IB Math AA SL/HL May 2026 Paper 1 Q5 Worked Solution (Sneak peak)

Added: 2026-05-16

[00:00:00]All right, we're doing a quick peek here at one of the questions from the May 2026 paper one exam for IBA and this is standard level time zone 3, so the Americas. Okay, we'll have the full video for work solutions hopefully coming out on Monday, Tuesday next week, but I wanted to do one question um before the weekend just to give everyone a peak of what we've got coming on the channel. Okay, so this was question five on the SL exam. I believe it was also on the HL exam and it was question uh it was also question five on the HL exam.

[00:00:31]Okay, so it's a nice straightforward two mark problem to start. We're doing a derivative here. So it's asking us find dydx. Okay, we should know this derivative or at least the start of it.

[00:00:41]The derivative of x to the 4 is simply 4x cubed. Okay, and then this sin 4x right here, we're going to use a chain rule for the derivative of this guy.

[00:00:52]Okay, so that's going to be minus the derivative of s is cosine. We'll leave the inside function alone. So that's cossine 4x. And then we'll multiply by the derivative of 4x, which is simply 4.

[00:01:04]Okay, we can clean this up and actually take out a common factor to complete our answer to part a. So 4 * x cubed minus cossine 4x. Okay, that links us nicely into a part B, which is hence or otherwise find this complicated looking integral. Okay, so this integral will come from a U substitution and that U is going to be exactly what we used in part A for the derivative. Okay, so this allows us to speed through the derivative and the u substitution pretty quickly with our 4 and then x cubed cosine 4x is our u. We know from our u sub practice, we go ahead and solve this for dx. So we multiply both sides um by dx. I'll go a little slower on this one.

[00:01:50]So dx * 4 x cubed minus cosine 4x should equal du. And this means that du is equal to sorry dx is equal to du over 4 * x cub minus cosine 4x. Okay. We'll substitute this back into our original integral.

[00:02:15]Okay, so this will turn our original integral into the integral of u 7 * x cub - cosine 4x. That should kind of connect for us that that's going to link up with our substitution for dx, which is du over 4 x cub minus cosine 4x. We can do a little bit of algebraic cancelling with these x terms to the left. And this gives us the integral of u 7 / 4 du. We can compute that integral by increasing the power u to the 8th.

[00:02:49]And then we divide by this new power. So multiply by 1 over 8. And because we're doing an indefinite integral, we have our arbitrary constant there, the plus c. So this is 1 over 32 something to the 8. that something is going to be ru which was x^ 4 minus sin 4x all raised to the 8. So that's actually quite a lot of work there on that problem just to get those three marks. So tough SL problem kind of a medium HL problem uh for question five. Okay. So if you enjoyed this sneak peek, keep subscribe to the channel. Watch out for future videos next week where we'll go through the full paper one and paper twos from time zone