Mathematics Grade 10 June 2025 Paper 1 KZN Term 2 Test @mathszoneafricanmotives

Added: 2026-05-13

[00:00:02]Hello guys uh welcome back to med zone African motives. We are still on our revisions and this is for our grade 10 mathematics and we are going to consider uh the full paper revision. So this was actually uh mathematics paper one a common assessment task. All right which was written in uh June uh 2025. All right. As a common test. So that was it from our KZNA natal province. So as we do understand we must go through the instructions and information that you're given that is uh this paper consist of five questions. Go through that, answer all the questions and also the calculations must be properly shown and the most important part your answers are supposed to be correct to two decimal places unless stated otherwise. Uh so you must go through the information and as of the question paper for those who need the question paper just go through the description of the video. you go through the description of the video or you go uh or you check the comment section. All right? Or you're going to check the comment section or you'll be able to download the question paper and the memo or you can also check in the drive. All right. For mathematics grade 10, just check on the description of the channel where you see grade 10. uh then you can check uh that one in the drive or simply the comment section that is the easier way for you to download uh the question paper and the memo all right so without wasting your time I'm going through the questions uh that was the first one question one consider the expression that we are given here write down the value or values of n. So we need the value of values of what of n for which 1.1 k is equal to zero. All right. So let us consider this is our expression that we are given that k is = 3 n over what the square root of n + 5. So the first condition we are told that k is supposed to be equal to what to zero.

[00:02:43]That's our 1.1 in place of K. Let us substitute uh that value that we're given which is what a zero and we determine or we calculate the value or values of n. The question we must consider there is on the values or value of n but on a condition the condition is that k is supposed to be equal to zero. So let us uh calculate the value of n. This is just going to remain as it is. As we can see we have got an equation where we are supposed to solve for n.

[00:03:29]Solve for what? For f the value or values of n. How can you solve this is a fraction and 0 is same as what? 0 over one as a fraction. So meaning to say we can simply cross multiply uh to determine the value of n. So that's 0 * the square root. So guys take note 0 * 1 that's a 0 0 * 2 that's a 0. So meaning to say 0 * this square root you are going to obtain what? A zero which is equal to 1 * 3 n and that's 3 n. So in order for you to solve for n from what we are given simply divide by what by three from our multiplication we can simply divide by 3. So that's our n is = 0 / 3 that's a zero.

[00:04:26]So that's the condition that we are going to have for the value of what k to be equal to zero. n is also equal to zero that is the condition all right let us consider another part and c that was 1.1 then 1.2 We are still on the value of values of n but on a condition this time that k is what? Undefined from this statement k is undefined. K is equal to this fraction that we given there and we given that it is what? Undefined.

[00:05:18]So given a fraction a b for it to be undefined remember the denominator b must be equal to what to zero that is what undefined that is the condition.

[00:05:34]So you just consider what the denominator what is the given part of the denominator that is this part the square root of what? n + 5 as we have got this condition that b must be equal to z your denominator for it to be undefined. So therefore it simply means our denominator which is the square root of n + 5 must be equal to zero.

[00:06:05]Just like before you formed an equation where you must solve for n.

[00:06:15]write down the value or values of n. So we must solve for n. All right, let us solve guys. We're back to our sets where we can get rid of what the square root.

[00:06:28]Simply square both sides. So we're going to square that is the square root and the square neutraliz. So you're going to remain with n + 5 which is equal to 0 squ that's a 0. Okay, 0 * 0 guys. So to find the value of n simply subtract five both sides thus our n is equal to 0 - 5 which is what a -5.

[00:06:56]So we obtaining - 5 for k to be undefined. The denominator is the one that you must uh consider.

[00:07:09]All right, let us u have another part.

[00:07:13]All right, let us have another part. But that is the condition guys. We simply have to uh work out your basics. It's all about the basics. It's all about the basics. All right. Uh that's the most important part. Work with your basics.

[00:07:29]Let us consider 1.3 K is nonreal.

[00:07:36]non real.

[00:07:39]If we consider our numbers, we have got real numbers and nonreals that cannot be simplified.

[00:07:53]Cannot be simplified.

[00:08:00]We are talking about square root of negative values. - 3 square root of -2 something that cannot simplify square root of a -5 square root of a3 cannot be simplified.

[00:08:25]So our condition from the given expression on 1.3 we are considering that this cannot be simplified that is what nonreal remember when it was undefined the value of n was equal to -5 when it was what?

[00:08:57]Undefined meaning to say the denominator is equal to what to zero.

[00:09:07]So if we consider nonreal we have to consider where we have got what the square root because I'm talking about square root of negative values.

[00:09:21]The numerator is okay. There's no square. We do not have a square root there. is n not square root of n. The square root is in the denominator.

[00:09:32]So for which values of n are we going to obtain under the square root? For which values of n? Remember at -5 this condition here for n is = -5 it was when it is what undefined when you're saying this is equal to zero. So whenever we are going to substitute a minus5 take a look if you put a minus5 here you're going to obtain a zero that is for the denominator to be equal to zero n is minus 5 that is what we obtained before. So meaning to say in your own consideration when n is5 you obtain what a zero. What about for nonreal values?

[00:10:26]Meaning to say it is always a negative that you're going to obtain under the square root.

[00:10:34]Meaning to say it's not only one value of n. They are now values of n comparing to -5. Are we going to talk about these values to the positive or we going to talk about these values to the negative for us to obtain what? a negative. Those are numbers such as -6, -7,8.

[00:10:58]Those are the ones that will give us what? Negative value.

[00:11:03]Negative values.

[00:11:08]If you are to substitute -4, -3, -2, you're going to obtain positive values.

[00:11:14]Let's take -4 in place of n plus what? + 5.

[00:11:22]This is going to give us 1 and square root of what meaning to say it's now a real number that you have -3 + 5 this is going to give us a two - 2 + a 5 this is going to give us a three. So you see these values to the positive as we are going to the positive to the right side they are going to give us real values something that can be simplified even by a calculator but from -6 going this side to the left these values will give us -6 in place of n + what + 5 this is going to give us1 meaning to say we have got square root of a negative something that cannot be simplified.

