a 9 375.

[00:35:12]Our first term is 128x into 1 - I'll prefer to use 0a 5 instead of a half. Okay. To the 8 we dividing it by 1 - 0 5.

[00:35:27]Okay. So if I have this as a 15 comma so because when I simplify this I'm going to get a 0a 5 and then multiply that side when I multiply that side let's just write it again multiplying it by 0a 5 gives us 1.8 x into 1 - 0a 5 to the 8. Okay so let's simplify this. So when I multiply this I'm getting um 225 over 32. Some people want to see it for themselves. Okay. So 15 comma 9375 ultiplied by 0a 5.

[00:36:08]That's 22 it's 255.

[00:36:12]Did I enter the right numbers? It's 15 comma 9375.

[00:36:21]Okay. So it's 255 over 32. Okay. 2 55 not 225.

[00:36:29]Oops. Let's clear this. Hello.

[00:36:33]Now it's misbehaving this thing. Okay. So, it's 255.

[00:36:42]Okay. So, 255 over 32 is equivalent to 128x open brackets.

[00:36:50]Let's see if we have um 1us saying 0a 5 okay to the exponent of 8.

[00:37:02]It's giving us 255 over 256.

[00:37:06]So 255 / 256.

[00:37:10]Okay. Then I'm going to divide both sides by these brackets. Okay. So 255 over 32 divided by the bracket 255 over 256 should give me 128x.

[00:37:24]Let's try to punch this in. So 255 over 32.

[00:37:32]Okay, I'm dividing it by 255 over 256.

[00:37:41]What does that give me? I end up with 8 is equal to 128x. Then I'll divide both sides by 128 over 128.

[00:37:52]Okay, let's see. So if we have 8 over 128, we'd end up with 1 / 16. So our x is 1 / 16. Now if x is 1 / 16, the question is asking for the length of the longest piece. And we say the longest piece was 128x. So meaning 128 * 1 / 16.

[00:38:22]Let's see 128 * 1 / the 16. This gives me 8. So the longest piece is 8 m. All right.

[00:38:35]Question number four is no 3.4.3 is done. So I believe that's that 3.4 is kind of tricky. Okay, but it's doable.

[00:38:46]Question number five, which is the exponential f of x is 1 / 3 to the x.

[00:38:50]Determine the inverse. Okay, inverse is going to be the log. So log x with the same base, which is a third. Okay, and then uh sketch the graph f and the inverse. Show the intercepts blah blah blah blah blah. So for us to get the y intercept, how many marks is it? It's six marks. Okay. So for y intercept, okay, we're saying x needs to be a zero.

[00:39:15]So 1 / 3 to the 0 gives us a one. Okay.

[00:39:19]So an exponential meaning that we have a one there.

[00:39:23]And then you can use any other point of your choice. Let's try to use um should we use a negative one? Yeah. If we try to use 1 over 3 to the1, let's see what we're going to end up with. So 1 over 3, okay, to the -1, this gives us a three. So when x is -1, y is a three. So I'm going to need just those points. So -1 is here. Three is up somewhere there. and then connect this to the intercept. We know that the x-axis will be an asmtote.

[00:40:06]Okay, let's include this point as -1 and 3. Call that f. And then we need to plot the graph of the inverse. So the inverse just switch points. Okay. Now someone will say why are you taking this kind of shape? Remember we have 1 / 3 which is a fraction. As long as you have a fraction then your graph should be a decreasing graph which is in that format. If this was greater than zero remember v to the x. If the b here was greater than zero then the graph would be increasing that sort of direction. Okay. So for the inverse we switch the y was a one meaning that our x must be a one. Okay. We have a point of -13. We're going to have a three somewhere there and then a one down here for negative. So that is our point meaning that our inverse is just moving in this sort of direction.

[00:41:08]Okay. So as long as the shape is correct, we have no stress with you guys. Okay. So that is how the inverse and the original is supposed to be sketched cuz they told us at least one point must be indicated. So we have 1 and 3 and then we also have this is going to be 3 and -1. It's not a must to use just these points. You can use any other points of your choice. Okay? It's entirely up to you. Okay? But as long as you've showed us a point and you switch them around. 4.3 they want the values of x for which the inverse is above3.

[00:41:47]So meaning that if that's a -1, we're talking about a -3. So if we have this as a -3, definitely the graph is going to continue and continue and meet up at that point. What will be the x value somewhere there where they meet? So we know that this region okay moving upwards up until the yaxis that is the region of interest where the graph is above 3. So we need to find the x value here.

