IB 2026 Maths AA HL Paper 1 May TZ3 FULL EXAM - Worked Solutions

Added: 2026-05-20

[00:00:01]Okay, welcome back to the channel and here we go with higher level analysis and approaches the paper one from May 2026 from time zone 3 which is the Americas. Uh we're going to go through full work solutions here and we're going to go at an HL speed. So let's get right into it with question one.

[00:00:18]All right, question one was a uh kind of statisticsy problem involving a sample of 60 students from a school of 600.

[00:00:27]Okay, we got the breakdown on the number of students in each grade right here.

[00:00:31]And we're told we're doing a stratified sampling technique. So, we're trying to match the proportion of our sample to the uh different groups in the population. Okay. So, we start off with two marks to find the number of 12th grade students in the sample. Pretty simple calculation here for HL. This is going to be the number of 12th grade students, which is 170, over the total number of students, 600, times the size of our sample, which is 60. Uh, the trap here was trying to get you to multiply these together when you could just reduce this number down. Okay, you're going to get 6* 170 over 60. You can take off another zero here as well as cancelling the sixes and essentially give yourself 17 students without really doing any complex calculations there. Okay, the question continued given us this box and whisker diagram. We get one mark to state the median. We know the median lies right in the middle of this box where the line is. So it's very easy to say here for one mark that the median was equal to $80. Okay, the money comes from uh the sample measuring the allowance that these students got. Okay, we're then told that the interquartile range was 24 and asked to write down the value of a. So we know the difference between uh 66 and a should be 24. Okay, so 66 plus 24 that should give us the median which I believe would be uh 80.

[00:02:00]Oh, sorry that would be the value of a.

[00:02:02]The IB was then actually pretty generous when they gave us both the outlier formulas. So Q1 minus 1.5 * IQR and Q3 + 1.5 * IQR. We had to determine the smallest possible allowance to not be an outlier. So we're using the lower bound outlier formula. And we're just substituting in here. So 66US 1.5 * 24. This is 66 uh - 36. 1.5 * 24 should give us 36 and we subtract those two and that should give us 30. 30 would be $30, sorry, would be the smallest allowance not considered an outlier. Okay, so a very wordy question one to get us going on higher level paper one. Question two is a triangle problem, a non-right triangle problem where we got this picture given to us and we were given that cosine of angle ABC is 1/5. Okay, so we're going to call this angle with vertex at b theta to help us work through this.

[00:03:01]We're asked to find AC. So that's the length of this missing side. We'll call that X. We dive straight into some cosine rule here. Okay, so x^2 = 5^ 2 + 6^ 2 - 2 * 5 * 6* cosine of theta which was given to us. So that's time.

[00:03:21]We got a one unknown equation here. So we can just start cranking numbers.

[00:03:26]Okay, we'll do a little bit of simplifying here. Canceling the 1/5 and the 5, which will give us a simply a minus12. We combine all this together.

[00:03:35]X^2 is 49, meaning X must be seven. And I believe the units were centimeters. So that gets us our first three marks for AC. Okay, it's okay if you introduce a variable and don't use exactly the notation they do. IB will still take this as correct.

[00:03:52]All right, second part, part B. Part one, we're asked to show that s of ABC angle is 2<unk> 6 over 5. Okay, so lots of different ways you can do this. I'm going to decide to go with the Pythagorean identity. So 1 = cosine^ 2 + sin^ 2 theta. Okay, I know cossine theta is 1/5. So this is 1 + 5^ 2 = sin^ 2 theta. This obviously means sin^ square theta is 24 over 25.

[00:04:25]We take a square root of both sides to solve for sin theta. The<unk> of 24 is 2<unk> 6 and the<unk> of 25 is 5. Okay, so that gets us to the answer we wanted.

[00:04:37]That completes our little show that proof right there. The last part is a hence. So using that value probably find the area of this triangle. Okay, area of a non-right triangle is 12 5 * 6 * s of the angle in between which for us is that theta. So this is 12 * 30 * 2<unk> 6 over 5. Okay, pretty easy number crunching here and we can take all this down and get ourselves to 6 <unk>6 cm squared for our final answer. Okay, so question two, pretty familiar.

[00:05:15]Nothing crazy going on there. All right, question three is a binomial expansion problem that starts with just showing the value of a binomial coefficient or a combination. So we're asked to compute or at least show that 8 choose 3 is 56.

[00:05:30]Okay, so if we just use the combinations formula, we know this is 8 factorial over 8 - 3 factorial. So that's 8 factorial over 3 factorial * 5 factorial. Okay, we clearly don't want to figure out what 8 factorial is without a calculator. So, we're going to use the properties of factorials to cancel this down. Okay, it's simply going to be 8 * 7 * 6 on top. That'll cancel the 5 factorial. And then 3 * 2 * 1 underneath from the 3 factorial. Okay, 3 * 2 * 1 is just 6. So, we cancel all those out. 8 * 7 gives us the 56 they wanted. And they're kind of stingy with the one mark right there.