[00:12:12]All right, let's take -7. -7 + 5 that's what -2 square root of what? A negative those are nonreal values n realus.

[00:12:28]So for these values -6, -7, -8, - 9 and so on and so on.

[00:12:37]So we are not having a single value.

[00:12:42]We are not having a single value. There are so many values that we are referring to. So comparing to -5 where it is what undefined we are saying the values of n these -6 -78 they are what less than five they're found on the left side of what of -5 so they are less than5 so n must be less than5 for us to have what non real for k to be nonreal It means the square root will be a negative for any value any choose any value guys that you want. Any value that is less than5 -6 -789 anyone you're going to see that the number that will remain under the square root here is going to be a negative value and thus non real. So that is the condition that we're having there. All right. So the answer was going to be n is supposed to be what less than five. Let us consider another section which is now question number two and 2.1 given to simplify the following expressions fully simplify.

[00:14:10]All right. As you can see this is expansion of brackets 2.11 we are given. All right let us consider this uh that is 4 p. Okay let me just try to use here a little bit.

[00:14:26]So this is 4 p into 2 - 5x. Simply we have to expand our brackets. All right, that is uh distribute by the term outside of what?

[00:14:41]Outside of the bracket, which is the 4 p. You're supposed to multiply each and every term inside uh the bracket. So you must multiply to positive2. Then you multiply to what? -5x.

[00:14:55]These are the two terms that we have inside the bracket. So that is the consideration actually that you're going to have when you are working with this.

[00:15:05]All right. Uh so let us just expand 4 p * 2 uh that's 4 * 2 the constants together or the coefficients 4 * 2 is going to be 8. Then we've got p remaining. Then we have got 4 p multiply -5x 4 * -5 that's a negative positive to a negative that's a negative so that's - 20. uh px. So you can write as -20 px or you can write it as -20 xp. All right, this is just one and the same xp or px. This is one and the same.

[00:15:48]So anyone uh that was it for two marks.

[00:15:53]All right. So this is it guys. Let us consider the other part. And like I said guys, this is just an expansion. 2.12 you are given -2 into 2x + 3 into 3x - 2. So this is just simply from our foil method when you've got two brackets you can expand uh using uh the foil method working the first outer inner and the last. So 2x is going to multiply first term here. So you just ignore this -2.

[00:16:30]All right. So that's -2. Open a bracket for the expansion of these. All right.

[00:16:37]So that's -2x I mean uh 2x * 3x is going to be pos6. 2 * 3 that's 6 x^ 2 x * x.

[00:16:48]All right. Then 2x to -2 uh that's -4. 2 * -2 that's -4x. We are done with the 2x. We move on to the next term which is pos3. So that's pos3 * 3x and that is going to give us what? Positive 9x. All right. Then pos3 * -2 3 * -2 that's -6.2 is an outer term from the brackets there. So of these brackets we consider this as the outer term which is to expand everything in this bracket. But before that you can even just try to simplify here. As you can see we have got what like terms. So we can just try to collect uh the like terms the 4x and positive 9x. These are like terms. We've got the same variable and same power on that x not x². x² is not a like term to these because of what the power it carries the power of two. All right. So you consider these two uh -4 + 9 that's going to be pos5x - 6. Then with the -2 outside expand multiplying each term -2 * 6 that's -12 x^ 2 -2 and 5 that's -10 x -2 and -6 that's going to be pos2 you can even use uh the calculator for multiplication in case that you're confused with this positive negative. So that was the expansion 2.12 for three marks you've expanded your brackets then simplify where you've got like terms uh so many ways so that's why I said guys you must download uh the question paper and the memo so that you can go through uh the sections and see other ways that you can actually use uh in terms of your simplification let us consider 2.13 Again it is expansion of brackets that we are having on this part. 4x + 1 uh squared. All right. Minus we have got a bracket here 3x + 1 another bracket 3x - 1. So if you to consider properly we can see that we have got the squaring of a binomial here. So we simply consider uh the squaring binomial. All right. And here these are just same terms but only that the other one here has got a positive and here there's a negative. So that's a difference of two squares. So we have a difference of what? Two squares.

[00:20:00]That is the condition. So this part where you're squaring a bracket 4x + 1 another bracket 4x + 1 you expand uh the same way that we expanded these two.

[00:20:17]the same way two brackets or you can simply consider uh what we had before from the squaring of a binomial that a + b 2 is equal to a 2 + 2 a + what b^ 2.

[00:20:37]So what you simply have to consider here is what is your a and what is your b. So this here is our a 4x and our b is what is 1. So if we are to consider this it simply states that we are going to have a squ which is the first term being squared. What is our first term? It's 4x being what? Being squared. So that's 4^ 2 which is 16 x^2.

[00:21:10]All right. So that's 16 x^ 2 we move on plus 2 a b this part. So that's plus 2 a b which is 2 * our a. The first term 4x * b our b is what is 1. So that's 2 a b 2 * a * b which was going to give us 8 x 2 * 1 * uh 2 * 4 * 1 that's 8 so it's going to be 8 x positive 8 x then the last part is considered as what b^ 2 the last term squared b is 1 so 1 2 is what is 1 so that's + one so this is the expansion that we are used to on the squaring of a binomial Okay. So from there you subtract. What are we subtracting here?

[00:22:07]The difference of two squares.

[00:22:10]So because of this subtraction must open a bracket because we're going to obtain an expression from the difference of two squares. We do understand that a minus b a minus b like this into a + b like this was taken from what? a 2 - b 2.

[00:22:34]So if we've got a bracket and another bracket same terms or same values but only that one has got a negative the other one a positive that's a difference of two squares. And take note of whether sign is positive or negative. It can be a positive here, a negative here. It's not a problem. So this is what we have.

[00:22:54]Our a is what? 3x. This one. This is our a 3x. And our b is what? Our b is 1. So from the difference of two squares concept, these two brackets were taken from what? A squar. A a is what? 3x squared.

[00:23:15]So 3 x^2 3 2 that is 9 x^ 2. All right.