[00:42:18]Okay, if the y is -3. So to make life easy, I'm going to switch this y value and make it x and substitute it into the original and see that f of -3.

[00:42:32]That's going to be 1 / 3 to the -3.

[00:42:36]Okay, this gives me let's see if I have this. Let me just delete that and put a three. This gives me a 27. Okay. So, meaning that if our x Okay, I don't know why I put here -3.

[00:42:55]It's supposed to be -1. So, if our x is a -3 somewhere here, the 27 must be somewhere there.

[00:43:04]Okay. So if we switch them around then this going to be a 27 and then this should be -3. So meaning the x values should be between 0 and 27. So rewriting that we're going to have 0 less x less 27.

[00:43:24]All right. So question four is also done. Not bad. Question number five still functions. So we have x^2 - 6 + 7.

[00:43:33]That's a parabola/ x - 4 that's a line -2x -4 that is also a straight line. So graph fis at q and t g and h they meet at a point r. Okay. And then they're telling us that the intercepts of g and h and s and q respectively are indicated. P is the turning point of f. And then the first question they want us to show with necessations that G and H are perpendicular to each other. So let's borrow a concept from paper two where we know that when lines are perpendicular the product of the gradients must be equal to1. So the gradient of h multiplied by the gradient of g should give us a1. If that is true then we are good to go. But from the equations we have remember lines are mx + c. Okay, mx plus c. So the coefficient of x will be our gradient. So for g we're going to have a half. For h we're going to have a -2. When you multiply this clearly you end up with a -1. So we're going to say that therefore therefore the graph of g is perpendicular to the graph of h.

[00:44:56]So that's what they needed. Multiply the gradients, you get your1 and draw the conclusion. They want the coordinates of R. Now we know that R is a point of intersection for the two graphs and at a point where the graphs are meeting the graphs are equal to each other.

[00:45:16]So meaning meaning that we're going to get the equation of h and equated to the equation of g.

[00:45:29]Okay. So let's see when we get the equation of h which is -2x - 4 I'm equating it to the g which is x - 4. So g of x is equal to h of x or the other way around. No problem.

[00:45:50]Let's collect the like terms. Let's collect the like terms. So I'm going to have -2x - x = -4 + 14 - 2x - a half is going to give me - 5 / 2x is = 10.

[00:46:10]Okay. Then let's divide both sides by the 5 / 2.

[00:46:17]We have 10 / 5 over 2. We end up with a -4. So the x is -4. Then we can substitute this -4 back into this equation or back into this n. So I'm going to say we know that this half x is the g. So I'll say g of -4 is a half * -4 - 4. Let's see. So a half which is 0a 5 * -4.

[00:46:57]Okay we subtract 4. What do we get? We get a -6. So meaning that our answer is so coordinates of r will be -4 and -6.

[00:47:07]Then they want the length of qt. Where is qt? It's saying that QT are basically the intercepts of G with the X ais and we know that those points with the Y must be zero. So let's get the equation of G equation of F and equate it to zero. So that's going to be -x^2 - 6 x + 7 = 0. Let's divide through by negative x^2 + 6 x - 7 is 0 vectors.

[00:47:42]That's x and x. I need a positive. So it's a + 7 - one. So from this bracket x is a -7 or x is a positive one. So meaning that the point q which is on the negative side okay will be -7. This will be a one.

[00:48:04]meaning that the distance is going to be 7 units plus 1 unit which must be something like 8 units. Alternatively, you can say that your Q is 7 and 0. Your T is 1 and Z. So QT or TQ becomes 1 - - 7 which is basically 8 units. It's not a big deal but then that's how it should have been done. And then 5.4 Four, there's a graph W. Okay.

[00:48:38]And the equation of a line that is obtained by shifting G perendicularly along H until it reaches Q. Meaning that we're going to move this line up until it meets Q.

[00:48:52]Okay, this is WX. So if they saying perpendicular, meaning that it must remain perpendicular at all points. We just sliding this line along there. So if that's a 90, this is a 90. It makes these two to be parallel. Remember the equation of this line G was given as a half X minus 4. So the gradient is a half. Even the gradient of this one must be a half because paral lines must have the same gradient. So since we know the coordinates of Q to be 7 and 0 or - 7 and 0, we can get the C.