[00:06:11]Pop B asks to find the value of m. So we need to read a little deeper here. Okay, we have this expansion 1 + mx all raised to the^ 8 and we're told the coefficient of x cub is 8 * the coefficient of x^2.

[00:06:25]So let's find those terms in this expansion. Okay, we would have 1 to some power and then mx all cubed. And we know that if that's a power three, the power on the one must be a five because those guys have to add to eight. And then the coefficient in front would be 8 choose 3. This should equal to 8 * again another coefficient 1 to some power and then mx all squared. We can work backwards to figure out the power on the one must be a six and the combination must be 8 choose 2.

[00:06:59]Okay. So working through this we've got 56 m cubed x cubed should equal 8 * something m^2. or actually we don't need the x cub because we're just equating the coefficient. So we'll get rid of those. So our job now is just to compute this 8 choose two. So we can do that one quickly. It's 8 factorial over 2 factorial * 6 factorial. We know that's going to simplify to 8 * 7 / 2 which should give us 28.

[00:07:29]So 8 * 28 m 2. Uh we can divide off an m 2 because we're told in the question that m cannot equal zero. So that'll simplify down to 56 m = 8 * 28. Right? We don't want to multiply that 8 * 28. So let's just divide off the 56. 8 * 28 over 56. And we'll go ahead and write that 56 as 8 * 7 to help us reduce that further all the way down to m= 4. Okay? Okay. And they asked us for a singular value and that's the value of m to get this completed. Okay.

[00:08:09]Question four is a trick question and we're asked to find the exact value of secant 75° and we need to express our answer in the form m + n where m and n are positive integers. Okay. So we're going to start by computing the reciprocal function of this which is cossine 75°. When we get that ratio we'll just flip it over to get seeant.

[00:08:27]Okay. So we're going to use a compound angle identity here. We're going to split this 75 into the sum of 45° and 30°. And then applying our compound angle identity, this is cossine 45° * cosine 30°us sin 45° sin 30°. Okay, these are all values we should know from our unit circle. So we'll go ahead and substitute in four numbers here. We won't use rationalized values knowing that we're going to reciprocal this anyway. Okay, so the two 45s are one over <unk>2. Cosine of 30° that should be roo<unk>3 over two. And then sine of 30° that's a half. Okay, so we're going to end up with roo<unk>3 - 1 over 2<unk>2. That's going to be the value of cossine 75°. Okay, again we'll reciprocal that to get secant 75°.

[00:09:21]That will give us 2<unk>2 over<unk>3 - 1. This is actually the value of secant 75°. is we now just need to manipulate it into the form they want. So we'll multiply by the conjugate of the denominator.

[00:09:36]So roo<unk>3 + 1, right? This will give us 2<unk> 6 + 2 on top. And in the denominator, it's going to give us roo<unk> 9 uh minus 1. So this is 2<unk> 6 + 2 all over two. That'll simplify down nice and easily to just uh actually sorry, I'm missing a root two there.

[00:09:57]It's a 2<unk>2 at the back. 2<unk>2 there. So this would be<unk> 6 plus <unk>2 for our value of secant 75° which is in the form they wanted. So that's how we grab those five straightforward marks right there. All right, question five we did in our sneak peek video a couple of days ago, but we'll do it again here. Okay, this is a simple derivatives question to begin. We're asked to find the derivative of this equation x 4 - 4 - sin 4x. Okay, this derivative should be pretty easy at hl level. Okay, we got derivative of a polomial. So x^ 4 will just turn into 4x cub. And then for this sin 4x, we're going to do a little chain rule.

[00:10:41]Hopefully we know our shortcuts for this paper. So this is going to turn into 4 cossine 4x. We're getting the derivative of the inner function uh multiplying the derivative of the outer function. Okay, so two marks of that. Pretty straightforward. Okay, now comes the trickier part which is the hence or otherwise do this integral. So it's pretty clear here. We're going to use a u substitution.

[00:11:05]We'll take out a common factor of four uh in a minute. We want our u to just be x to the 4 minus sin 4x, right? We're then going to compute its derivative.

[00:11:18]So, du dx is the result we had previously. This is where we'll take the four out. If I can actually factor that four out correctly. Okay. And we're going to see that this part is already matching up with the back part of this integrant. So, that's looking good. We now just simply solve for dx. So, dx is equal to du over 4 * x cub - cosine 4x.

[00:11:44]We can drop all this kind of side calculation into that integral and it'll turn into the integral of u 7 time x cubed - cosine 4x. And then we'll switch the dx out for du over what we just calculated 4 x cubedus cossine 4x. We get the cancelling of the x's that we need to tell us this substitution is successful. We can drag the four out in front as a 1/4. And our integral just becomes the integral of u 7 du.

[00:12:21]That's an easy integral to do. That's u to 8 over 8 plus our arbitrary constant c. And we can simplify this down and substitute back in our u value now that our integral is done. Okay. So our final answer here looks something like that. 1 over 32 x 4 minus sin 4x all raised to the 8 plus your constant c. Okay, so again, not too bad for HL. We're about to get into the trickier ones coming up. Okay, question seven is a tricky sequences and series question. We've got an infinite convergent geometric sequence. This tells us the absolute value of r is less than one. And we're told that the first three terms are u1, u2, and u3. And we're told for the sum of the first k terms is equal to half the sum of the infinite geometric sequence. So we're going to write that as an equation. That would be sk is equal to 12 s infinity.