[00:23:22]So that was going to give us 9 x² - b ^ 2 minus our b is what? Our b is 1. So that's 1 2 which is 1. So that is it from our difference of two squares. Or you can simply expand these two brackets properly the same way that you used two.

[00:23:43]You're still going to obtain uh same answers, guys. 3x * 3x 9 x^ 2 3x * 1 uh 3x 1 * 3x that's 3. Oh, sorry. I put both positive. The other one is supposed to be negative. Yeah. So 3x * 3x that's 9 x^ 2 * 1 - 3x. Sorry, that's -1. 1 * 3x positive uh * -1 that's a negative 1.

[00:24:15]Then collecting the like terms that's what that was going to be - 3 + 3 that's a zero. This cancels so you're left with what? -1. Exactly what we have here.

[00:24:27]This is the same thing for that expansion of 4x + 1 2. You can also have that consideration two brackets 4x + 1 another bracket 4x + 1 expanding proper the way that you're used to 16 x^2 + 4x 1 and 4x 1 and 1 that's 1 then you end considering the like terms 4 + 4 that's what 8 x + 1 the same answer that you obtained here from a direct expansion So these expansions simply used to make our calculations easier.

[00:25:12]But if you do not understand, expand the way that you used to. All right. Now that we've got a negative, what is left here is simply to consider expansion of brackets by that negative.

[00:25:26]* 9 x^ 2 - 9 x^ 2 - * negative here that is going to be pos1. Then collect your like terms. All right, collect like terms. Simply consider simplification of like terms where in our case we have got this part of x² and x². So that's 16 - what? - 9. 16 - 9. That is what you're going to have which is 7. So that's 7 x².

[00:26:02]8 x we do not have any part with x. So that's 8 x remains as it is. Then we have got 1 and 1. These are constants.

[00:26:12]One + one. And that was going to give us what? A two. So that is the condition here. Plus two. All right. So that is the conditional done. So after expanding the brackets do not forget to simplify further. So the simplification part in order for you to simplify you consider uh the like terms. All right. Collect like I just said terms here. These are what these are like what like terms. All right. So that was it on question 2.13.

[00:26:47]Then uh 3 2.2 with three marks that was factoriize.

[00:26:54]The previous part it was expansion of brackets. Now factoriize.

[00:27:02]We must have brackets at the end. But we have brackets already.

[00:27:09]So it means in these two brackets there was supposed to be a common bracket.

[00:27:17]There was supposed to be a common bracket x^2 into x -1 + 25 into 1 - x. So take note these two brackets are not the same.

[00:27:34]This is positive x. This is negative. So take note this is x.

[00:27:43]This is -1. This is positive one. These brackets are not the same. But we can rewrite one of these brackets in terms of the other from the concept that if you're given y - x. It is the same when we factor out a negative x becomes a positive y becomes a negative. The moment you factor out a negative, these two changes their sign.

[00:28:19]So if we to consider that because these two brackets are not the same and we want to make them to be the same. Let us rewrite the second bracket in terms of the first as this one.

[00:28:34]So we must rewrite 1 - x as x - one. It is possible from this factorization because the moment all right let me rewrite this and and say the moment you factor out a negative you have to say it's a negative that you factored out it means x becomes a positive from here x becomes what a positive one becomes what a negative this is the concept x - y it was y - x now it's x - So x is a positive. Now it changes the sign.

[00:29:13]But remember there was a positive 25 here outside of the bracket. So if you multiply these two it is going to give us - 255 25 * a that's - 255 into x -1.

[00:29:30]So these two brackets are exactly the same. Now x - y x -1 from this idea of your factorization guys we can rewrite one bracket as the other one once these two brackets are exactly the same as we can see we can now factor out remember it's what factoriize so you can factor out the common bracket you can factor out what the common bracket which is what x - one. So if you factor out this x - one as a bracket, what is going to remain? If we factor out x - one, you're going to remain with x² for the first part. And here if you factor out x - one, you're going to remain with what? A 25.

[00:30:29]You have factoriize but not completely as given there.

[00:30:36]Why? Because what we have inside the second bracket here can be factorized further because this is what? A difference of two squares.

[00:30:50]We've got a difference of what? Two squares.

[00:30:55]A difference of what? Two squares. This is what we have. A squus b. the one that we said it was taken from a minus b into a + b. So if we check x² this 25 here is same as what 5² a minus a^ 2 - b^ 2. So meaning to say it can be factorized further by having two brackets x and what and five x squar 5 squared.

[00:31:32]So one is going to add another one is going to subtract x - 5 another one x + 1 + 5 or x + 5 x - 5 the square root of x² it's x the square root of 25 is what is 5 but 1 is positive another one is negative you have what factorized completely where we do not have anything common any common term term or any common sign or any common number there inside of the the brackets. This is fully factorized.

[00:32:11]So you just also consider the memo uh so that you you consider other ways. There are so many ways actually that you can utilize uh the simplification or the factorization that we given there. All right. So this was now question number three and uh this was issue of solving for x right this was actually question number three where you must solve uh for x for from the given equations you're supposed to solve for x and 3.11 we are given this all right uh for three marks all right let me just try to write it here 3.11 you're given x minus - 3 which is equal to 18 /x and solve for what? For x.

[00:33:08]Solve for x. So as we have got a fraction, let us consider the first part here as a fraction. The whole part is same as what over one. We haven't changed anything because 2 over 1 is equal to 2. So x - 3 over 1 remains as what? x - 3. We haven't changed anything. But this allows us to cross multiply. So x is going to multiply x - 3. And 1 is going to multiply what? Is going to multiply 18. So distributing the bracket by x we're going to obtain x^ 2 x * -3 -3x which is equal to 1 * 18 that's 18. That's a quadratic equation where the highest power is what is 2. So remember the format of a quadratic equation a x² + b x + c must be equal to zero. So one of the sides supposed to be equal to zero. So as we have got these two terms, let us just take the 18 guys to the other side. So it is going to be what?

[00:34:22]18. It was a positive. So it's going to be -8 when taken to the other side of the equation. And thus solve remember by what? By factorization.