[00:49:31]Okay? Or we can even get the equation of the line. So they want us to get the Y intercept, which is the C. We're going to say it's a half X plus C. Let's use -7 and 0. So 0 is a half * -7 + C. So C becomes a 7 / 2. 5.4.2 two if the graph W meets the line of symmetry of F at M which is not shown. Okay, meaning that at the point where this is meeting this axis we are calling it M. They want us to get this distance B M. So step number one I think we need to find out at what X value are we having this turning point. So X is B over A that is one way. Alternatively, since we have the intercepts, we can get the midpoint. Okay, but I'll prefer to use this. So, our B was a -6 / 2. The A was -1. So, it's a 6 / -2, which gives us a -3. So, we have a -3 here. Now, what we can do? We can take or find the difference between the two graphs. get the graph of f minus the graph of w and then we substitute x with a3.

[00:50:58]Alternatively, we can say what is f of -3.

[00:51:03]Okay, so we substitute -3 into the equation of f which is with a -3 2 - 6 * -3 + 7.

[00:51:18]Okay, when you substitute that, what happens is that we end up with uh an answer. Let's see. Let's see. So, okay, but this is 3^ 2. We are subtracting 6 by -3. Oops.

[00:51:45]And then we plus the seven.

[00:51:48]what we end up we end up with a 16.

[00:51:50]Okay, that's a 16. We try that with a with a W. So W with -3. Remember we said it was a half of X - 7 / 2. Okay, I'm going to have a half.

[00:52:08]Okay, multiply by the -3 and then we subtract the 7 / 2.

[00:52:19]This gives us -5.

[00:52:22]No, something is off. Something is off.

[00:52:25]This was a plus, not a minus.

[00:52:29]Got a positive. Why am I put a minus there?

[00:52:34]Okay. So let's let's let's let's fix that mess because we can't get a negative answer there. Okay. So delete and plus. So that becomes a two. Okay.

[00:52:44]So meaning that um we have a 16 here and then we have a two there. So the difference what is 16 - 2? We get 14.

[00:52:55]Okay. So f of -3 - w of -3. So 16 - 2 which is 40 minutes.

[00:53:03]Okay then um that question is done.

[00:53:06]Perfect. We go to question number six which is a parabola and exponential sketch below as are given at that point and then AB intercepts of X and BC the turning point G is also in the vertical asmtote. Good.

[00:53:27]Write down the coordinates of E. So we know that E is the y intercept of this graph f and from the equation this number here that doesn't have a letter is always a intercept.

[00:53:43]Okay with the y intercept rather. So it's going to be zero and -15 since they wanted it in coordinate form. Then 6.2 they want the coordinates of b. Where do we find b? B is the x intercept for both graphs G and F. But we don't know the equation of G. So we're going to use F and we equate the F of X to zero.

[00:54:08]Okay. So which means it's going to be the X^2.

[00:54:14]Okay. - 2x - 15 into 0. Let's factor this. X - 5 X + 3 is into 0. So x is a 5 or x is a -3. So a okay is going to be a -3 0 and then our v which is on the positive side should be five and zero.

[00:54:39]Okay, that's what they wanted.

[00:54:42]They wanted the p. So our focus was just going to be that. And then they want the range of f. Let's see. Range of f. So when you talk of range, you're dealing with the y values. But the only thing we know about this g is the turning point is just the x value. So if we replace x with a one into the equation, we can get the y value. Okay? And that y value will help us in determining the values of y for which this graph is valid. So from this y value moving upwards. So I'll go down and say what is my f of 1.

[00:55:16]So 1 2 - 2 * 1 - 15. So this gives me -16. So the y should be greater or equal to -16.

[00:55:29]And then consider the hyperola. They want us to write down the values of p and q. Remember their symptoms are meeting at 1 and2.

[00:55:38]That point h that we were given. So if this is the vertical, we're going to switch and say p is going to be -1. the q must stay as a2 and then can we show that a is 8. So looking at now the new equation okay of g of x is going to be a / x -1 - 2 but the only thing we need is the a for us to get the a we need to get another point that is passing through that graph and the only point is the v which is 5 and 0.

[00:56:15]Okay we substitute that 5 and 0. Okay, so at V, which is 5 and 0, we're going to have 0 = A over 5 - 1 - 2. Move the two that side. It's going to be a / 4.

[00:56:31]Multiply both sides by 4.

[00:56:34]Okay, so this cancels that meaning that a is 8.

[00:56:40]All right, so that is cool. 6.4.3. Hence determine the equation of the axis of symmetry of G with a negative gradient.