[00:13:19]And we'll start building things with that in a minute. We're also told that u k + 1 so the next term not in our sum is 1 over 16. Okay. We're asked to show that u1 is 18. And here's how we do that. Okay. So we're going to start with this SK formula. This is U1 * 1 - R to ^ K. And this is then divided by 1 - R. We also have our S infinity formula which is U1 over 1 - R. And that's going to be multiplied by a half in a minute so that we can equate these. Okay. So u1 1 - r to the k over 1 - r should equal 12 u1 * 1 - r. Okay, we can multiply both of these expressions by 1 minus r. That'll get rid of the denominators for us.

[00:14:08]We're also going to multiply by a two in order to get rid of that part of the fraction. So, this will leave us with 2 u1 * 1 - r k is equal to u1.

[00:14:20]If we distribute into here, uh oh, we can actually divide the u1's off. So, 2 * 1 - r k should equal 1.

[00:14:31]Okay, this means 1 - r the k is equal to a half. r the k is equal to uh 12.

[00:14:42]So r is equal to the kth root of 12 whatever that is. Okay. And then we can use this uh 1 16th thing to help us find the first term. Okay. So u to the k + 1 that is u1 * r to the k + 1 minus one using the nth term of a geometric sequence.

[00:15:05]So this is 1 / 16 = u1 r the k. What is r the k? We have it right here. That's equal to a half. So u1 * a half is equal to 116th.

[00:15:19]Multiplying both sides by two this gives us u1 = 18. Okay. And there may be quicker and cleaner ways to do that, but that's the way I did it. So that's what we'll go with in the video. Okay. Part B wants to us to find u of 2k + 1. So we'll work that right here on the side.

[00:15:36]We'll use the nth term of a geometric sequence here. This would be u1 * r to the 2k + 1 - 1. We know u1 is 1/8. So this is r to the 2k. And we can do some sneaky algebra here to write this as r the k squ 2. Okay, that's going to be the same as simply substituting in what we have circled in red. So that's 12 squared. So 1/8 * 1/4 to give us our value here of 1 over 32 for u to the 2k + 1. So it's a little easier part B after a difficult part A. Okay, question eight is a complex numbers question. And this is where we get into the really kind of high level part of section A.

[00:16:21]We're given this complex number Z = 3 + K. We're told that K must be less than zero. So we're looking at a negative kind of imaginary part here. We're also told that um the angle for this complex number must be between negative pi and pi. Okay, we're going to start by kind of getting this idea onto an argan diagram. So we're thinking about a complex number with a real part of three and a negative imaginary part. So it's a complex number Z that's sitting in quadrant uh four here. Okay, we're told that Z to the 5 is a real number and asked to find the value of K in this strange form with a tangent function.

[00:17:01]Okay, so we'll start by applying the Demar theorem. Okay, we know that Z ^ 5 will equal R 5 cossine 5 theta plus I sin 5 theta. Okay, and we want the complex number in this form. So we can think about real and imaginary parts.

[00:17:20]Okay, we know the imaginary part of Z 5 will be r 5 sin 5 theta and we know that has to equal to zero because we're told in the question that this has to be a real number. Okay, so this means sin 5 theta must equal zero which tells us 5 theta must be some multiple of pi. Okay, so n pi where n could be any integer.

[00:17:46]Okay. So if we think about this 5 theta could be I don't know -3 pi -2 pi uh pi 0 pi 2 pi so on. There's an infinite number of options this angle could be.

[00:18:04]Okay. But we know that we want this angle to be in quadrant four. So when we divide off the five we're only going to keep the ones that would be in quadrant four. Okay.

[00:18:16]And this would be 2<unk>i over 5 and<unk> over5.

[00:18:24]Those would be the only ones that we keep because those are the only ones in quadrant 4. So let's cross out all of the others here. Okay. Anything greater than that that gets into quadrant 4 would not be in the range they gave us.

[00:18:37]Okay. So we have two angles here for theta, meaning we're going to get two values of k. And now we just need to figure out um the format for those angles. Okay, the key equation to connect this together for us is the equation for um the argument. Okay, so we know that um where are we here? We know that Z is equal to tan theta tan of our angle which is equal to the real part divided oh sorry the imaginary part divided by the real part. Okay. So the imaginary part for us is k over the real part which is three. If we solve this we'll get 3 tan theta equals k. And if we substitute in our theta values here, we should I believe be done. Okay. So k = 310<unk> over 5 and k = 3 10 -2<unk> over 5. And those are in the format we want, which is a + 10 b pi, where b is a negative number. That's what they told us. That b would beg - 1/5 and -25ths. Okay. Okay, so that's a tricky complex numbers problem as we're starting to get towards the end of section A.

