2025 Extension 2 Maths HSC Exam Explained! (Part 5)

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[00:00:07]Good day and welcome back to Australia's largest YouTube channel for part five of our journey into the depths of last year's HSC Mathematics Extension 2 exam.

[00:00:16]We are now into the second half of the exam where we are going to start see more mindbending questions. So, let's not wait any longer. Let's see what question 14 has in store for us.

[00:00:24]Question 14A was last year's recurrence relation questions. uh we are starting off with our integral labeled n which is cot the^ of 2n between these two angles.

[00:00:34]Now we're trying to show that this uh this integral is equal to 1 / 2 n -1 - i of n minus one. Okay, cool. So let's oh also they g us a lovely little hint that the derivative of cot is equal to minus cos^ 2. So that will be important. Okay, let's just figure out what we're trying to find in our working out here. Let's have a look at what I n minus one looks like. If we go to our expression and change the n to an n minus one, it looks like this. And so at some point in our working out, we're going to be looking for a cot the^ of 2 n minus 2. So why don't we just start there? Why don't we split this cot the^ of 2n into a cot the^ of 2 n minus 2 by filling in the gap here with a cot squared. Okay, this works out nicely because now when these two guys multiply, adding the powers together gets me 2 n - 2 + 2. So it gets me back to where I started, right? So we have a cot to the 2 n minus 2. And now what can we do with this cot squ? Well, we can reflect on our teachings in the advanced course and remember that someone probably tried to teach us that 1 + cot^ 2 is equal to cosec^ 2. Which means we can write this cot squ as cosec^ 2us 1. Now why is that good? Why have I done that? Well because I made a note of the question that we are trying to find a relationship between cot and cosec um with some calculus involved. So having a mix of cot and cosec in my integral is probably a really good idea.

[00:01:54]It's a really great idea actually because now when I basically do a massive expansion here and expand out this integral, I'm going to have this whole thing times cosec^ squ as one piece. Okay? And the second piece is going to be the whole thing just times by minus one. So it's going to give me minus the whole integral. And that's awesome because then on the end this is exactly what we already identified to be i n minus one. So we already have a minus i n minus one in our working out.

[00:02:19]We just now need to show that this front part, this integral is equal to 1 / 2 n minus one. And we've got our three marks. All right. So, how are we going to achieve that? Well, the question gave us a massive hint that we already identified that the derivative of um the derivative of cot is equal to cosec 2.

[00:02:34]Now, this is really cool because now down here we essentially have a function cot is our function to a power and it's kind of being multiplied by the derivative of that function which is cosec^ squ. So, when you see a function to a power being multiplied by its derivative, that's a massive hint that you should be using the reverse chain rule for this one. Okay, here's the reverse chain rule on the reference sheet. Basically says when you've got a function to a power, multiply by its derivative, you can just wipe away the derivative, uh, increase the power by one, and then divide by the new power and then just plus c. All right, so how do we get our integral ready to use this formula here? Well, we need our integral to have cot to a power, which we do.

[00:03:11]needs to be multiplied by its derivative which is negative cosec^ squ. So let's just put that cosec^ squ at the front.

[00:03:17]Give it a negative by chucking a negative out the front. And now it's all good. And now using the reverse chain rule, we can integrate this by increasing the power of our cot. So it's going to raise that to 2 n minus one. We can then divide by 2 n minus one and we can wipe away our derivative and everything's going to be all Gucci. So let's have a look at what happens if we do that. There we go. So we have cot to the power of um our new power which is increased by one. We've divided by our new power which is increased by one and now we just got to sub in pi on two and pi on four and hopefully get this thing here. Now if you are not careful if because if you if you type in cuz like right cot is tan upside down right but tan pi on 2 doesn't exist. However cot pi on 2 does exist. Okay the reason tan doesn't exist is because the bottom of the tan fraction is undefined. But that becomes the top of the cot fraction.

[00:04:06]Okay. Anyway, if you fiddle around with your calculator, you will eventually figure out that cot of 90 is equal to zero. And that's because cot is equal to um cos over s and cos 90 is equal to zero, right? So c pi on 2 is equal to zero, even though tan 90 is undefined.

