Help with some problems

Hello!

My name is Alexander Holmberg, I am new to Khan forums and have a few math questions regarding a test I am writing on Saturday. I was wondering if someone could help me with some of the questions from an older test. The test is originally written in Swedish and is called Matematik och Fysikprovet. The test concists of one physics part and one math part. No calcuators are allowed on the test. This means that there could be a faster solution, but it is not possible to use it. I have not found a good way to translate the questions, so they are listed below.

I would be happy with any kind of help, if it is only one question or part of a question that is fine by me. I have tried different strategies to solve these questions and I have asked many people.

The first question I am wondering over:

If m # n = m(n−1) for all integers n and m, then n # m for all integers n and m is equivalent to
a) m # n; b) (m+1) # (n−1); c) (m-1) # (n+1) d) None of a) - c)

Correct answer
c),

Comment:
I do not know exactly why or what this type of math is called or how to solve it.

The second question I am wondering over is:
All positive integer solutions to the equation x^(x^2 - 3x) = x² is given by

a) x1,2 = 3 ± sqrt( 17) ÷ 2 b) x = 3 + sqrt( 17) ÷ 2

c) The equation does not have any positive integer solutions d) None of a) - c)

Correct answer:
d).

Comment:
I tried solving this first by using the pq formula and got a). Then I realised that (a^b)^c = a^(bc). This time I got c) as the answer, which is also incorect.

The third question I am wondering over is
Name the biggest integer number a so that all integer solutions

3x² + ax - (a² - 1) = 0
are integers and positive

Correct answer
-(2sqrt(39)÷13).

Comment:
I am really uncertain as to how to solve this question. I first tried pq and got x = -a÷6 ± sqrt(a²÷12 + 1). I thought that a should be -12, since sqrt would disapear. This was totally wrong, I am thinking now that it may have something to do with complex numbers. However, I am uncertain how to solve the question.

The fourth question I am wondering:
Solve the inequality ((x-1)÷((x+1)(sqrt(6 - x - x²)) ≥ 0. Give the summation of all of the solutions.

Correct answer:
-1

Comment:
I tried solving this question by solving each parenthesis individually.
x1 = 1 x2 = -1 x3 = (1 ± sqrt(13))÷2. The summation of these numbers is (1 ± sqrt(13))÷2, which is totally incorrect.

The fifth question I am wondering is:
The height against the base in a isosceles triangle is 4 cm long and the area is 12 cm². Calculate cosy, where y is the angle between the triangles two long legs.

Correct answer:
7÷25

Comment:
I barely know where to begin when attempting to solve this question. I devided 12 by 4 to get the base? Then I calculated the long side using Pythagoras; sqrt(25). This was of course incorrect.

The sixth question I am wondering is:
The parallelogram ABCD has the side lengths 5 cm and 3 cm, and the angle between the diagonals is 45°. Calculate the area of the parallelogram and give the answer in mm²

Correct answer:
800 mm²

Comment:
I tried to solve it by doing 2sqrt(9 + 25) * 5, which was incorrect.

That was it!

Thank you for taking your time and reading through!

ianturcanwhite

26. maj 2021 00:22
Redigeret

For the first question, it was a bit weird, but it took me about 10 seconds to get it when I revisited it (I hope). n # m is an operation like n + m.

It gives you the definition of n # m which is n(m-1) for all n and m.

So by the definition, m # n = m(n-1) = mn - m

So you just use this definition for each answer and use distributive property and compare it to mn - m.

For the 2nd question I did this:

x ^ (xx - 3x) = xx, x > 0

take the logarithm of both sides and rewrite using the logarithm rules

(xx-3x)log(x) = 2log(|x|)

since x > 0 rewrite:

(xx-3x)log(x)=2log(x)

if log(x) = 0 or (xx-3x) = 2, then the equation is true.

The solution to log(x) = 0 is 1, and the solutions to xx-3x = 2 are (3 +- sqrt(17) / 2

So the only positive integer solution is 1.

For the 3rd question, I did this:

3xx + ax + aa - 1 = 0

Multiply both sides by 12 and do some rewriting and you get

(6x)(6x) + 2a(6x) = 12 - 12aa

Complete the square

(6x + a)(6x + a) = 12 - 11aa

Since x and a are assumed to be both integers, then that means 6x + a must be an integer. then that means 12 - 11aa must be and integer square greater than or equal to zero.

12 - 11aa >= 0

Do some rewriting and you get

aa <= 12 / 11

If a is an integer, then the only integers that are solutions to that inequality are -1, 0, and 1.

You could try 0, which results in

36xx = 12

4xx = 1 which has no integer solutions

If you try +- 1, you get

6x +- 1 = -1 or 6x +- 1 = 1

Both equations have an integer solution of 0 while both of their other solutions are not integers. Since 1 > -1, the solution is "a" = 1.

Help with some problems

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