[00:56:52]Now since the asytos are still meeting at 2 and 1, the equation is -x + c. For us to get the c substitute -2 is -1 + c. Where there's x, I'll put a one. Where there's y, I put -2. So move this one that side. So it's -2 + 1 is c. So C becomes1. So our equation is going to be F minus one. Okay. Whichever format you use to obtain it and as long as the two marks are in the back. All right. So that is 6.4 then 6.5. They're telling us that it is further given that the X values of C and F are -2A 41 and then 0a 41. So meaning that this is a 2 4 1 this f is a 0 4 1. Okay. What do they want us to do?

[00:57:51]Determine the values of x for which g of x - f of x is greater than zero. Let's move the f that side. So g of x is greater than f of x. So where do we find the graph of g to be above the graph of f? So you can see this region here that the graph of G is above F. Okay. But also in this region here we can see that G is still above and then after that G is below. So meaning that the values or the points that they gave us which will be the -2A 41 is less X is less 0a 41.

[00:58:30]And then we also saw that the other condition is between one. the simp and b which is the five. So we're going to say 1 is less x is less five. Okay. So that is how I would have worked out 6.5.

[00:58:47]And then um 6.6 6.6 they're telling us we have a graph which is h of x okay is given as -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 two graphs will not intersect. So starting by the graph of g, let's remind ourselves it was 8 x -1 - 2. Now the h of x is -2x + k. So we're going to look at it in a way that let's assume they are intersecting.

[00:59:25]It simply means that the two graphs should be equated to each other. So h of x should be equated to the g of x. Okay.

[00:59:33]Then if they are not intersecting then that's where the k comes in with the nature of roots meaning that our discriminant needs to be less than zero.

[00:59:44]So let's start with the equating part.

[00:59:46]So h is -2x + k is equivalent to 8 / x -1 - 2. Okay. I need to get rid of this denominator. I'll multiply through by xus one.

[01:00:02]Okay, so we say we're going to multiply by x -1 x -1 x -1 and then x -1.

[01:00:13]Opening this going to be -2x * that we're going to get -2x^2 -1 and -2x is going to be a positive 2x.

[01:00:25]Okay.

[01:00:27]and then the k multiplied by the x, we're going to end up with a kx.

[01:00:36]Okay?

[01:00:38]And then the k and the one is going to give us k.

[01:00:50]Okay. So, we're saying that's going to be a negative case, but this is going to be kx.

[01:01:02]Okay, let's just fix that.

[01:01:10]Okay, so we say that's going to be kx - k and then this is going to cancel that.

[01:01:17]shall be left with 8 - 2x + 2. Okay, so let's regroup with the like terms.

[01:01:32]So when we regroup this, I'm going to have um a -2x^2 this and that. I can move this 2x to that side. I'll end up with a 4x plus a kx.

[01:01:48]This is going to be - k.

[01:01:53]This is 8 + 2 is a 10. Move it that side. You're going to get -10 is equal to zero. I want to create the a, the b, and the c automatically.

[01:02:04]Before I do that, let me factor this out.

[01:02:08]So, -2x^2 + um I'll say 4 + k that is x and then - k + 10 cuz there's a negative is equal to zero. So that I have a I have b and then I have c. So we can divide through by not a big deal. It's not a big deal.

[01:02:33]Okay. So let's say b ^ 2 - 4 a c we say it should be less than zero.

[01:02:41]Okay, let's start from there. So what's our b? It's 4 + k or k + 4. We square that - 4 the a is a -2 now. And then our C is a K + that I don't know how I'm going to use that.

[01:03:06]Okay. So I'll say times K + 10.

[01:03:12]Okay. So opening this we're going to get a 16 + 8 K + A K 2. But I know that -4 * -2 is going to give us 8. And then 8 with this negative it will end up being a8. So -8 * k I'll get -8 k8 * 10 is8 gives me a zero. And then I see that the 8k and the 8k will cancel. We're left with k^ 2. 16 - 80 is going to be -64 is = 0. K - 8 k + 8 = to zero. And then either k is going to be 8 or k is going to be -8.

[01:04:06]Then what we do next? Remember these are like uh critical values. Now someone's going to ask me when did we keep equal signs? Remember this was uh less oops so all this should be less than zero okay the k8 and the8 are more like um critical values okay less and then if that's the case if that's the case meaning that um if our k is less than that meaning that it's going to be negative 8 is less k but also less than 8. Okay. So that is the last question of six. We move on to question number seven. First principles calculus.

[01:04:56]So we know that the limit as f of x + h minus f of x. Okay. H ts to zero. We are dividing by h. We already have our f of x. Meaning that f of x + h is going to be 4 x + h 2 - 3.

