# Has the math abi really gotten heavier?

“Heavy” is always such a term.Everyone finds other things “heavy” – “heavy” is subjective. And “too heavy” is always used in the theme in the sense of “unfair”. I think it makes sense to differentiate a little bit here. It was demanding. But it should also be remembered that it is part of the baccalaureate, the highest education school leaving certificate in Germany – the “elite”, if you like.

I wrote the Abitur this year in Bavaria myself, but I also have to say that I was never a problem child in mathematics and – without wanting to splash – almost all credits were rated with 15 points.Nevertheless, I believe that I can assess the tasks quite objectively.

The biggest turmoil has been in Bavaria, where the petition has already attracted more than 70,000 supporters.If you want to get an idea of the Bavarian maths abitur, you can see the tasks here:

The Abitur this year was quite demanding.Sometimes I understand the protests (more on that later), but on the whole I still found the Abitur personally fair.

I would like to briefly comment on some of the tasks that have often been discussed:

Enter the term of a [math-mathbb-R-[/math-defined and reversible function [mathj[/math an, which meets the following condition: The graph of [mathj [/mathand the graph of theinverse function of [mathj[/math have no common point.

Here you had to think for a moment.However, if you know how the graph of a function is related to that of the reversal function, you almost have it. Candidates would have been here [mathj(x) = ex[/math and correspondingly [mathj-1-(x) = x[/math (or vice versa).

There is a Bernoulli chain with the length [mathn[/math and the probability of hitting [mathp[/math.

Explain that for all [mathk’ 0, 1, 2, …, n'[/math’ the relationship [mathB(n; p; k) = B(n; 1-p; n-k)[/math applies.

Here, trying out would have been the way to the goal.Just set up the formula for a corresponding Bernoulli chain and then consider it:

[mathB(n; p; k) = sbinom n.k. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

[mathB(n; 1-p; n-k) = ‘binom’n’n’n-k’, ‘cdot (1-p)’n-k”,cdot (1 – (1-p))’n-(n-k)'[/math

The two terms can be easily transformed into two equivalent terms with appropriate simplification without large computing – this proves the claim.Evidence is provided in every lesson and is nothing new, even if the specific evidence procedures have been removed from the curriculum.

Justify that the following statement is correct: There are infinite levels that do not contain a point whose three coordinates match.

Mathematical understanding is required here.For it is also a required competence to transform the required information into mathematical constructs from a given task and to show something about it. Here the crux of the matter would have been to recognize that all points whose three coordinates match are on a straight line and that there are infinitely many planes that are parallel to it and thus do not contain a point of these straights. This is not an easy task, but it is quite feasible if you deal with it a little bit.

Given are the two balls [mathk_1[/math with center [mathM(1|2|3)[/math and radius [math5[/math and [mathk_2[/math with center [mathM(-3|-2|1)[/math and radius [math5[/ math.

The cut figure of [mathk_1[/math and [mathk_2[/math is a circle.Determine the coordinates of the center and radius of that circle.

I actually found this task quite complicated.Not in execution, but in the imagination. Because cut figures are not part of the curriculum and therefore very intimidating for students who are not so good at math. You have to bite your way through. But this is also part of an Abitur. With 3 BEs, the task is also fairly evaluated.

For the value of [mathk[/math shows Figure 3 (see next page) of the corresponding graph with its repentance.

In this coordinate system, the two axes are scaled differently. Determine the missing numerical values on the mark strokes of the y-axis using a suitable slope triangle on the reverse beam and enter the numerical values in Figure 3.

A new task that many have not yet seen.A “transfer task” as they are often so beautifully called in teachers. It requires the scaling to be entered by scaling the corresponding y-change on a reverted slope (from which the slope was previously calculated). Here one did not get any further with the usual cooking recipe learning, but had to apply the knowledge – but this does justice to an Abitur.

In the factual context, describe an event whose probability is associated with the term Can be calculated from the ‘math’s’s”s””””limits_’i=5’8B(25; ‘frac’1′; i)[/math’ to be calculated.

Here one could point out that the sum sign is actually not part of the curriculum.I say quite frankly: understandable. Even if one could have tasled down the meaning of the expression in context (the summation of the individual binomial distribution – thus the cumulative binomial distribution), this is a task where I would be on the side of the protesters if this happens.

But now probably the most controversial task in the debate: the lottery owner with the river preference.

In the case of a lottery, the prize is advertised that each lot wins.

The lots and the corresponding prizes in kind can be assigned to three categories, which are referred to as “Danube”, “Main” and “Lech”. In the lottery there are four times as many lots in the category “Main” as lots of the category “Danube”. A ticket costs 1 euro. The owner of the lottery pays 8 euros in the category “Danube” for a non-cash price in the category “Danube”, 2 euros in the category “Main” and 20 cents in the category “Lech”. Determine how large the share of the “Danube” lot must be if the owner wants to make an average profit of 35 cents per lot.

Here, many seem to have been confused that there was no information on the number of individual lots.But it didn’t have to be, because no number had to be calculated, but only a share. Who then has established the quantity relations between the various shares has already won. Depending on the number of Danube lots, this would simply be [mathd [/math(share of Danube lots), [math4d[/math (share of main-lose) and [math1-5d [/math(the remaining share – the difference to 1 or 1100%).

For energy reasons, the distance between the two places where the two drilling channels meet the water-bearing rock layer should be at least 1500m.

Based on the model, decide whether this condition is met for each possible second drill channel.