[00:20:00]Okay, strap in here as we tackle the last problem in section A, which is an eight mark proof by induction question.

[00:20:06]Okay, so we want to show that this uh sum of all these fractions is always greater than or equal to 72 for all positive integers greater than or equal to two. Okay, so the first thing we do is define this statement. We'll use the notation SN to represent this statement they gave us. And we start every proof by induction by proving the basis case.

[00:20:27]Okay. So we need to prove or show that.

[00:20:30]Okay. We need to prove or show s of two is true. That's our first case. Okay. So s of two we want this sum. We want n to be equal to two. So that means this sum will be 1 over 2 + 1. And this sum will end when we get to 1 over 6. Okay. Oh, sorry, not six. 1 over 4. So, our first fraction is 1 over 1 + 2. It's then 1 over uh 2 + 2, which is the 1/4 we need. So, it ends right there. And this should be greater than or equal to 7 12ths. Okay.

[00:21:10]Well, if we simplify this, it's 1/3 plus a 4th.

[00:21:15]that should be greater than or equal to 712ths. If you add those two fractions, it's 712ths is greater than or equal to 712ths, which is true. So therefore, the basis step s of two is true. Okay, quite a tricky basis step to start with, but we'll keep going here. We then get to um the assumption step. So we want to assume that s subk is true where this k variable is any positive integer greater than or equal to 2. Okay, so this is just a setup step for the induction step in set three. Okay, so we'll write out what this means. 1 over k + 1 + 1 over k + 2 all the way up to 1 / 2 k should be greater than or equal to 72. Okay, that's all we have to do in the second step is just make an assumption. Step three is where the work is. This is the induction step. We need to show that S of K + 1 is true. Okay, we [snorts] don't have to redefine K because we defined it earlier. But let's write out what this is. Okay, so this is 1 over k + 1 + 1 + 1 over k + 1 + 2 all the way up to 1 / 2 * k + 1. And we need to show that's greater than or equal to 7 12ths.

[00:22:40]Okay. Well, if we simplify this a little bit, it's 1 over k + 2 plus 1 over k + 3. And then we'll write out a few more terms at the end. The very last term would be 1 over 2k + 2. Meaning the previous term would be 1 over 2k + 1 and the previous term to that would be 1 over 2k. So we're trying to connect it to what we did in uh part two or our assumption step. And if we look we see this almost lines up. We've got the 1 over k + 2 1 over k plus 2. We'd have the 1 over k + 3 and the dot dot dot.

[00:23:18]We've got the 2k or the one over 2k right there. So, we've got a couple of extra terms circled in green here as well as a term that's missing, which is this 1 over k + one that we've circled in purple. Okay. So, we're going to try and match these up a little bit. Okay.

[00:23:36]What we're essentially going to do is we're going to take the result from part two that we assume to be true and rearrange it a little bit. Okay. So what that means is that we will get uh the following. We will get 1 over k + 2 plus 1 over k + 3. This is all coming from step two. We should probably tell them that. So this is from 2 + 1 2k. This should be greater than or equal to 72 minus 1 over k + 1. So we've just moved the purple term to the right side here. From here we can use part three a little bit or we can form part three. Okay, by simply adding the two green terms on both side. Okay, so 1 over k + 2 is 1 over k + 3 and so on and so on plus 1 over 2k + 1 over 2k + 1 + 1 over 2k + 2. This should be greater than or equal to 72 minus 1 over k + 1 + 1 over 2k + 1 + 1 over 2k + 2. Okay, now we'll just rephrase our question a little bit. We have the left side here exactly matching the left side we have on our part three. So we're at this stage where our left hand side now matches what we had from part three. And we're just trying to show that this is greater than or equal to 712. So that means this part at the back, which I'm going to uh underline in yellow. Okay, this part at the back must be greater than or equal to zero in order for this to be true because that means we're adding something to 72. Okay, so this is going to simplify to showing that -1 over k + 1 plus 1 over 2k + 1 + 1 over I'm actually going to factor a two out of here. 2 * k + 1. We just need to show that that's greater than or equal to zero, which will mean that that stuff on the left is greater than or equal to 7 over 12. Okay, so we have some like terms here in the first and the last term. Okay, we can combine these together. It's negative one of these things plus a half of those things. So that's minus one of those things or sorry 1/2 of those things. So this will turn into minus one over 2k + 1 is greater than or equal to zero. We can move one term over to the other side and we're going to actually end up kind of cross multiplying here. Okay, knowing that these are positive numbers because k is greater than or equal to two. So we don't have to worry about switching our inequality around. Okay, we'll get 2k + 2 is greater than or equal to 2k + 1. If [snorts] we subtract off the two ks here, we get 2 is greater than or equal to 1. That's a true statement. So therefore, this proves s subk + 1 is true. Okay, by mathematical induction. Okay. And there's a longer version of the conclusion that the IB likes you to write. We can figure that out in the mock scheme, but for the sake of this video, we're just trying to show kind of the algebra of how these problems are solved.