[00:34:35]You simply have to factoriize uh that's have two brackets equal to zero. uh from the coefficient of x² being 1. This is going to be x and x. Then I'm going to consider what the products of -18 or factors of -8 which adds up to -3. All right. So what are these factors or products of -8? Factors.

[00:35:05]All right. So we need factors of what?

[00:35:07]products of -8.

[00:35:12]18, right? Uh there are so many factors 18 and 1 8 -8.

[00:35:20]Uh we have got -6 and three. All right. I also left -9 and 2, 9 and -2, -6, 6 and -3. So many. But of these factors, which of the two when you add, you're going to obtain -3?

[00:35:39]That's -6 and uh and positive3.

[00:35:44]If you add these two, you're going to obtain a -3. - 6 + 3, that's what - 3.

[00:35:51]But if you multiply, it gives us what?

[00:35:54]-8. So that's -6 + what? + 3. So each bracket must be equal to zero in order for us to solve from the zero concept the zero there. So that's each bracket x - 6 must be equal to 0 or x + 3 must be equal to 0. So what is the value of x - 6 to the other side is going to be positive. All right pos3 to the other side will change into a negative. So those are values of what of x from our given equation. All right. Then 3.12 uh that's an inequality that you're given. And for the values of x which are real numbers uh and also represent your answer on a number line that was 3.12.

[00:36:50]All right. So let us see guys we given uh 2 - x is greater than or = -3 but less than zero. All right. So let us consider the constant of this this two here to this side and to the other side.

[00:37:09]It means it was going to be -2 here -2 there. All right. -3 - 2 that's -5.

[00:37:18]uh then you're going to remain the moment you do this you are going to remain with what ax the moment you transpose uh that you're going to remain with thex all right which is -x less than 0 - 2 that's a -2 all right to get rid of the negative you have to divide by what all right and remember the division by a negative reverses or changes the signs that that you're given. The inequality signs that you're given here must change or reverse because you divided by what?

[00:38:02]A negative. So you must change them. So this is going to face the other side like this has to change. All right.

[00:38:11]Negative negative that's going to give us a positive five. All right. This one also changes. So it is going to face uh the other side like this. Negative negative that's a positive. All right.

[00:38:24]So that is our solution. But this solution must be presented. All right.

[00:38:29]Remember the presentation must be x is greater than a less than b. x greater than a but less than b. So our x is greater than what? As we can see guys, x greater than it is open to this to greater. It is pointing to x is greater than 2. So you write it that way x is greater than 2. But the same x as you can see it is pointed by five. It is less than what? It is less than or equal to 5. So that is the final presentation that you can have and not forgetting the number line from what we have uh from the values of x. All right. Uh we only these are positive values. So there's no need of us presenting a zero but we can just show a zero there. Two and a five those are the most important. So x uh these are the x values at two as we can see it is greater than you're just going to represent with a circle right at five because there's an equal to you present a what a shaded circle so greater than less than just a circle greater than or equal to less than equal to a shaded what a shaded circle then you join just like that.

[00:39:55]So this is it. You have represented on a what? On a number line you now have a number a number line just like that. So it's all about uh revision. As you consider this part just work out uh as many questions as you can. All right. So we're done with this part. Let us move on to 3.2. two. It is also part of our equations. But this one is a word problem that we are given. And we are being told here that on 3.2 a father is twice as old as his son.

[00:40:42]So that's like the current age or what we have now current.

[00:40:50]So we told that the father is what?

[00:40:54]The father is twice as old as his son.

[00:40:58]So the the age is referred from his son.

[00:41:05]So let uh the son's age be what? Let the sums age be equal to X because the father is referred from from the son. So if we have the son's age, we can have the father's age. Let's say the son is 20 years, then we know that the father is simply twice 2 * what? 20 from the son's age. It is easier that way. So if the son's age is x, what about the father's age?

[00:41:43]So the father's age is going to be twice two times that twice. It means what? Two times what you're given. If the son's age is x, so it means 2 * x, which is what? Which is 2x. This is the presentation of their ages. Now currently but we given now another statement after that 12 years ago 12 years ago what does it mean 12 years back we are referring back 12 years back the father was three times the son's age. So in our case guys, we do not know the son's age.

[00:42:40]So let's consider the statement that is that states 12 years ago. 12 years back.

[00:42:47]So meaning to say 12 years back we are going to notice that the son's age will be equal to what? We are going back currently or now it's x. So 12 years back it means we must subtract from the current age from the age that he is at right at the current moment that's 12 years back the son's age was going to be x - 12. So this is the the condition.

[00:43:25]What about the father's age 12 years back? It is the same thing.

[00:43:32]Now the father is 2x. So meaning to say 12 years back we're going to subtract from the 2x. So it is 2x - what - 12.

[00:43:44]That is the condition of what? 12 years back 12 years ago.

[00:43:51]But we are told that in this or in their given years of 12 years ago, the father was three times the son's age. So we can form an equation from that statement.

[00:44:10]Form and what an equation from that statement because you're told that the father was what? Three times the son's age. 12 years back 12 years ago. The These are the ages, but they're in terms of x. So, the father's age 2x - 12 must be equal to 3 * 3 * the son's age. So, that is 3 * what? The sun's age. For those 12 years big, what is the sun's age? It is x - what? It is x - 12. three times the son's age.

[00:44:50]That's for 12 years ago.

[00:44:55]The father was three times the sun's age. So that's a statement. From that statement, we form an equation. From that given statement, we form an equation where we can solve for what? Where we can solve for x. and solving for x we are determining how old is the sun now currently the sun is what x years so we simply have to solve for what uh solve for x how old is the sun now the sun is what x years old so simply solve for what for x from the given equation so those are the word problems you form an equation from the statement that we given in terms of the unknown value. In our case, we chose x. So, we're going to use x. All right.

[00:45:54]So, let us solve guys. Simply expand brackets. 2x - 12 is = 3 * x. Then 3 is going to multiply -12. So, 3 * x + 3x. 3 * -12. That's - 36. We can collect our like terms in this case considering the like terms that 3x to the other side is going to be a 2x - 3x is equal to - 36.

