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This is seemingly easy, but it is extremely hard.

This one is the Collatz conjecture, which remains unsolved for years. 

Link: https://en.wikipedia.org/wiki/Collatz_conjecture

EDIT: Sorry, I read the question wrong. But it is still good to read about the conjecture.

f(11) = 34, so f(34) is defined.

f(34) = 34/2 = 17, so f(17) is defined.

f(17) = 3(17) + 1 = 52, so f(52) is defined.

f(52) = 52/2 = 26, so f(26) is defined,

f(26) = 26/2 = 13, so f(13) is defined.

f(13) = 3(13) + 1 = 40, so f(40) is defined.

f(40) = 40/2 = 20, so f(20) is defined.

f(20) = 20/2 = 10, so f(10) is defined.

f(10) = 10/2 = 5, so f(5) is defined.

f(5) = 3(5) + 1 = 16, so f(16) is defined.

f(16) = 16/2 = 8, so f(8) is defined.

f(8) = 8/2 = 4, so f(4) is defined.

f(4) = 4/2 = 2, so f(2) is defined.

f(2) = 2/2 = 1, so f(1) is defined.

f(1) = 3(1) + 1 = 4

Now we ended up in an infinite loop of 4, 2, 1, 4, 2, 1, etc.

Let's count how many numbers did we encounter.

For \(x\in\{11,34,17,52,26,13,40,20,10,5,16,8,4,2,1\}\), f(x) is defined.

Therefore there are at least 15 integers in the domain of f.

Thank you!  I searched up a proof on Google and I think I understand now! :-)

This means

\(n^2 < 3^5\\ n^2 < 243\)

Because 15² = 225 < 243, and 16² = 256 > 243,

The largest n such that n²⁰⁰ < 3⁵⁰⁰ is n = 15.

https://web2.0calc.com/questions/please-help_16171#r1

Wow that was so fast thank you!

When x <= 3,

f(x) = 0

2x + 1 = 0

x = -1/2

When x > 3,

f(x) = 0

8 - 4x = 0

x = 2 

but x has to be greater than 3, so this is not actually a solution.

The sum of all values of x such that f(x) = 0 is -1/2.

If you and your dad do not understand any part of my solution, feel free to ask me here. :)

Wish you happy problem-solving! :D

We express x^2 + y^2 and x^3 + y^3 in terms of (x + y) and (xy).

\(x^2 + y^2 = (x + y)^2 - 2(xy)\\ x^3 + y^3 = (x + y)((x + y)^2 - 3(xy))\)

Now, let x + y = t, xy = u.

\(\begin{cases}t^2 - 2u = 7\\t(t^2 - 3u)=10\end{cases}\)

Substituting the first equation into the second,

\(\begin{cases}t^2 - 2u = 7\\t(7 - u)=10\end{cases}\)

Manipulating the second equation, \(u = 7 - \dfrac{10}t\)

Substituting, 

\(t^2 - 14 + \dfrac{20}t = 7\\ t^3 - 21t + 20 = 0\)

By factor theorem, (t - 1) is a factor of the left hand side.

\((t - 1)(t^2 + t - 20) = 0\\ (t - 1)(t - 4)(t + 5) = 0\\ t = 1 \text{ or }t = 4 \text{ or } t= -5\)

Therefore, the possible values of (x + y) are 1, 4, and -5. 

Thank you so much!!! This really helps. :3

If Fred starts on 1,

There's a 1/2 probability that it moves to 0 directly, and a 1/2 probability that it moves to 2.

Once it moves to 2, it must move back to 1.

And when it moves back to 1 again, there are again 1/2 probability that it moves to 0, and 1/2 probability that it moves to 2.

This repeats again and again.

So that means there are 1/2 probability it reaches 0 in 1 step, 1/4 probability it reaches 0 in 3 steps, 1/8 probability it reaches 0 in 5 steps, etc.

Because the e_1 = sum of all possible (probability * steps),

\(e_1 = \dfrac12 + \dfrac34 + \dfrac58 + \dfrac7{16} + \cdots = \displaystyle\sum_{n = 1}^\infty \dfrac{2n -1}{2^n}\) --- (1)

Now, we consider (e_1)/2. (The standard way of evaluating an A.G.S. (arithmetico-geometric series))

\(e_1 \cdot \dfrac12 = \dfrac14 + \dfrac38 + \dfrac5{16}+\cdots\) --- (2)

Subtracting (2) from (1),

\(\dfrac{e_1}2 = \dfrac12 + \left(\dfrac{2}4 + \dfrac28 + \dfrac2{16}+\cdots\right) = 1 + \left(\dfrac14 + \dfrac18 + \cdots\right) = \dfrac32\)

\(e_1 = 3\)

Therefore, if Fred starts at 1, then it is expected to reach 0 in 3 steps.

By a similar idea as above, we can get \(e_2 = 2\cdot \dfrac12 + 4\cdot \dfrac14 + 6\cdot \dfrac18 + \cdots = \displaystyle \sum_{n = 0}^\infty \dfrac{n + 1}{2^n} \)

By a similar approach, we can evaluate e_2 to be 4.

Therefore, (e_1, e_2) = (3, 4).

Name the points. Let the point on the bottom left corner be A, that on the bottom right corner be B, and the one on top be C.

Let H be the point on AB such that CH is perpendicular to AB.