[00:26:58]So section B starts with a little trig equation that we need to solve. f ofx= g ofx on this interval. Okay, so we'll go ahead and set that up. That is sin 2x = 1 - cosine 2x. Okay, we're going to go straight for our double angle identity.

[00:27:15]So 2 sin x cossine x = 1 - and then 1 - we'll go uh cosine no we'll do the one with the sign because we want the ones to cancel. So 1 - 2 sin^ 2 x [clears throat] okay distributing here we will get 2 sin x cossine x is equal to 2 sin^ 2 x the ones cancel. It's then simply a case of bringing this sin^ 2 x over the same side as the other trig functions and taking out a greatest common factor. You can also divide by a two at this point if you like.

[00:27:55]Okay. So we'll actually do a 2 sinx out front. That's our common factor leaving us a cosine x minus sinx. All of this equaling zero allows us to apply our zero product property. So sinx equals 0 or cossine x= sinx. Okay, this is now unit circle work and we have an interval of 0 to pi inclusive. So s would equal 0 at 0 and pi. S and cosine would equal each other halfway through the first quadrant at pi over4. So those will be our three solutions to this trig equation.

[00:28:35]Part B wants us to make a sketch of these two graphs on the same axes and label the coordinates of intersection.

[00:28:43]Okay, so we should be pretty familiar with growing drawing trig graphs at this point. The main thing we want to do is make sure we get things like period and amplitude correct. So we'll start with f ofx which was sin 2x. We know the period for this function is 2 pi over b. So in this case 2 pi / 2. So it's going to have a period of pi. Okay, that's the only transformation happening to this function. So we'll break up our x-axis accordingly. So we can get five points on this amplitude of one going up to one and down to negative one. And it's a standard sign function. No reflections, no translations.

[00:29:22]So we should feel good about drawing five points here and a nice smooth sine wave going through there. Okay, so that'll be our f ofx. Okay, g of x is a little more complicated. It's 1 minus cossine 2x. We may choose to rewrite that as cosine 2x - sorry negative cosine 2x + 1.

[00:29:47]So this is a reflected cosine function that's been shifted up one unit. Okay, it has the same period. So 2 pi / 2 still gives you pi. So, it's still going to complete its cycle when it gets to pi, but it's now going to have a midline at one, meaning it's going to max out at two and hit its minimum at zero. Okay? That reflection means it starts at its lowest point. So, it goes low, middle, high, middle, low.

[00:30:15]Okay? And we can draw our cosine wave going through right there. This green is our g of x. And now, they did ask us to highlight the intersection point. So we'll do that. Those would be at 0 0.

[00:30:29]This is pi over 4 and 1. And then the last intersection point is obviously pi and zero. Okay. So that'll get us our four marks for that trig graph.

[00:30:41]Okay. The last part of this question is talking about area and it's the [clears throat] area enclosed by f and g between a and b and we're told a and b are both positive. Okay. So if we go back to our previous picture, it's talking about the area between these two functions, but between the two positive x intercepts. Okay? So not using the intercept at the origin. It's this green area that I'm coloring in right here.

[00:31:07]Okay? So clearly this picture gives us our bounds as well as which function is on top and which function is on bottom.

[00:31:13]It's g of x minus f ofx. So to find this area, we're going to do an integral from p<unk> / 4 to pi and then it's going to be 1 - cosine 2x - sin 2x dx. And we'll just get to work on this integral.

[00:31:31]So this integral is pretty easy, especially if you know your kind of reverse chain rule shortcuts. The integral of cosine 2x, that one would be 12 sin 2x. And then the integral of sin 2x that would be negative cossine uh 2x / 2 or 12 of that. That's our indefinite integral. To make it definite, we'll tuck our limits on the back. So pi over 4 to pi. Now it's just substituting in. We'll substitute in the pi first. So pi minus 12. This will be sine pi / 2.

[00:32:11]Sorry, not pi / 2. That'll be a 2 pi + 12 cossine 2 pi. And then when the pi over 4 goes in, that's pi over 4 - 12 sin p<unk> / 2 + 12 cosine pi / 2. Okay, more trig function evaluation here. Sine of 2 pi is 0.

[00:32:33]Cosine of 2 pi is 1. Sine of pi / 2, that's one. Cosine of pi / 2, that's zero. So this will all turn into pi uh plus half plus another half coming from the double negative minus the pi over4 here. So we'll get our final answer for this area 3 p<unk> over 4 + 1. And that's the first question in section B done. All right. Question 11 on higher level was a vectors question.

[00:33:04]And we're told we've got this plane pi with this equation and we've got a parallel line that does not lie on pi.

[00:33:11]Okay. And we're told some information about that line. So we're going to start with a picture here. We always want a picture on our vectors question. So here is our plane pi.

[00:33:22]Excuse my poor drawings. We've got the normal vector for pi which we know comes from the coefficients when our equation of our plane is in cartesian form. So this is 2 1 and a.

[00:33:36]And then we have our line. I'm going to put it underneath the plane right here, but it could be on top, but it's essentially running parallel to the plane. And we know a point on this line that it passes through 71 1. And we also know the direction vector for the line, which I'll draw in green.