[00:46:27]Then -2 to the other side is going to be a positive. So that's a pos2. Then that means 2 - 3 that's -1x. So -x = -36 + 12 that's -24.

[00:46:44]So simply divide by that -1. X is = what? 24. Remember X represents the sun's age.

[00:46:56]Now at a now condition the sun's age is what is X. So therefore we have answered the question. So that's uh the sun's uh the sun is what is uh 24 years now currently we are not asked about the father's age it is the son answer the question so exam it is not what what you know it is about what you being asked answer the person properly understand the question the given information you form an equation from there all right let us consider 3.3 another part of the question where we have got a statement and another statement and another part that we are being asked to determine uh from these statements uh if x cubed. So a statement that we given as if it must be used some some x cubed + 1 /x cubed is given that it is equal what? It is = pos2 1 / x cub is = 12. All right that's an equation and x^2 + 1x^2 is = 7. All right, let's let us just try to figure out x^2 + 1 /x^2 is = 7. Then find this.

[00:48:42]All right, determine the value of x² + sorry this is just x. All right, the value of x + 1 /x.

[00:48:54]Now they addit three there. All right, so take note. We are supposed to just working with this just like x^2 + 1x^2 x cub + 1x cub. So let us just find x + 1 /x from that answer we simply have to add what to add 3. All right. So that is the whole idea. How can we determine x + 1 /x from what we're given here? Uh from the first one.

[00:49:27]All right. I mean from the first one we can factoriize. Let's try to factoriize because we need x + 1 /x. The second one cannot be factorized. This one cannot be factorized. We do not have a sum of two squares. A 2 + b 2 cannot be factorized.

[00:49:49]All right. So here factorization cannot happen. But the first one can be factorized because we have got the factorization of what? Sum of two cubes.

[00:50:02]Sum of two cubes. This can be factorized. Uh thus x cubed here x cubed plus 1 / x cub it is the same as 1 / x to the exponent of 3 like this. So it is the condition of a sum of what? two cubes where you've got all right let me just write it down where you've got x cubed + y cubed being factorized as what x + x + y all right into x 2 + x that is going to be negative this one changes into axy + y 2 so it's all about the first and last term condition the first term being x, the last term being y. In our case, our first term is what?

[00:51:02]So that is what you just have to consider guys. Your factorization is just factorization that you're having here. So in our case, uh this was going to give us what? Let me see.

[00:51:15]All right, this is not going to help.

[00:51:17]Let me just rewrite let me just rewrite it here. x cub + 1 / x cub that is the same like this to the exponent of 3. So that's x + y. The first term plus the last term. So our first term is x and the last term is 1 / x. It's x.

[00:51:45]The last term is 1 /x. So that's the condition that we must apply x + y.

[00:51:53]Simply add the first and the last term.

[00:51:55]So first and last term x + 1 /x and we're going to have x² which is the first term squared.

[00:52:08]So our first term is x. So that's going to be squared x²us x y. So that's the first * last first * last that's our x y. So first term it's x last term 1 / x. So if you multiply this is going to give us a one. This is just one one. Remember denominator and numerator concept this was going to give us the one. So this x y was going to give us -1 + y^ 2 that's our last term squared. Our last term is what? 1 /x. But it must be squared. So that's 1 2 which is 1 x² 1 / x². So it's + 1 / x².

[00:52:59]So that's we managed to factoriize from the sum of two cubes. We're used to that. So just factoriize then consider from that to say what we are being asked here uh like I say they added a three. So let us focus on finding x + 1 /x which is exactly what we can see here this one x + 1 /x. So the rest must be replaced from these given algebraic identities that we are given on top.

[00:53:40]We must obtain we must obtain from this.

[00:53:44]Remember x cub + 1 /x cube this is the same this one.

[00:53:50]So we can replace it by its equivalent which is the numerical value which is 12.

[00:53:57]So we going to have 12 the numerical value for this from here is equal to x + 1 /x the one that we want to determine here times the bracket here in this bracket we have got x² and 1 / x² which must be replaced by its equivalent. We have an equivalent for this x + 1x^2 = what? 7. So this whole part can be replaced by what? A seven.

[00:54:41]x + 1x^2 is = 7. Then a -1 was going to remain as it is. Remember what you said that you want to find the value of x + 1 /x.

[00:54:55]All right. So that's 12 is = x + 1 /x * what 7 - 1 that's a six. This is a product guys. So you can simply divide by what? Divide by 6. Divide by 6. Uh thus we're going to have the numerical value of what? x + 1 /x. x + 1 /x. That's 12 / 6 which was going to give us what? A 2. So numerically we obtained x + 1 /x. But our question was not about x + 1 /x. It is also adding a 3 + a 3. So this is what I was saying. Simplify this. Then add a 3 on your answer from x + 1 /x. We are simply going to add a 3. and a three both sides because it's an equation. So whatever that you do on the left hand side must be done on the right hand side. So therefore x + 1 /x + what + 3 is = 2 + 3. The two that we had originally here you add three.

[00:56:18]So that is the condition we going to obtain what a positive five. So this is it.

[00:56:25]x + 1 /x + 3 must be equal to what? Must be equal to 5.

[00:56:34]That is how you play around with these uh typical questions. Uh so revision is most important guys. Revision just have to consider your basics from that. These are just basics of your revision.

[00:56:50]factorization and so on and so on that is very very important. All right, let us consider another part which was now on question number four.

[00:57:04]So question number four you're given uh this is now under our functions. All right, we're given f ofx.

[00:57:12]All right, our f ofx which is um minus that's a negative here. Uh take note.

[00:57:19]Yeah, that's - x^2.

[00:57:22]All right. So, our f ofx is = - x^2 + 4.