[00:04:22]And for the second limit, cot of pi on 4. Well, tan 45 is 1. So cot 45 is going to be the reciprocal of 1, which thankfully is still one. So these two limits get us really nice answers for our numerator. We're just going to be doing um we're going to be doing 0 take away one.

[00:04:38]Now this is pretty much finished off because on the top we get negative 1 but there's a negative out the front. So 1 makes pos1 / 2 n -1 - i n -1. And that was our target to achieve for three marks. And that's the band five part of this question done and dusted. We can now try the easier part of the question which is part two which says hence or otherwise calculate I subscript 2. Okay.

[00:05:01]Hey, the way I like to handle these questions in my recurrence relations is I like to start with the smallest possible integral and I like to build up. So I'm going to start by calculating I Z. So going back to the integral and making the power of N equal to Z, I get C to the power of Z and anything to the power of 0 is just one. So starting off with I just integrating one between pi on 4 and pi on 2. So 1 with respect to theta integrates to theta and then I just sub in pi on 2 minus p<unk> on 4 and I end up with pi on 4. Now for I1 using my recurrence relationship I'm going to do I1 is equal to 1 / 2 * 1 - 1 minus I and we already figured out the value of I before that is pi on 4. So for my fraction here I end up with 1 / 1. So I just get 1. And so the value of I1 is 1 - p<unk> on 4. Now I can finish off the question and get I2 by using the recurrence relation once more. I can do I2 is 1 / 2 * 2 - 1 take away the value we just found which is this guy right here. So this fraction turns out to be 1 over 3 I want to say. Yep. 1 over 3 um take away 1 - p<unk> on 4. You simplify that and you end up with positive p<unk> on 4 - 2/3 as your one mark answer for part two. And so really important that you guys recognize that for questions like this. If you couldn't figure out part I there was nothing in the working out of part I that helped me do part two. Okay? You can you should always be able to do these second parts and pick up a mark or two for these really scary recurrence relations. So even if you can't do part one, please don't give up.

[00:06:31]Still have a crack at part two because it can be a relatively straightforward mark or two.

[00:06:37]Okay, that'll do it for question 14 A.

[00:06:39]Let's move on to question 14B. We're back in mechanics land now. And we have the acceleration of a particle is given by 32x X^2 + 3. X is displacement blah blah blah blah blah. Initially, the particle is at O and has a velocity of 12 m/s in the negative direction. Okay, so I've got my acceleration. I'm just going to write down my initial conditions. So, I've got um initially um position is zero and velocity is negative. That's going to be really important later on. Okay, part I. Show that the velocity of the particle is given by this. So, it's asking us to go from our acceleration equation to our velocity equation. So, um normally when you're just dealing with time and you have your acceleration, you can just get to your velocity by doing the integral.

[00:07:19]But here's the thing. We're not working in time. We're working in displacement.

[00:07:22]So, it's really important that you guys know that when you take your acceleration in terms of x and you integrate it, you don't get your velocity. What you get is your velocity squared on two. Okay? So, very very important. Integrating acceleration with respect to x gets you velocity squared on two. Okay. Now, how do we integrate this? Um, if we wanted to, we could just expand it out and do run-of-the-mill integration, but I'm going to be a little bit fancy and once again use the reverse chain rule. So inside of my function here, I have a derivative of 2x. So I'm just going to have a 2x out the front of my expression by putting a 16 out the front. Um because now I can just integrate this by taking this interior function. I can just increase the power to two divide by the new power of two and just wipe away this derivative of 2x. So that's what I'm going to do. I've got x^2 + 3 to the^ of two / two with a 16 chilling out the front. And that's equal to v ^2 on two.

[00:08:15]But don't forget your plus c. While we're here, we should probably figure out what that plus C is by using some initial conditions. So, initially our velocity is -12 and our position is equal to zero. So, we can sub in v equal to -12, which gets us 144 on the left hand side. Subbing in zero on the right looks like this. We're going to get um 8 lots of um eight lots of 9, which is 72.

[00:08:37]Hey, we end up with 72 = 72 + c. So, that's really convenient. It means that c is equal to zero. Now we can go back to our equation over here and we can multiply both sides by 2 to get v ^2 = 16 x^2 + 3^2 which is almost the target.