[00:33:56]That direction vector should be 0 2 1.

[00:34:00]Okay. All right. Now let's see what they want us to do. They want us to show that a is equal to -2. Okay? And that's going to come from a dotproduct. So we're going to do the dotproduct of n and the direction vector which we'll call d. So n do d should be equal to zero because we know that normal vector is perpendicular to the plane. And we know the line is parallel to the plane. So kind of a transitive property there that if one thing is perpendicular to another thing, it's also perpendicular to something that that's parallel to. Okay.

[00:34:32]So this will give us an equation to solve. This is 21 a dotproduct with 0 2 1 should equal 0. So that should give us 2 * 0 + 1 * 2 + a * 1 should equal 0.

[00:34:48]This solves very easily to give us a= -2 just like they told us it would.

[00:34:55]Part B has an sorry, part two has an unusual command term in here which is deduce. Okay, deduce the range of possible values for B. So B is sitting here on the right hand side of our plane equation. And this is actually sneakily difficult if you don't know what to do here. Okay, the only other information we have is that point A is not on the plane. Okay, so the only thing B cannot be is the value that would make that point on the plane. Okay, so how do we find out what that value is? We simply substitute the coordinates of A into our plane equation and then figure out what B would be for that to happen. Okay, so we substitute this in using our new A value. Okay, this would be uh 14 uh minus 3 which gives us 11. So this is the one value B cannot be. B cannot equal 11 because if B does equal 11, A is on the plane. Okay, any other value A will not be on the plane and we'll be good here. Okay, so that's actually quite a tricky uh thing to figure out especially in a timed exam environment.

[00:36:06]In part B, we're told that B will equal two. So, our equation of our plane will become 2x uh what was that? 2x + y - 2 z should equal 2. Okay, we'll rewrite that so we don't have to look at the AB version again. And we're told that we have a line L2 now that passes through point A and has this direction vector.

[00:36:29]Okay, so we can add that to our picture.

[00:36:31]Uh we'll do it in yellow. It's a line that's going through A.

[00:36:37]Okay, this will be our L2 right here.

[00:36:41]And we know it intersects the plane at a point which we'll call N. Okay, so not to be confused with the normal vector.

[00:36:49]This is N. We need to find the coordinates of this point of intersection to get this part of the problem done. Okay, so we'll start by writing the equation uh the vector equation of the line L2.

[00:37:03]So that's R2 equals some vector plus lambda times the direction vector.

[00:37:08]Direction vector we were given 1 1.

[00:37:11]Okay. Uh we know it passes through a. So we can use 711 for the position vector there. Now here's the key part. We can define the coordinates of n in terms of one variable rather than xyz. We know that n is 1 7 plus lambda comma -1 plus lambda comma 1 minus lambda. Okay, just using the equation of the line L2 then.

[00:37:38]Okay, we know this must also sit on the plane. So we can substitute this into our plane equation and then solve for lambda. Okay. So 2 * 7 + lambda + -1 + lambda - 2 * 1 - lambda should equal 2.

[00:37:56]So this is 14 + 2 lambda - 1 + lambda minus 2 + 2 ler = 2. We combine all this good stuff together. We should get five ler that's a 13 - 2 that's an 11. So + 11 = 2 telling us lambda = -9 over5.

[00:38:20]Okay. And then we substitute this back into our uh coordinate for n. So n will be uh 7 + -9 over 5 which should give us 26 over 5. It's then -1 + lander. So that should give us -14 over5. And then finally it's 1 minus lambda. So that will give us a positive 14 over5. Right?

[00:38:47]So quite a lot of busy work there with those fractions. But that's how we get the coordinate of n.

[00:38:53]Part C wants us to find the acute or it tells us the acute angle between L2 and pi. Okay, we'll add that to our picture.

[00:39:01]That's getting kind of busy. That's called alpha. And we're asked to find the exact value of sin alpha. Okay, so we know that we can find the angle between a line and a plane using the following formula. It's sin alpha is the dotproduct of the normal with the direction vector divided by the magnitude of both these vectors. Okay, maybe I'll call this d2 to emphasize it's the direction vector for that second line. We use sign rather than cosine for this formula. So it's not given in the booklet. You can do a bit of kind of a geometry and right triangles to figure that out. But this formula will give us what we need here.

[00:39:39]So we need the dotproduct of n and d2.

[00:39:43]Okay, n for us was uh given to us or we found that. Let me just find where that is. That is 2 one. And then our a value from earlier that's -2. Our second direction vector that's 1 one.

[00:40:02]This is divided by the magnitude of these two vectors. We can compute those real quick. Okay, 2^ 2 + 1 2 + 2^ 2.

[00:40:10]That's going to be the square<unk> of 9.

[00:40:11]That'll turn into a nice little 3. And then the magnitude of the other vector is just 1 2 + 1 2 + 1 2 giving us <unk>3.