[00:57:30]Then we are being asked to sketch on 4.11. Sketch the graph of f. Clearly indicate all the intercepts with the x's on the graph. All right. We do understand for this type of a graph guys which is taken from - x² + or minus c or it is x² plus or - c uh it is symmetrically about the yaxis symmetrical about the yaxis and c is the y intercept so in our case from what we had given the four that we are seeing here is our y intercept where the graph crosses the yaxis where the graph crosses or passes through the what? uh the y-axis we know that is going to be at what at pos4 and the shape of the graph from a negative all right from that negative the coefficient of a I mean of x² that coefficient when it is a negative it means our graph is going to be concave down facing down like this remember all that information then we can simply find uh the x intercepts because the y intercept cept is at what?

[00:58:52]At four or you can say when x is = 0 y= what? y= 4. Now for the x intercepts y is supposed to be = 0. So that's our f of x must be equal to z. So that's your remaining with - x^2 + 4 being= 0. Solve for x - x^2 4 to the other side that's a negative cancel negative x^2 is = 4. So to find our x introduce the square root and that x was going to be what? Plus or minus 2. The square root of 4. Uh that's plus or minus 2. So your graph is called the x intercept. These are the x intercepts that we have at pos2 and -2.

[00:59:39]Let's just say -1. Uh let's say we've got -1 -2 uh positive 1 2 uh 1 2 3 4 5 and so on and so on. So that was uh the condition. Remember the y intercept at what? Pos4. This is your y intercept. X intersects there are two values -2 and what positive2 -2 and pos2 in the xaxis and remember the shape it is concave down facing down so you're just going to have your sketch uh passing through -2 to the y intercept uh right to the y intercept at four remember it's going to face down so it's going to go back to this pos2 two like that symmetrical about the yaxis. So that was our function f from what we are given.

[01:00:42]We've got all intercepts of the x's minus 2 and two the x intercepts for the y intercept. All right, just like that. So from the same f of x 4.12 write down the range of f. So remember the range you're referring to the y values they are limited from where to where the y values that's the range. So y is limited to four going down. Y is supposed to be less than or equal to 4.

[01:01:17]It is limited. It cannot be found above that. So it is less than or equal to 4.

[01:01:24]Or you can simply uh say this y is supposed to be taken from negative infinite negative infinite up to positive what? Up to pos4.

[01:01:37]Fromative infinite up to what? Up to pos4. Anyway that you're going to take uh from there is applicable for any value of x that you're going to substitute.

[01:01:51]That was the condition. Then uh the values on 4.13 determine the value or values of x for which f of x is greater than zero. So f of x to be greater than zero that's above the x-axis where the graph is what above the x-axis. That's where we are referring to our f ofx to be greater than zero because y is equal to zero in the what? In the x-axis. So f ofx to be greater than zero that's above the x-axis. So for which values of x?

[01:02:27]For the values of x in between -2 and two the critical values those roots or solutions uh the x intercepts uh remember it was -2 and what and positive2. So our graph to be above the x-axis it it happens in between it is happens what in between those values. So that's x has to be what? Greater than uh greater than -2 but less than what but less than two. So that is the condition of our x values or simply our x is being taken from -2 to what to two uh like that. So that is how you can present your answers uh in that manner.

[01:03:18]All right. Then another part describe the transformation from f right. So take note about this.

[01:03:26]The transformation from f to h in words if hx is equal to this right. What transformed f to what?

[01:03:38]f to h. All right. So here we can simply look into the given functions and see how one uh is transformed to another.

[01:03:53]Actually how f is transformed to what?

[01:03:57]To h.

[01:04:00]How f was transformed to h. Our f of x is -x 2 + 4. Then this is our hx. Our hx is = into x -1 into x + 1. So we can't actually see we cannot see properly the transformation because here we've got brackets. This one it is in a simplified manner. So let us just try expand our brackets uh rewrite in this format so that we can properly see uh the transformation which took place. x -1 x + 1. This is a difference of two squares from what we had before. A^ 2 - b 2. So it was taken from x^ 2 - 1 2 which is simply x^ 2 - 1. So that's our hx multiplying by the negative outside the bracket. It was going to be * x^ 2 - x^ 2 *1. That's a positive 1.

[01:05:06]So let's compare our hx is - x^2 + 1. If you take a look from our f of x the first part here - x^ 2 - x² exactly what we had before here exactly what we had. So we ask ourselves what actually happened now to obtain this hx from f of x from f ofx f to h it means h was taken from so writing hx from what from f of x it is going to be written as - x^2 + 4. So from this f of x, this is our f of x guys. This is our f of x.

[01:06:00]What happened in order for us to obtain this one here? Because that is the only difference that we have. This is - x².

[01:06:08]Also this is - x². So to obtain this one, what happened? We simply subtracted a three from four. 4 - 3 gives us what?

[01:06:20]Gives us a one. So our hx as it was taken from f ofx it simply means this is this whole party this whole party that we see represents f ofx originally here this is our f of x as it was then we simply subtracted a what a three from there so that we can obtain what 1 4 - 3 gives us what a 1. So it simply means our hx from f ofx it simply means what f ofxus 3 this is f of x remain. So what is this transformation guys? f of x - 3 the original function we are subtracting three from the original function that's the graph simply was shifted three units going down f ofx minus 3 4 - 3 these are y values that are being affected the graph uh it it is just like this all right another thing that you can simply use um without this without this part here like to say it was f of x - 3. All right, you can also do it this way.

[01:07:36]Another way uh that you can tell f ofx it was at pos4.

[01:07:45]f of x was at what? Pos4. It was like this at pos4.

[01:07:51]This was our f. Our h it's - x^2 + 1.

[01:07:55]What simply changed is the y intercept.

[01:07:58]The y intercept is now at what? The the y intercept is now at pos1. This is our this is our h. So we are saying what happened from f to h. What happened? We simply moved from 4 to one. We are moving what?

[01:08:20]Horiz. I mean uh we are moving down vertically. So this is a a shift of how many units? 4 - 1 that is what? Three units. Exactly the same explanation that I was having a translation or a shift of what? Three units downwards. This is what we can see from our f of x. So we can tell it from the graph to say all right the y intercept you just work with the y intercept. The y intercept is at what pos4 for f. The y intercept for h it's at what? It's at positive one. So what happened from f going to h from from from this is your starting point f to h you are moving three units uh going down that is our transformation. All right so we can say uh therefore we can simply say uh that's a shift. We have got a shift. That's a shift.