[00:08:52]We're trying to get to this remember. So we have v ^2. So now we're just going to get v by taking the square root of both sides. And when you take the square root of the right hand side the squared goes away. The 16 turns into a four. But you've got this plus minus. All right?

[00:09:05]And you've got to figure out and show that you're figuring out whether it's a plus or a minus. So, we clearly want the minus because that's what the question is telling us to work towards here. But how do we explain why it's negative and not a plus? Well, once again, we're just going to use our initial conditions. Our initial conditions was that when x is equal to zero, velocity is -12. So, when we put zero in here, we're going to get three, right? But we need this to give us an answer of -12. So, therefore, it tells us the thing out the front needs to be a -4. Otherwise, we don't get this initial condition. Blah blah blah blah blah. velocity is negative when x equals z is is sufficient for you to jump from that to that and pick up the two marks for your show that part one is done and dusted. We have our equation proofed.

[00:09:45]Now we move on to part two which is find the time taken for the particle to travel 3 m from the origin. Okay, so we're trying to find the time taken to travel some distance. So now that we have our velocity, it seems like we want to go another step, integrate once more and figure out our displacement equation. So let's do that. Let's write our velocity as the derivative of displacement with respect to time. And now if we want to integrate this, we're going to have to do some flipping, right? Because the right hand side is in terms of x, which is the top left variable. So taking the reciprocal of both sides or just moving some stuff around, you know what I mean? So we're multiplying the the dt across the right.

[00:10:22]Um dividing this x^2 + 3 over to the left. However you want to do it, you know, moving it around. As long as you get the x's with the dx's and the number with the dt, everything's going to be fine. Now we can go ahead and integrate both sides. and we're looking at the left hand one and we're like, how do we integrate x^2 + 3 on the bottom of a fraction cuz we can't use a logarithm.

[00:10:40]We can't use a logarithm, but we can use inverse tan. So, we know that um we have x^2 + a^2 on the bottom of our fraction here. So, on the top, you want to have the square root of this number here. So, I want there to be a square root of three there. So, I'm going to make that happen by putting a 1 over roo<unk>3 at the front. And now, this guy is set up to use inverse tan integration. So left hand side is going to integrate to inverse tan of x on<unk>3 and the right hand side is integrating with respect to t. So we're just going to get -4t and a constant. So there we go. 1 over<unk>3 tan inverse x over<unk>3 = -4t + c. We should probably figure out what that constant is by once again using our initial conditions. So initially we have um time equal to 0 and x equal to z. So we get tan inverse of nothing which is nothing. So we get nothing equals nothing plus c. So we get C is equal to nothing as well. So we go back to our equation and we're trying to find here's a little trick to this question which uh which made it a band four actually. We are trying to find the time taken for the particle to travel 3 m from the origin. It doesn't say whether we are to the right of the origin or to the left of the origin. It just says 3 m from the origin. Now we are meant to look once more for like the fifth time at our initial conditions and say at the start when t equals z we were starting from the middle and we were moving with negative velocity. So we're moving to the left. So from that we can infer that the time taken for us to travel 3 m from the origin means we're looking at when our x gets to -3 because we are initially moving in the negative direction. We're going to get to -3.

[00:12:12]We're not going to get to positive 3. So all that inference says hey we need to figure out um when trying to find the t when x is equal to -3. So let's go to our equation over here and let's sub in x= to -3. Now how do we figure out this with tan inverse? Well little tip you can use here is that tan inverse and sin inverse are both odd functions. So when you have a negative inside of that function you're allowed to just yink that out the front. And very conveniently 3 / roo<unk>3 is just equal to roo<unk>3. Or you could just chuck this in your calculator and ignore everything I'm saying. But I'm trying to teach you guys how to think, not how to push buttons, right? Okay. Tan inverse of roo<unk>3, that's going to be 60° or pi on 3. So if we make some room over here, we're going to get -1 over roo<unk>3 *<unk> over 3= -4t. That's a cool little rhyme. I'm doing a lot of rhyming today. I don't know why. All right, let's um multiply this together and just divide by -4. And that's going to get us our t for our two marks. So we're going to get um we're going to get<unk> over 3<unk>3, but then we divide that by -4 and it becomes positive and the bottom becomes 12. So for two marks we have t is equal to pi over 12 <unk>3 seconds. That's question 14b. All done.

