Year 12s: what will be the first way you research your future options? >>

A few simple problems

Scroll to see replies

Reply 20

Zacken

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.

Reply 21

constellarknight

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.

Reply 22

EricPiphany

Original post by constellarknight

Exactly. Like I said in my previous post, Indeterminate is not making much sense, which is apt given his username.

Spoiler

(edited 8 years ago)

Reply 23

math42

Original post by Indeterminate

Enjoy

1 Problem

2 Problem

3 Problem

4 Problem

5 Problem

oops I screwed my constants up lol, think this one is right now..
Is Q2 something like..

Spoiler

(edited 8 years ago)

Reply 24

EricPiphany

Problem 3

\displaystyle 2(e^{-1}-e^{-2})

The function

\displaystyle f(x) = x - 2(e^{-1}-1)e^{-x} - 1

for

\displaystyle x \ge 0

, and

\displaystyle f(x) = -x -2e^{-x-1} +1

for

\displaystyle x < 0

This is a really lovely question. :smile:

(edited 8 years ago)

Reply 25

atsruser

Original post by Indeterminate

Enjoy

5 Problem

Is question 5 correctly stated? Because I'm not enjoying it too much at the moment.

Reply 26

username1560589

5 is nice. Currently thinking about 3.

Reply 27

EricPiphany

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

\displaystyle \frac{-1}{x(x+1)}

works.

(edited 8 years ago)

Reply 28

A Slice of Pi

Original post by 1 8 13 20 42

oops I screwed my constants up lol, think this one is right now..
Is Q2 something like..

Spoiler

Or equivalently...

Spoiler

...but yours is much neater

Reply 29

username1560589

Original post by EricPiphany

Did you get 3 for problem 5?
I just went looking for a function g(x), I think

\displaystyle \frac{-1}{x(x+1)}

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.

Reply 30

A Slice of Pi

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.

Reply 31

atsruser

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.

Reply 32

drandy76

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

Reply 33

atsruser

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.