[01:09:21]We have got a shift or a translation. A shift or a translation of what? Three units. Three units. The minus three uh three units uh downwards. Going down.

[01:09:39]Three units going down. That's the presentation of the uh transformation.

[01:09:46]What happened from the original function f of x to give us hx? Write them in the same format, the same format. Expand brackets. From that you can tell what actually happened from one to another one according to what you're given. So that is it. Or you can simply work with the graph. Just sketch the graph. Then you can see all right it's simply uh it was simply shifted three units going down from 4 to 1 that is the idea then 4.2 we are given to determine the equation of the function kx all right that's 4.2 to our kx is equal to a / x + q with the horizontal as simple to and y = 3. So y= 3 is an as simple to simply put it in place of q. These are the y values. So it means kx = a / x + what + 3 the y value. So what we need here is to find the value of a. How can we find the value of a? They gave us another statement.

[01:10:55]The line f ofx is = x + 2 intersects with the graph of what? Of k. This graph of k and there's another line. They do intersect. Where they intersect is at point 2 4. That's a point of what? A point of intersection. Remember guys, the point of intersection simply means is the same. This point 2 4 lies on kx and f of x at the center it lies on these two graphs and the point of intersection the two graphs are equal.

[01:11:43]So we are given this.24 as an advantage to use because this.24 as it lies on f ofx it also lies on what? On kx. So if it lies on kx we can use it to our advantage because we are saying we are given the value of x and the value of y. So we can find uh a from there by substituting the point that we're given. We have to find uh the value of what of a. Since y is 4, that's our kx. So 4 is = a over x. Our x is what? Our x is 2 plus what? + 3. Solve a. We've got an equation. All right. Uh take the three to the other side. That was going to be a -3. 4 - 3 that's 1 is = a / 2. This is same as over 1. So you can simply determine the value of a by what? Cross multiplying 1 * a that's a 1 * 2 that's a 2. So if a is equal to 2, you can substitute there.

[01:12:59]You have got the equation of our kx. So, kx is equal to a which is 2 over what x + what + 3 just like that.

[01:13:13]That's the equation and we are dealing with the equation of what a hyperbolic like that. Uh so the question here was just to find an equation determine uh the equation of the function. All right.

[01:13:24]So just use that information guys properly. All right. Question number five. You are given f ofx and gx and they are not drawn to scale. It is just a a a sketch that you're given here. Uh that's our f the straight line and the g which is the hyperola as we are given and that's the x and the yaxis. So this point will be at zero where these two meet. All right. Then you're given that uh these two graphs you have got r and s as points of what intersection. So that's r and s are the points of intersection. R here and this point s all right then also we're given dbq is a line perpendicular to the x-axis.

[01:14:24]So that's dbq.

[01:14:27]This line is what? Perpendicular to the x-axis meaning say it is found at 90°.

[01:14:34]So throughout this line at Q we given the value of x is -1. So it means throughout this line x is = what lines parallel to I mean lines perpendicular to xaxis are given as x= x=0 x= 1 x= -1 like that. So that line since we have got the value of x here uh this one to this one what is the value of x is -1 throughout the line all right x is equal to what -1 and also we are given uh right with the d and b on g and f respectely the coordinates of what q is 1 0 uh so that is the point where we are taking this Q1 0. So like I say this -1 is going to be throughout the line. All right. So from there you are now asked to determine uh 5.1 uh that is the equation of the line of symmetry of G with a negative gradient. So as you to consider our G as our hyperola the equation of the line of symmetry which is also referred to as the axis of symmetry with a negative gradient its equation is given as y = - x + c. For a positive gradient y = x + c positive x negative gradient negativex.

[01:16:18]So what you need is to find the value of C. But what happens is that whether the line has got a negative gradient or it has a positive gradient, these two they pass through the uh what we refer to as the a simple to simple to. So for the as simple to we can see that our x from what we are given all right from our f ofx uh I mean from our gx our gx is taken as -2 / x + 2. So x here is simply at zero then y is at what at 2. So that's 02 as our as simple to say at x is = 0 in the yaxis and y = 2 y = 2 we do not know somewhere there that's part of what of the simple to any part of these two lines the line of symmetry whether it has a negative or it has a positive gradient they pass through the simple at that point where x is zero and y is two from the equation you must be able to note your asimple tote. So you use that uh to the advantage uh the part of what of our as simple to given the x and y value. So this will help us to find the value of what of c from y which is two our x is what it's a zero there. So find what?

[01:18:01]Find uh the value of C. Solve C. All right. So that's it. 2 is = 0 + C.

[01:18:08]That's C. So therefore Y = - X + what? C - X + 2.

[01:18:17]That's the equation of the axis of symmetry, the line of symmetry of G where we have got what? A negative gradient. All right. Another part of the question it was simply to 5.2 to determine the coordinates of R and S for six marks. So the coordinates of R and S guys are here we've got R and S. We can simply find this from the two graphs since they are intersecting.

[01:18:50]These two graphs they are supposed to be equal at the point of intersection f of x and gx are simply what equal. So that's how you can simply find what uh that's how you can simply find these coordinates because at the point of intersection uh the two graphs are equal f ofx must be equal to what to gx at the what? at the point of intersection.

[01:19:20]So if that is the case, let us just equate the two graphs. - x + 1 is supposed to be equal to gx - 2 / x + y + 2. That's an equation that you formed.

[01:19:35]So from this equation, simply solve for x. All right. Let us consider this is pos2. We can simply take it to the other side of the equation. So it's going to be -2 there. So it's - x 1 - 2 that's -1 = - 2 /x. So we have removed this.

[01:19:59]This can be written as a fraction. So that's over one. So we can simply cross multiply. All right. Just like that. You can simply cross multiply. So x is going to multiply - x -1 1 * -2 -2 expanding brackets that's -x^ 2 -x is = -2.