[00:40:20]So doing the work here, we have sin alpha is 2 + 1 + 2 / 3<unk>3. So I believe this is going to turn into 5 over 3 <unk>3 and that's the exact value of sin alpha for four marks. In part D, we're given that the distance from a to n is 9<unk>3 over 5. And we want to find the distance between L1 and the plane pi. Okay. So to do this, I'm going to draw a new diagram and I'm going to draw the plane as if we're looking right down the edge of it.

[00:40:54]Okay, so our eyes are sitting right on top of the plane and the plane is going into the page or into the screen. Okay, so think about this blue line [cough] as actually being that plane pi and then underneath this or even on top of it, I don't really know which one. Okay, we have our line L1. So, we had our point A on here that was 7 uh A was 71 1 and we had our point N which had that horrible coordinate with all the fractions that was sitting on plane pi.

[00:41:27]We also found the angle between uh the plane and this line L2 and L2 passed through both A and N. So, in the previous part of the question, we found the value of sin alpha. Alpha would be this angle right here between the uh plane and L2. Okay. But we can use that angle to make a right triangle here that will use the distance will give us the distance between uh the line and the plane here. And it's pretty easy to see when we draw it from this angle that this alpha comes with us. Okay. So we can do a simple bit of right triangle trigonometry to say that sin alpha should equal the distance we want x over the distance from a to n which we were just given. Okay. So x is equal to a n * sin alpha. a n was given to us. That's 9<unk>3 over 5. And sin alpha we just found that's 5 over 3<unk>3. Okay. All this cancelling along the way is another good sign. time we're on the right lines. Okay, so we should end up with simply having this distance between the two things as three units.

[00:42:40]Okay, and that tracks with a small amount of credit we're given here only two marks to complete this bit of work.

[00:42:48]So the last part of this problem, part E, wants us to find the vector equation for the line formed when L1 is reflected in pi. So again, this kind of sideways view will help us here. We've got our plane pi going horizontally as well as our line underneath it. Now, I've drawn on the line underneath, but I don't know if it's actually underneath. Okay, so what do we need to finish with here? We need the vector equation of a line. So, r equals something plus lambda something. Okay, we know that the direction vector can stay the same because that's not changing if this line flips up. So, all we need is a known point on this new line, which we'll go ahead and kind of draw in purple here.

[00:43:26]Okay, so think about this purple as our new line after the reflection.

[00:43:31]Okay, we know that it has the same direction vector because it's going the same way. So we can bring that with us.

[00:43:37]That's still our d vector right there.

[00:43:40]But we need a point on this new line that we'll call a primed. Okay, now we're clearly just going to reflect a in the plane and we're going to do that using the normal vector, right? And if you remember earlier, we did actually find the magnitude of the normal vector.

[00:43:57]The magnitude of the normal vector was three, which magically lines up with what we just found in the previous part, which was the distance between the plane and the line was also three. So that normal vector is kind of magically the right length for us. What we need to figure out now is which side of the line is the plane on in relation to that normal vector. Okay. So, how do we do that? In order to do that, we are basically going to take the normal vector and add it to this point A and see if that gets us onto the plane. Okay. So, we're going to call this other point here, I don't know, let's call this P. Okay. So, we're going to see if O A, let me write it with correct vector notation, O to A plus the normal vector N, this will equal O to P. And we're going to ask the question, is P on the plane? If it is, then we can just add another normal vector to get to a prime.

[00:45:02]But if it's not, we'll need to reverse the direction of that normal vector in order to get on the plane. Okay, so the coordinates of A were given to us. Those were 71 1. Okay. And the normal vector we pulled that off. That was two uh 2 1 -2. So if we add those two vectors together, we'll get nine.

[00:45:28]Uh what will that be? That'll be zero.

[00:45:30]And then this last one be 1. So is this point on the plane? So we'll plug it into the plane equation. 2 * 9 + 0 - 2 * -1. That should equal 2 if it's on the plane. Okay, this is going to be 18 + 2.

[00:45:53]Okay, that's clearly not equal to two.

[00:45:56]18 is not equal to zero. So that tells us this normal vector is currently going the wrong way. Okay, I'll redraw it on the picture here. the normal vector is actually pointing down based on how I've drawn it rather than up. Okay, so not a big deal. All we need to do I'll draw it the correct length too. All we need to do is reverse the direction of that normal vector in order to put two of them on a in order to get to a prime. Okay, so let me put that in vector notation. O to A prime should equal O to A minus 2 * the normal vector. Okay, that minus is going to change the direction on that vector to get us where we need to be. Okay, so this is 7 -1 1 - 2 * 2 1 -2. Okay, if we do this, we'll get a vector from O to A prime. And that o to a prime should be what is that? 7 - 4 that's going to be a three. 1 - 1 - 2 that's a neg -3. And then 1 + 4 I believe that should be a five. And we'll drop this right in here in our vector equation for this line. And get that problem with a tricky little finish for five marks. Done there. Tough problem to end that question. All right. Our last question on this paper one is a differential equations problem in the calculus kind of topic.