[00:13:21]Okay. On to question 14 c. This is um a really tricky one. A lot of band five part this question. Um roots of unity always throw people for a massive loop.

[00:13:29]So let's go um let's go slow and steady.

[00:13:31]We have a complex number um such that 1 plus w blah blah blah blah blah is equal to zero. Show that w is a seventh root of unity. Okay let's start off with that. So we have 1 + w + w ^2 up to w6 equal to z. All you got to do to show that this is a 7th root of unity is you need to multiply both sides by w - 1.

[00:13:50]The reason for that is that you are expected to know and um it's in my videos hopefully you've already seen this but um the left hand side this is the factorized version of w ^2 minus one. Okay you do w minus one at the front and then you've got power of six power of five power of four all the way down to power of zero. Okay, so the left hand sides are both equivalents. And it's really important that you know that otherwise you can't access this band five mark. The right hand side when you times it by zero, you just get zero. And now you have W7 - 1 equals Z. So you get W7 equals 1. And that by definition is a root of a seventh root of unity. And that's the first mark done and dusted.

[00:14:27]Okay, on to part two now, which is where in my opinion things get really tricky, but this is ironically the band four parts. Clearly I don't know what I'm talking about. The complex number alpha is www ^2 w4 and it's a root of this equation which has um real coefficients and alpha is complex. Okay, good to know. Find the other roots in terms of positive powers of w. Okay, so this one little trick to apply. All we need here is um the three magic words of the conjugate root theorem. We have a polomial that's got one root and we're trying to find the other one. So the two roots are always going to be in conjugate pairs. So all we got to do is find the conjugate of w ^2 and w ^ 4.

[00:15:05]Now a cool little fact about conjugates is when you're doing the conjugate of a sum like this, it's the same thing as the sum of the conjugates. Okay? So imagine there's a line there and my computer's not glitching out. We've got w conjugate, w ^2 conjugate, and w4 conjugate. Okay? And something else you should hopefully really know is this cool little relationship between um roots and conjugates, especially when we're talking roots of unity. Here is a dodgy little picture I made of the seventh roots of unity, right? 1 2 3 4 5 6 7 6 7. And now if you look at the first one, which is W, right? It's just right there. That is actually the conjugate of this one down here, which is W6. So if you take the, you know, W1, take the conjugate of it, which flips it across the real axis, it becomes W6. Now if you look at w ^2 which is the next one uh w ^2 if you take the conjugate of it it becomes this one here which is w5.

[00:15:58]So our roots of unity are all happening in conjugate pairs with this one being its own conjugate. So basically the short version of this is that w conjugate is equal to w6. W ^2 conjugate is w5 and w4 conjugate over here turns into this one which is w3. Right? So sorry there's w4 and conjugate is w3. So taking these conjugates gives us w6, w5, w3 and that's the other root of x^2 blah blah blah in terms of positive powers of w for two marks.

[00:16:30]Okay, now we can do part three which is find the numerical value of c. So we just figured out the two roots of um of this quadratic. How can we use that to figure out the constant? Well, the constant is related to the product of the roots. Yeah, we know that when we do alpha * beta, we get c / a. Now, conveniently, a is the coefficient of x^ 2, which is just 1. So, if we take our alpha, which is just here, and we multiply it by our beta, which we figured out in part two, this should give us the value of c / 1. So, here we have c equal to alpha * beta. Now, we're going to do a really, really big expansion. So, we're going to do um w * all three of these. We're going to do w ^2 * all three of these. and we're going to do W4 times all three of these. So, we have nine terms and we've got a lot of clean up to do. Now, some facts that we can use to do our cleanup is we can use the fact that as we already established in part one, if we do 1 + w up to a power of 6, that should be equal to zero. That should nullify some terms for us. But we don't have a lot of these. I don't see a w ^2 anywhere. I don't see a w. All of our powers are too big. But we can fix that because, as we already established, w to the^ of 7 is equal to 1. So, we got a couple of those. And what's cool is you can take this equation and multiply both sides by w and you can build your higher powers and link them back to earlier powers. So multiplying both sides by w gets us that w8 is equal to w1 and etc etc. So w9 will be w ^2. Yeah. So all the powers that are bigger than seven. You can just reduce them by seven and it's the same thing. So we can go to this eight and we can change this to a w. We can change this 9 to a w ^2. We can change this 10 to a w3. And everything's all good.