[01:20:24]Remember the format here of a quadratic equation the power being 2 ax^2 + b x + c is = 0. One side must be equal to zero. So moving the minus2 to the other side, it was going to be a positive. So that's pos2 is equal to 0. We can solve for x, but we can at the same time avoid that x² be negative. Simply divide by throughout your equation. So this will be positive x^2 positive x pos2 to -1 that's a -2 = 0. So how can I solve for x? Uh by what? By factorization.

[01:21:11]We simply need two brackets that we are going to consider. Uh since the coefficient of x² is 1. This was going to be x and x here. So what are the factors or products of what? -2 which will give us the coefficient of x.

[01:21:32]And what is the coefficient of x? It simply means that's a one. If you're not given the number that's a one, positive one. So we need factors products of -2.

[01:21:42]What are these factors of what? -2. -2 and 1. 2 and -1. But of these which part when we add we going to obtain what?

[01:21:54]Positive one. That's 2 n minus one. If we add here this is going to give us what? A positive one. So that's 2 pos2 and what -1. So let us find the value of x + 2 must be equal to zero from the zero product or x -1 must be equal to zero. So those will be the values of x. 2 to the other side will be a negative minus one to the other side is going to be a positive. So you've got two possible values of x. But the question here it is not about the values of x.

[01:22:32]All right? Is about the coordinates having these points r and s. So according to uh this guys this are yaxis remember. So to the right those values are positive. So s is supposed to be at a positive value. This s must be at a positive value. So of these values which one is positive? + 1. So s must be where x is one. And this area here must be on a condition where our x is what?

[01:23:09]Negative. And that's oh sorry for that.

[01:23:15]And that's a what a -2. That's a condition for our our there at what? At -2. All right. So that was the condition guys. You solve for x from the two functions that they are equal at the point of intersection f ofx and gx. They are supposed to be what?

[01:23:40]They are supposed to be equal. Now that you have got these values of x to find the y values simply substitute these into our given functions. Remember these two functions they are equal. So uh thus to find the exact point or the coordinates for r there when x is what?

[01:24:06]-2 when x is -2. Substitute in any one of these two equations because they are the same. So y is equal to I'm going to use f ofx that's - x. Our x is -2 + what + 1. So y is going to be negative. That's a positive2 + 1 which is 3. So that's 3.

[01:24:31]When x is -2 at what? At r there. So r x is -2. Y is what? y is pos3.

[01:24:43]All right. Then we do the same thing for the point s. But s here guys is an advantage because it lies in the what?

[01:24:52]In the x-axis and in the x-axis we know that y is equal to zero. So at the point where s is there's no need for you to substitute. You can substitute is not a problem. But you can see guys x is 1. So y is going to be zero in the x-axis. Or you can substitute here 1. So it's going to be -1 + 1. That's a zero. So these are the two points R and S.

[01:25:19]Those are the two points R and S. So you see understanding of functions sometimes makes it easier in terms of the simplification on its own because some of the things we need to substitute. You can tell from the graph that you're given or from the function that you're given. 5.3 the length of DB. Uh that was three marks. The length of DB uh DB remember we told about these two guys that they are actually on a straight line and this straight line that we are talking about is perpendicular to uh to the x-axis like I said before that is uh the equation of the line uh throughout that we given is x= -1 one.

[01:26:11]So what you simply need here when you are to find the distance or the length of DB all right that's 5.3 to find the length of DB it is simply the difference since this is a vertical line you work with the y values the y at d minus the y at what at b so we can use that as y1 and y2 y1 1 y2 y1 - y2 or you can say the y value at d minus the y value at b that's the length of what uh that's the length of uh db the y value at d minus the y value at b. So the y value at d here this is where we have got our our g. This is the graph of g. This this g this is g. So we need the y value.

[01:27:14]What is the y value there? What is the y value for a condition where x is what?

[01:27:21]Minus one. That is the that is it. That is the only that is the only thing that we're talking about guys. So from gx where x is = -1 the y value the one that we are referring as y is the value of f of y I mean g of what this is from gx so that's g of1 substitute -1 in gx and find the y value that's the idea so that was going to be uh -2 over the value of x is what? -1 just like that + 2. So this is going to give us 2 + 2 which is what? Uh 2 + 2 which is 4. So the y value at d the y value or y1 the y value at this point is what is 4.

[01:28:21]That's it.

[01:28:23]That's the y value minus y b the b value here at point b or y2.

[01:28:32]This part now lies on what? On f.

[01:28:38]B lies on f. So where it lies? It is when x is -1. But on what? On f. It lies on f. So that is the condition that you now have for yb. It is simply for that same value of x which is -1 but you are substituting in f of x. So that's f of -1 which is simply in f of x substitute a what? A1 there. Uh that was going to be 1 + 1 which is 2. So the y value at b or the y value this point here is what is two. So the difference is the length of DB between these Y values that is uh the length of DB. The Y at point D is what is four minus the Y at point B which is what? Which is two. So the difference 4 - 2 is going to be 2 units.

[01:29:36]So that is the length of what the length of DB or the length of BD. So even if you use y2 minus y1 you must take note at the end when you're referring to the length you write it as the magnitude the positive value even if you have got a negative value when you're calculating the length you must write your answer as what as a positive. So that is the whole idea of uh determining the length.

[01:30:06]They'll be just playing around with you guys. Find the y values, subtract them.

[01:30:11]If the line is a vertical, if it was a horizontal line, you work with the x values x1 x2. So the length is going to be x2 - x1.

[01:30:24]For this one, a vertical line y2 - y1. That is the length. So that is the whole idea. So whenever you've got a vertical line, you work with Y values. A horizontal line, you use what? X values to determine the length of a line. The length of a line.

[01:30:44]So that was it for this question paper.

[01:30:46]You just need to work out as many question papers as you can. Uh as we do understand guys, revision is most important. All right? Just try to do as many question papers that can help you.

[01:30:59]And also check out uh the playlist for these uh June revisions. All right, just go through the playlist and see other revisions that we have done before because these questions as you can see they're just similar guys. So that was it for 50 marks. Uh we're going to look into other questions uh till we meet again.