[00:47:27]Starts us off nice and easy. We just need to find a derivative. So we'll do a little chain rule here or chain rule within a chain rule. dydx the derivative of tangent should be secant^ squ. We'll leave the inner function alone. So 1 minus e 3x. Okay. We'll multiply that by the derivative of the inside which is -3 e to [clears throat] the -3x. Okay. uh a little bit of uh manipulation needed here with a secant squar identity. So secant squar is equal to 1 + tan^ 2. So this is 1 + tan^ 2 of 1 - e 3x.

[00:48:05]Okay, all of that's multiplied by -3 e -3x. We know we don't really need to manipulate that bit. It's matching up with the answer already. Okay. And then we can just drop in our original y here to get the result they wanted cuz y is equal to tan 1 - e 3x. Okay. So we can see here this part is matching exactly what we were given in the question.

[00:48:30]Okay. So that's how we get going here for four marks. So part B asks us to show that the second derivative d2yd dx^2 is equal to9 when x= z. So we'll start off by calculating the second derivative. d2y dx^2. This is going to equal a product rule. So we'll start with -9 e 3x that's the derivative of the first. Leave the second alone plus leave the first alone. -3 e 3x take the derivative of the second which is implicit. So that's 2y dydx.

[00:49:02]We know that x = 0. We can find y at that point. So this is tan of 1 - e 0.

[00:49:09]So that's tan 0 which we should know is zero. So basically x and y are both going to zero. So when we do that d2y dx^ 2 -9 e 0 I took that derivative wrong right there. I've got an extra plus one. Let me get rid of that. Um 1 + 0 2 here + -3 e 0 2 * 0 * dydx. We don't even have to find what dydx is here.

[00:49:38]This all simplifies down to simply -9.

[00:49:41]So d2y dx^2 is just equal to9 as they requested us to show for five marks.

[00:49:55]All right. Part C. They tell us that this function is now being used for Mclloren series and we need to find that series up to the x squ. So for Mclloren series we need f of 0 frimed of 0 and fprime of 0. Well we just found fprime of 0 is -9. We also found f of 0 was zero. So we just need dydx when x = 0.

[00:50:18]Okay. So if we use the result from the first part part a this is -3 e 0 * 1 + 0 2. So this will clearly simplify down to -3. Okay. From here we can build our Mclloren series. We can say f ofx is approximately equal to 0 - 3x - 9 / 2 factorial x^2. We'll simplify that a little bit and just make this -3x - 9 / 2 x^2.

[00:50:51]And that's how we do that pretty straightforward part C right there.

[00:50:58]All right. Part D connects to part C. If we wanted to with the hence or otherwise we're asked to find a limit. So I think you could apply Latal's rule here but we're going to go ahead and try and do it with the hence. So using our result from the previous part. So we can replace the top part of this with our Mclloren series. It would have a bunch of other terms on the top but all those terms would have a factor of x in it which allows us to take out a greatest common factor of x.

[00:51:26]Okay.

[00:51:29]That greatest common factor then cancels with the x in the denominator. And we can evaluate this limit by simply substituting in -3 - 9 /2 * 0 plus a bunch of other terms which zero out giving us a limit of -3. Okay. And three marks for that one.

[00:51:51]All right. Last part of the exam is part E. We're told that the answer to part C.

[00:51:55]So let's remember what that was. That was -3 - 9 / 2x. Okay, that's equal to the first two nonzero terms in the expansion of this. Okay, I'm going to rewrite this as x * a + bx to the -1.

[00:52:11]And a and b are told to us to be rational numbers and we need to find the value of those guys a and b. Okay, so we're going to do the expansion of this guy using um the binomial theorem extension formula. Okay. So in this case if we compare it to the formula in the formula booklet a is equal to a b is equal to bx and n is equal to1 for us it's a negative one exponent. Okay. So all of this will multiply out. Okay. to the following. It'll be a to the one open parenthesy one minus it'll be b uh over a oh sorry bx over a. Okay, the minus is coming from the negative term right there. And we actually don't need any more terms here because we've got that x sitting in front which will bump the power up for us. Okay, so when we multiply this out, it's a -1 * x minus Um what do we got there? A1 bx^2 all over a. So this is essentially 1 / a * x - b / a^ 2 x^2. And if we equate that to our expansion with the McLaren series.

[00:53:29]Okay. Uh I'm missing some x's in there.

[00:53:32]That should be -3x - 9x^2. If we start equating those, we can see 1 / a is equal to -3 and 9 /2 should equal b / a^2. So we can solve the first one for a, a is equal to 13. The second one will solve as b = uh 9 / 2 a 2. If a was 13, b is 9 / 2 * 1 9th. meaning B will simply be a half. Okay, so two fractions to finish as they said there would be.

[00:54:12]And that completes this paper one for May 2026. Higher level time zone 3.

[00:54:17]Overall, a pretty fair paper hitting a lot of parts of the syllabus, but I think this is a good one from the IB.

[00:54:23]So, thumbs up for them. If you like the video, you've made it this far, please go ahead and give it a like. Feel free to subscribe to the channel as we'll get the paper two and hopefully the paper three up very