[00:18:07]Yeah. And those w7s as we already established we can change those to ones and it's simplifying really quickly. Now question is really well designed because in our setup now we have as we're hoping to get a 1 + w ^2 + w3 4 5 6 those things all add together and give us zero and all we'll be left with is this 1 in the middle and this one on the end. So we have 1 + 1 + 0 and that is equal to two. And that's how you get this last band five mark question 14 C.

[00:18:37]Okay. And one last question for question 14. This is another scary band five inequalities question and I think it's really fun. Um once again, this is one of those cases where I did it myself. Um it took quite some time and then I found a quicker and better solution online that I'm going to present to you now because I'm not trying to show you um necessarily always my solutions. I'm trying to show you the best solutions that I um can present essentially.

[00:18:59]Anyway, I'm I'm just rambling. Let's get on with some maths. We have positive real numbers A, B, C, D, and we have um their fractions. And these are consecutive terms in an arithmetic sequence with a common difference of k.

[00:19:10]k is greater than zero. Show that b + c is less than a plus d. All righty. Where are we going to start? Well, we have um we have four consecutive terms in an arithmetic sequence and they're all positive. So straight away that tells me that 1 / a is the smallest fraction and 1 / d is the largest fraction. Now, what's useful about that is that if 1 / a is the smallest fraction, that tells me that a is the biggest number and d is going to be the smallest number. Okay?

[00:19:37]Because imagine like you know this was like 1 over a th00and then a is going to be a th00and. So a is going to be the biggest. Anyway, if you think about this for more than like 5 seconds, it makes sense. So we have a is the biggest number. A is bigger than b is bigger than c is bigger than d. Okay. Now we need to set up our k. K is the common difference between our terms. So if I do like this term um take away this term that gets me k right and I can do a similar trick with these two. I can say 1 / d take away 1 over c um is equal to k. Now this is the important setup. We have these two set up for k. We're basically going to try and um fandangle these together to build an equation and then we can mix it with this to turn it into an inequality to build this for three marks. So I'll show you how it works. It's really cool. We're going to take this um expression here and we're going to combine these two fractions into one. So we're going to multiply this one by a and this one by b which gives us a minus b over a b. Now what that means is we can multiply that across and we end up with k a b is equal to a minus b. And I promise you this is leading somewhere cool. We can do a similar thing with these two fractions.

[00:20:39]We can cross multiply and get c minus d over cd is equal to k. That gives us that c minus d is equal to um kcd. Now we need to build an expression relating K A B and K C D. So we can link these two together and build our inequality.

[00:20:55]So what do we know about A B and C D?

[00:20:57]Well, I know that A is bigger than B and B is bigger than C and C is bigger than D. So this logically tells me that if I take these two AB and I multiply them, it should be bigger than C * D because these two pieces are bigger than these two pieces. So we have AB is bigger than CD. If I multiply both of these by k, I get k a b is greater than k cd. And now I can build these together because k a b is just a substitute for a minus b and um and kcd is a substitute for c minus d. So I can chuck that one on the left.

[00:21:29]I can chuck that one on the right. And what am I trying to get? I'm trying to get a um I'm trying to get b and c being um being smaller than a plus d. So I've got the a here. I can add this D across and I can add this B across and then I've got A plus D is greater than C plus B, which is exactly what I'm trying to find. I've just got to turn it around to make the smaller part on the left. And so we say therefore B + C is less than a plus D. And that's how we get three out of three marks. But there are a number of different ways of um solving this inequality. So if you found another way that you think is really cool and it's a bit more straightforward than this, let me know in the comments and I'll be delighted to hear it. Anyway, that wraps up question 14D and this video. So, I'm going to take a break and go have a little sit down in a quiet corner before I have before I have a crack at um question 15. Anyway, thanks for making it through to the end and hope you tune in for the last two pieces where things are going to get even worse. But um that'll do for now. Bye for now. Not forever.