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# A few simple problems watch

1. (Original post by constellarknight)
Exactly. Like I said in my previous post, Indeterminate is not making much sense, which is apt given his username.
We all make silly mistakes from time to time. He's deleted his post a few minutes back. Don't harp on about it.
2. (Original post by Zacken)
We all make silly mistakes from time to time. He's deleted his post a few minutes back. Don't harp on about it.
Alright whatever.
3. (Original post by constellarknight)
Exactly. Like I said in my previous post, Indeterminate is not making much sense, which is apt given his username.

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How do you get the spanking smiley?
4. (Original post by Indeterminate)
Enjoy

Problem 1

Let

and be positive integers (not necessarily distinct)

Prove that

is divisible by

without using the remainder theorem

Problem 2

Find

Problem 3

Let be a function satisfying

for real

You are given that

Find the value of

Problem 4

Find

Problem 5

A rational function is defined as follows

Find the value of

oops I screwed my constants up lol, think this one is right now..
Is Q2 something like..
Spoiler:
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-arccos(1/(sint + cost)^2)
5. Problem 3

The function for , and for .

This is a really lovely question.
6. (Original post by Indeterminate)
Enjoy

Problem 5

A rational function is defined as follows

Find the value of

Is question 5 correctly stated? Because I'm not enjoying it too much at the moment.
7. 5 is nice. Currently thinking about 3.
8. (Original post by morgan8002)
5 is nice. Currently thinking about 3.
Did you get 3 for problem 5?
I just went looking for a function g(x), I think works.
9. (Original post by 13 1 20 8 42)
oops I screwed my constants up lol, think this one is right now..
Is Q2 something like..
Spoiler:
Show

-arccos(1/(sint + cost)^2)
Or equivalently...
Spoiler:
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...but yours is much neater
10. (Original post by EricPiphany)
Did you get 3 for problem 5?
I just went looking for a function g(x), I think works.
Yeah. I just wrote f(x) = ax^2 + bx + 1, since f(0) = 1, then integrated. You get some logarithmic terms that only go away if b = 3 and since g(x) is rational, b must be 3.
11. (Original post by morgan8002)
Yeah. I just wrote f(x) = ax^2 + bx + 1, since f(0) = 1, then integrated. You get some logarithmic terms that only go away if b = 3 and since g(x) is rational, b must be 3.
Very well reasoned.
12. (Original post by morgan8002)
Yeah. I just wrote f(x) = ax^2 + bx + 1, since f(0) = 1, then integrated. You get some logarithmic terms that only go away if b = 3 and since g(x) is rational, b must be 3.
I'm an idiot - I didn't even try to integrate - I assumed that there would be some very slick theoretical approach.
13. (Original post by atsruser)
I'm an idiot - I didn't even try to integrate - I assumed that there would be some very slick theoretical approach.
same, by the time i was getting around to integrating i lost motivation, really wished i had stuck with it now
14. (Original post by drandy76)
same, by the time i was getting around to integrating i lost motivation, really wished i had stuck with it now
In fact, I'd got as far as writing out the integrals ready to integrate and didn't bother - I went to watch TV instead. I've been led astray by the complexity of Indeterminate's other problems, I think. Still, it's a useful lesson - you need persistence as much as ability to solve problems, and if you don't follow a lead, you'll never know if it was the right one.
15. Impressed by the response to this thread!

I see that #4 is the only one that remains, so here's a hint
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16. Looks like I'll have to wrap things up

Solution 4

Let denote our integral.

Multiplying through by gives us

Now let so that

Now

and so

Now back in terms of

17. (Original post by Indeterminate)
Looks like I'll have to wrap things up

Solution 4

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