If the title made you think if of having a stroke, this thread is to discuss two of the problems presented in this year's high-school exit exams. In specific, for higher level mathematics (the proper one, not social science mathematics). So you should be able to solve these. Also it's worth mentioning that the problems have been slightly refitted. Rest assured, the contents are exactly the same.

Problem 1:

Take [tex: f(x) = (1-ax)^{\frac{-1}{2}}], where ax<0, [tex: a \neq 0]

a, Prove using induction that
[tex: f^{(n)}(x) = \frac{a^n(2n-1)!(1-ax)^{\frac{2n+1}{2}}}{2^{2n-1}(n-1)!}], [tex: n \in \mathbb{Z}^+]

b, Show that the Maclaurin series (not the Taylor series) for f(x) up to an including the [tex: x^2] term is
[tex: 1 + \frac{1}{2}ax + \frac{3}{8}a^2x^2]

c, Show that
[tex: (1-2x)^{-\frac{1}{2}}(1-4x)^{-\frac{1}{2}} \approx \frac{2+6x+19x^2}{2}]

d, Knowing that the expansion of the Maclaurin series for f(x) is convergent for |ax|<1, indicate what restriction must be put on x for the previous approximation to be valid.

e, use [tex: x=\frac{1}{10}] to make an approximation for the value of [tex: \sqrt x] expressed in terms of [tex: \frac{c}{d}] where [tex: c,d \in \mathbb{Z}^+]

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Problem 2:

Consider the following ODE:

[tex: \frac{dy}{dx} - y cosec(2x) = \sqrt(tan x)]

a, Use Euler's method with a step size of [tex: \frac{\pi}{12}] to make an approximation for the value of y when [tex: x = \frac{5\pi}{12}]. For added spice, you may not use your calculator here.

b, Show that
[tex: \frac{d}{dx}(\frac{1}{2}ln(cotg x)) = -cosec 2x]

c, Show that [tex: \sqrt(cotg x)] is an integrating factor for this ODE

d, Solve the ODE for y

e, Consider the results from (d) and (a), and that (d) is valid for [tex: 0 < x < \frac{\pi}{2}].
(i) Calculate the exact value of y for [tex: x = \frac{5\pi}{12}]
(ii) Considering the slope of the curve, indicate why (a) is not a good approximation
(iii) Write the reason why the approximation for y is lower than the actual value

f, Using the ODE, deduce why the solution has a positive slope for [tex: 0 < x < \frac{\pi}{2}]

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I think both are fun to do. Problem 1 focuses on a practical application for a primitive tool for approximation of square roots. Problem 2 is exploring an ODE and testing the student's knowledge on when can approximations be applied.

Generally speaking, the sections for the problem are supposed to guide the student onto a certain conclusion. However, for problem 2 I partially cut that out. So here it is, for parts (d) to (f)
[spoiler: the solution for the ODE is [tex: y = x \sqrt(tan x)]]

Also [tex: f^{(n)}(x)] is the nth derivative of f(x).

If on the contrary you aren't struggling, impose a 30 minute timer for each problem. If you're interested in the points for each section it is:
1: (a) 8 points
(b) 2 points
(c) 4 points
(d) 1 points
(e) 5 points
2: (a) 3 points
(b) 4 points
(c) 4 points
(d) 5 points
(e) 3 points, 1 for each section
(f) 2 points

I will publish what I believe to be the solutions in a few days in order to give time for solutions. Latex can work inside of spoiler blocks, but it's a hassle. So anyone attempting I recommend you don't look bellow this post. The thread is also open to anyone wanting to discuss any aspect of the problems, or questions (for example if I cut out too much information). And even though there is proof by induction and an ODE, this is high-school level, albeit on higher level. So [bold: you should be able to solve this.]

Problem 1.

Why even have a gay "a" parameter?

Take f(t) = (1-t)^(-1/2)

a, we have to prove instead that the n-th derivative is (2n-1)!/(2^(2n-1)*(n-1)!)*(1-t)^(-(2n+1)/2)

First derivative is just (1-t)^(-3/2)/2 and base case for n = 1 is 1!/(2^1*0!)*(1-t)^(-3/2) which is obviously the same.

Taking the derivative of (2n-1)!/(2^(2n-1)*(n-1)!)*(1-t)^(-(2n+1)/2), and conveniently multiplying and dividing by 2n, one gets (2n-1)!/(2^(2n-1)*(n-1)!)*(2n+1)/2*(2n)/(2n)*(1-t)^(-(2n+3)/2) = (2n+1)!/(2^(2n+1)*n!)*(1-t)^(-(2n+3)/2) which is what one gets if plugging n+1 into the original formula.

b, for t up to t^2 it'd be 1/0!+1/2*1/1!*t+3/4*1/2!*t^2 = 1+1/2*t+3/8*t^2. Substituting t=ax gets the desired result.

c, substituting above for t=2x and t=4x it's just a matter of multiplying (1+x+3/2*x^2)*(1+2x+6x^2). Expanding and ignoring the terms with exponents > 2 leads to 1+2x+6x^2+x+2x^2+3/2*x^2 = 1+3x+19/2x^2 = (2+6x+19x^2)/2

d, both |2x| < 1 and |4x| < 1 must be true so |x| < 1/4

e, if t = 1/10, then 1/sqrt(1-1/10) = 1/sqrt(9/10) = sqrt(10) / 3
Using b, it can be approximated as 1+1/2*1/10+3/8*1/100 = 843/100, so sqrt(10) is approximately 3*843/100 = 2529/100 = 3.16125
[spoiler: sqrt(10) = 3.1622776...]

Too lazy for problem 2.

>Why even have a gay "a" parameter?
No idea to be honest. I suppose it's for part (e). In which I made a little typo. It was supposed to be calculating the value of [tex: \sqrt 3]. However, the overall solution is the same:
[tex: 3^{\frac{1}{2}} = (1-\frac{a}{10})^{-\frac{1}{2}}]
So [tex: a = \frac{20}{3}]
Substitute into (b) to get [tex: 1 + \frac{1}{2}\frac{20}{3}\frac{1}{10} + \frac{3}{8}\frac{400}{9}\frac{1}{100} = 1.5 \approx 1.7320508...]
And in case anyone's curious, With terms up to x^3, we get [tex: \frac{43}{27} \approx 1.593], for x^4, [tex: \frac{1067}{648} \approx 1.647], so it is indeed slowly converging.

Though this was my mistake. The rest seems fine, except for (a), where I don't exactly understand where the a^n went and where did the minus come into t's exponent.

P94920
>I don't exactly understand where the a^n
Right, I kind of ignored that when I decided to just use t and not ax but a similar reasoning gives you the missing a^n.
>and where did the minus come into t's exponent.
u wot
(1-t)^x -> -(1-t)^(x-1).
You start with an exponent of -1/2, you're only going to get lower and lower values as n increases.
I didn't say anything because I thought it was kind of obvious but you actually made a typo with (1-ax)^((2n+1)/2) by missing the minus sign.

P94925
>Right, I kind of ignored that when I decided to just use t and not ax but a similar reasoning gives you the missing a^n.
Alright.
>u wot
To my excuse, where I have this on, the gap between the minus sign and the fraction is close to being a print error. Checked problem 2 and I don't see any other typos, other than some ignored constraints in the definition. This is embarrassing, should I redo the OP?

P94927
>This is embarrassing, should I redo the OP?
No, embrace the cringe.