We calculate angle A in terms of n using Law of Cosines.

\(\cos A = \dfrac{(n + 1)^2 + (n + 2)^2 - (n + 3)^2}{2(n+1)(n+2)}\\ \cos A = \dfrac{n-2}{2(n+1)}\)

By this, AH = \(\dfrac{n - 2}2\).

Using Pythagorean theorem on triangle AHC,

\(\left(\dfrac n2 - 1\right)^2 + n^2 = (n + 1)^2\\ \left(\dfrac n2 - 1\right)^2 = 2n+1\\ \dfrac{n^2}4 - 3n = 0\\ n(n - 12) = 0\\ n = 12\)

The perimeter is (12 + 1) + (12 + 2) + (12 + 3) = 42.

It is the sum of geometric series.

It looks like this:

For -1 < r < 1, and an arbitrary a,

\(a+ar+ar^2+ar^3+\cdots+ar^n+\cdots=\dfrac{a}{1 - r}\)\(\)

Notice that \(\triangle RYX \sim \triangle XZQ\) by AA postulate.

Let x be the side length of the square. Then

\(ZQ = 12-x\\ RY=8-x\\ YX=x\\ XZ=x\)

By similar triangles,

\(\dfrac{RY}{YX} = \dfrac{XZ}{ZQ}\\ \dfrac{8-x}{x}=\dfrac{x}{12-x}\)

Solving this equation gives the required answer.

Ans: Side length = 24/5.

As an exercise, you can try to show that when PQ = a, PR = b, the side length of the square is \(\dfrac{ab}{a + b}\).

Np ;)

Noticing the repeating pattern is equal to x.

\(\log_a(x\color{red}\log_a(x\log_a(x\cdots))\color{black}) = x\\ \log_a(x\color{red}x\color{black}) = x\\ \log_a(x^2) = x\\ 2\log_a(x)=x\)

By a simple trial-and-error approach,

(a, x) = (2, 2) is a solution.

Therefore the required positive integer a is 2.

Thank youuu!!

For all x < 0,

\(f(f(x)) = x\text{ (given in the problem)}\\ f\left(-\dfrac1{2x}\right) = x\)

But -1/(2x) > 0, so \(k\left(-\dfrac1{2x}\right) = x\)

Now, let t = -1/(2x). x = -1/(2t).

\(k(t) = -\dfrac1{2t}\\ \boxed{k(x) = -\dfrac1{2x}}\)

No. The 2 shouldn't be distributed to each fraction, and you did the "q" one wrong.

Q2)

By triangle inequality, \(|x| + |y| \geqslant |x + y|\)

By induction, we can generalise this to \(\displaystyle\sum_{i = 1}^n |x_i| \geqslant \left|\sum_{i = 1}^nx_i \right|\).

Therefore 

\(|a- b| + |b-c|+|c-d|+\cdots+|m-n| \geqslant |a-n|\\ |n-o|+\cdots+|y-z|+|z-a| \geqslant |n - a| = |a - n|\)

Adding these inequalities gives

\(|a-b|+|b-c|+|c-d|+\cdots+|m-n|+|n-o|+\cdots+|x-y|+|y-z|+|z-a| \geqslant 2|a-n|\\ 20\geqslant 2|a-n|\\ |a-n|\leqslant 10\)

The maximum value of |a - n| is 10.

Thank you for replying!  I do not really know how you got the first equation, I might be missing something really simple, but if you have the time, could you explain it to me? Thanks!

You made 2 errors:

4*4⁴  ≠  ( 4*4 )⁴

^{It's not   -1   it's  +1}

Use the Pythgorean Thearom and find the hypotenuse remember a^2+b^2=c^2

divide RQ by 2 and divide PQ by 2 then use the Pythgorean Thearom to find ZX

Q1)

\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac B{x + 1}\\\dfrac{x + 7}{(x- 2)(x+1)} = \dfrac{A}{x - 2} + \dfrac B{x + 1}\\\)

Multiply by \((x - 2)(x + 1)\) on both sides.

\(x+7=A(x+1)+B(x-2)\\ x+7=(A+B)x+(A-2B)\)

Comparing coefficients, \(\begin{cases}A + B = 1\\A - 2B = 7\end{cases}\)

Solving gives \((A, B) = \left(3,-2\right)\)

You should know that 

\(p^a\cdot p^b = p^{a + b} \qquad (1)\\ \dfrac{p^a}{p^b} = p^{a - b}\qquad (2)\\ \left(p^a\right)^b = p^{ab}\qquad (3)\\ (pq)^a = p^a q^a\qquad (4)\)

Using these three formulae,

\(\quad \dfrac{\left(pq^3\right)^3\cdot \left(4p^2q\right)^2}{\left(2pq^2\right)^3}\\ \stackrel{(4)}{=} \dfrac{p^3 \left(q^3\right)^3\cdot 4^2 \cdot \left(p^2\right)^2\cdot q^2}{2^3 \cdot p^3 \cdot \left(q^2\right)^3}\\ \stackrel{(3)}{=} \dfrac{p^3q^9\cdot16\cdot p^4\cdot q^2}{8p^3q^6}\\ = 2\cdot \dfrac{p^3\cdot p^4}{p^3} \cdot \dfrac{q^9 \cdot q^2}{q^6}\)

Can you do the rest with identities (1) and (2)?