[01:05:15]into x + h. Let's substitute limit h ts to 0. We have 4 into x + h 2 - 3 into x + h. We are subtracting 4x^2 - 3x. This is all over h. So limit as h ts to zero. Open this 4K cap outside 6 2x h 2 - 3x - 3 h we are subtracting 4 x^2 + 3x remember at the end of the day f must cancel out as you move on. So limit as h ts to 0 is 4 x^2 + 8 xh + 4 h 2 - 3x - 3 h - 4x^ 2 okay we are plusing 3x all this is over h. So let's cancel out what needs to be cancelled. So we know that the 3x is going to cancel the 3x and then the 4x^2 will cancel the 4x^2. We are only left with the limit as h ts to 0 8 xh + 4 h 2 - 3 h / h. So let's take out the h or factoriize the h. Decided to put an extra step there or what? I don't know.

[01:06:51]This is 4 hus 3. So the h and the h will cancel. Our next step is to substitute h with zero. Meaning that our answer should be 8 x - 3 which is fine. Okay.

[01:07:06]7.2. They want us to determine the dx of this. Let's take out what is common. So it's going to be dx. You factoriize x - 3 x + 3 because that's a difference of two squares. You're dividing by. Let me take out a negative sign down here to be x - 3.

[01:07:26]Okay, so this is going to cancel that.

[01:07:29]Introduce the negative back. You'll have -x + 3 and the derivative of this gives us -1.

[01:07:39]Okay. Then question 7.3.

[01:07:42]If you're given y as this, can we show that dy dx is the square for roo<unk> y?

[01:07:49]So let's open this up first. It's going to be 2x - 3 * 2x - 3.

[01:07:58]So opening up again, we're going to have 4x^2.

[01:08:03]That and that is -6x -6x + 9. Meaning that y is 4x^2 - 12x + 9.

[01:08:14]And then our dy dx is going to be 8 x - 12. This 9 will fall away since it's a constant. Okay? And then what do we see that's common there? We're going to have a 4.

[01:08:29]Okay? Cuz 8 x and 12. We have a 4. We shall have a 2x - 3. This is our dy dx.

[01:08:37]But we're going to say if we are to make y the subject here, you need to take square roots on either sides. Meaning that root of y is the same as 2x - 3. So where is 2x - 3? Just replace that. So meaning that our dy dx is going to be 4 multiplied by the square root of y.

[01:09:03]Okay. So that is how 7.3 was going to be worked out. those sweet 12 marks. At least there are no school square roots things there. Okay, question number eight, which is the cubic functions.

[01:09:16]Determine the coordinates of the intercepts of f with the axis. So let's start with the y intercept.

[01:09:22]We know that x is always a zero or you can take the constant term which is going to be zero with a four. And then for the x intercept it's like we are solving this equation. we need to use the try and error method. So which number can we substitute here to give us a zero? If I use one and that's going to be 1 - 3 + 4 that's 5 - 3 which is a 2 which is not zero. Okay.

[01:09:53]And then uh If we try 2, it's going to be 2 cubed, which is 8 - 2 2 is 4. 4 * 3 is a 12 + 4. So 4 + 8 is a 12 - 12, which is 0. So meaning that x being a 2 is a root. Okay. So for x intercept we said y is going to be zero. So x is a 2 is a root and then we can use I prefer using the synthetic approach b c and d coefficient of x cubed is a 1. x² is a -3. x we don't have x there so it's a zero. The constant is a four. Okay. Then we break this down. You're always told you shade this off. And then we drop that first coefficient down.

[01:10:58]Okay, then let's do the splitting using the root that we found. X being a 2. So 1 * 2 is a 2. -3 + 2 gives us a -1. -1 * 2 is going to be -2. 0 + -2 is a -2. -2 * 2 is a -4. You get a 0. Meaning that we have x^ 2 - x - 2 as our quadratic.

[01:11:25]If that is true, we can factor this out.

[01:11:29]X - 2, X + 1 is 0. So either X is a 2 or X is a 1. So the coordinates they want will be 2 0 and 1 0. So the first one we got was a 2 0 also meaning that we have a repeated route which translates into it being a turning point.

[01:11:54]We shall see in 8.2.

[01:11:55]turning point you always get the first derivative. So our f was x cub okay - 3x^2 + 4. So first derivative equated to 0.

[01:12:12]It's going to be 3x^2 - 6x. 4 is a constant. It falls away.

[01:12:17]Equate that to zero. Let's take out 3x.

[01:12:20]That's going to be x - 2. So either 3x is a zero or x - 2 is a zero. So if that is the case x is a zero or basically x is going to be a two but they want the coordinates. So meaning that we need to substitute 0 into the original. We end up with a four. Remember the y intercept. So that is 0 and four. Or we can substitute two and we end up with a zero just like what we did by the first part that the f of two gave us a zero to make x - 2 a vector. So the coordinates here will be two and zero. All right.

[01:13:09]Then u 8.3 they want us to sketch a net graph. So with the net graph we need axis and turning points. So we started with the intercepts. The y intercept is a four. like a pos4. So I'll indicate a positive4 there. The x intercepts we only got -1 and a two. So I'll indicate a -1. I'll indicate a two somewhere there. And then we have two turning points which are also being represented by that 04 is somewhere here and then 2 0 is somewhere there. Now since the a is positive which means we're going to have a frown and then a smile. So meaning that the first turning point is going to be a maximum and then this is going to be a minimum. Okay, if that is true then let's just connect.

[01:14:00]Okay, connect turning point drop it down turning point move in that direction.

[01:14:10]Okay, so that is your graph of f. So the intercepts are indicated as well as the turning points.

[01:14:20]Then for which values of x will f be concave up. So concavity upwards the f of x needs to be greater than zero. So our first derivative was 6x^2 - 6x.

[01:14:35]I don't know. Was it 6x? No, it was 3x^2.

[01:14:39]was 3x^2 - 6x.

[01:14:43]Okay, 3x^2 - 6 x. Okay, so meaning that our first derivative or our second derivative is going to be 2 * 3 which is 6x. We are subtracting 6 is greater than zero. Move the six side and then divide both sides by six meaning that x is greater than one. All right. And then 8.5 use your graph in 8.3 determine the values of P for which this graph will have just one distinct root. Use the graph. So for one distinct route it's either we move this graph beyond four.

[01:15:19]Okay. Which means our P should be will that make sense that we are moving the graph above four. Let's see.

[01:15:39]Okay. So from the graph what we said was that um if we drop this four units down okay we shall have one positive. If you shift it slightly above if the p is greater than zero meaning just even if it's one unit cuz this is the turning point and that is also a turning point.

[01:16:03]So we're looking at um 0 and four. So what are we going to do? We're going to say that either our P must be greater than zero or P must be less than -4.

[01:16:20]I think those are the solutions that we can use. And then question number nine, a school fundraising committee is planning a metric dance for the grade 12 learners. The weekly profit in runs depends on the number of teachers involved and there's a relationship.

[01:16:33]Determine the profit when only five teachers are involved. So it's -2 into 5 2 + 160 * 5 + 19 600.

[01:16:45]So we're looking for the Okay. So the calculator is misbehaving.

[01:16:54]Calculator is misbehaving. But it's fine. Let's see. When you substitute this into the manual calculator, I'm getting 2315.

[01:17:07]Okay, that is our profit when five teachers are involved. Then calculate the number of teachers that will need to be involved in fundraising to achieve a maximum profit. So for maximum profit, we need the first derivative.

[01:17:20]So that's going to be -2 and 2 which is -4x + 160 should be greater to zero. And then we going to need to move the 160 that side. So it's -4x.

[01:17:35]Okay.

[01:17:38]Is = 160. Divide both sides by -4.

[01:17:43]X becomes a 14. And then uh 9.3 determine the maximum profit. So P40 we said it's a 2 * 40 2 + 160 * 40 + the 19600.

[01:18:06]So this gives me a profit of 22 800.

[01:18:12]All right. So that is 9.3 and I think that brings us to the end of the paper. Thank you very much guys. Okay and that is how the paper was supposed to be and as according to my observation I believe it was a fair paper. Okay, it is possible for learners to get total in this paper.

[01:18:36]Okay. So for those that are yet to write paper one, we wish you all the best in your preparations. And then those that are planning on writing paper two, still we still wish you the best in um your preparation. We shall be dropping videos for the preparation of paper two just soon. Bye-bye and good luck.

[01:18:59]Hello great wolves. will come to this session where we shall be discussing the proposed memorandum okay for the grade 12 mathematics paper one for the health province okay so this paper was written today and the memo is yet to come out but in the meantime we can together go through it and compare and see what exactly was expected of the learners okay so time is 3 hours the max basically 150 we 30 pages plus an additional page. Okay, the information one. Okay, instructions are we have nine questions that we need to work with and from the nine we need to answer all of