MATH1081 Topic 1 Revision mega-thread (1 Viewer)

[Drewk]

Okay so as we have a topic 1 test this week in our first tute and some of us are still struggling with a few things here and there (including my self) i thought this would be helpful. We can collaborate on answers to past papers and hopefully get any concepts explained to us by each other.

Please upload your own answers and/or check ones done by others

Ill start of with answers to 1st 2 tests and as ppl upload more answers i can add them here:

2008 Test 1A(Explained really well in this vid)

2008 Test 1B:
Q1
Q2 is the exact same as in 2008 Test 1A so see the vid
Q3

2009 Test 1A:
Q1

 

[HeroicPandas]

For question 1, part (ii), you need to give reasons why it is T/F

 

[Drewk]

For question 1, part (ii), you need to give reasons why it is T/F

Oh yeah right but, i dint but u can just see form part i)

 

[HeroicPandas]

Oh yeah right but, i dint but u can just see form part i)

It is obvious, but write out proper answers to prepare for test

 

[Drewk]

It is obvious, but write out proper answers to prepare for test

Yeah k fair enough, for the one that is false u can just say {{a}} is a sub set of A3 not an element right?
could u also check my answer for 2009 Test 1A Q1 before i add it to the main post

 

[Drewk]

Also how can you prove Q2 2009 Test 1B is invective or subjective ? i think form the graph it does fail the horizontal line test but how would u write out a proof for this one??

 

[HeroicPandas]

Yeah k fair enough, for the one that is false u can just say {{a}} is a sub set of A3 not an element right?
could u also check my answer for 2009 Test 1A Q1 before i add it to the main post

yep

ah... i dont think that is clear enough for me lol

To be clear,

A = {n, s, w}

P(A) = blah (from above)

Therefore, A U P(A) = {empty set, n, s, w, {n}, {s}, {w}, {n,s}, {n,w}, {s,w}, {n,s,w}}

P(A U P(A)) = 2^(|A U P(A)|) = 2^11

 

[Drewk]

yep

ah... i dont think that is clear enough for me lol

To be clear,

A = {n, s, w}

P(A) = blah (from above)

Therefore, A U P(A) = {empty set, n, s, w, {n}, {s}, {w}, {n,s}, {n,w}, {s,w}, {n,s,w}}

P(A U P(A)) = 2^(|A U P(A)|) = 2^11

Yep cool ty

 

[HeroicPandas]

Also how can you prove Q2 2009 Test 1B is invective or subjective ? i think form the graph it does fail the horizontal line test but how would u write out a proof for this one??

Range (f) = (-infinity, infinity)

f is 1-1 iff for all y ∈ R, there exists AT MOST one x ∈ R such that f(x) = y
So it's either yes or no. Clearly from a quick sketch, it is NOT 1-1, so we need to provide a counter-example (numerical example)

So when y = -4, 2x^3 + 3x^2 = 0 --> gives 2 x values, hence not 1-1 (i drew a sketch of f and easily found a place where for that one Y-VALUE, there exists 2 - to sketch, draw 2x^3 + 3x^2 then shift curve down 4 units)

f is onto iff for every y ∈ R, there exists at LEAST ONE x ∈ R such that f(x) = y and the range (y-values that exists for f, in this case R) of f equals the co-domain (particular set in which you are defining the function, in this case R) of f

So since range(f) equals co-dom(f), then f is onto

f is bijection, iff f is 1-1 and onto -----> f is not bijective


Oh for range and domain, here is an example:

f : [0,1] -> [-1,5] defined by f(x) = x

Range is [0,1]

Co-domain is [-1,5]

It should be clear that RANGE IS the subset of CO-DOM


If I made a mistake, please tell me!!

 

[Drewk]

Range (f) = (-infinity, infinity)

f is 1-1 iff for all y ∈ R, there exists AT MOST one x ∈ R such that f(x) = y
So it's either yes or no. Clearly from a quick sketch, it is NOT 1-1, so we need to provide a counter-example (numerical example)

So when y = -4, 2x^3 + 3x^2 = 0 --> gives 2 x values, hence not 1-1 (i drew a sketch of f and easily found a place where for that one Y-VALUE, there exists 2 - to sketch, draw 2x^3 + 3x^2 then shift curve down 4 units)

f is onto iff for every y ∈ R, there exists at LEAST ONE x ∈ R such that f(x) = y and the range (y-values that exists for f, in this case R) of f equals the co-domain (particular set in which you are defining the function, in this case R) of f

So since range(f) equals co-dom(f), then f is onto

f is bijection, iff f is 1-1 and onto -----> f is not bijective


Oh for range and domain, here is an example:

f : [0,1] -> [-1,5] defined by f(x) = x

Range is [0,1]

Co-domain is [-1,5]

It should be clear that RANGE IS the subset of CO-DOM


If I made a mistake, please tell me!!

im really bad with these Qs so help me out a bit here how did u get the range as [0,1] ?
and for the 2 values u get u go like: x^2(2x+3) then x=0 or x= -3/2 right??

 

[HeroicPandas]

im really bad with these Qs so help me out a bit here how did u get the range as [0,1] ?
and for the 2 values u get u go like: x^2(2x+3) then x=0 or x= -3/2 right??

The function is f(x), so if you input blah, you get blah

Input [0,1], you get [0,1], hence that is the range

yep, continue with that working out or you will lose marks


sorry, but i have to go now, i'll help you tomorrow - good luck with this

 

[Drewk]

The function is f(x), so if you input blah, you get blah

Input [0,1], you get [0,1], hence that is the range

yep, continue with that working out or you will lose marks


sorry, but i have to go now, i'll help you tomorrow - good luck with this

K i got confused and thought u said range was [0,1] for this specific Q not for a some general example all g

 

[Drewk]

Also help me out with 2010 2B Q2 plz

 

[Drewk]

For 2009 Test 1A
Q2
i) is it fine to write "f is 1-1 iff for every y an element B, there is at most one x and element of A such that f(x)= y" and " g is 1-1 iff for every y an element of C, there is at most one x an element of B such that f(x)=y"
part ii) i said "suppose g o f(x) = g o f(y), prove x=y..... =g(f(x))=g(f(y)) from definition of composition.... as f and g are 1-1 (given in Q) then x=y hence g o f is 1-1"

Q3 not sure bout my 1 line to 2nd line step tho

 

[HeroicPandas]

For 2009 Test 1A
Q2
i) is it fine to write "f is 1-1 iff for every y an element B, there is at most one x and element of A such that f(x)= y" and " g is 1-1 iff for every y an element of C, there is at most one x an element of B such that f(x)=y"
part ii) i said "suppose g o f(x) = g o f(y), prove x=y..... =g(f(x))=g(f(y)) from definition of composition.... as f and g are 1-1 (given in Q) then x=y hence g o f is 1-1"

Q3 not sure bout my 1 line to 2nd line step tho

i) Yep correct, I think you should write 'one-to-one' instead of '1-1'

EDIT: Also include that for x, y in A, if f(x) = f(y), then x = y and likewise for function g

ii) Show a bit more working out

Start from g(f(x)) = g(f(y))

f(x) = f(y) as g is one-to-one

x = y as f is one-to-one

hence g 'circle' f is 1-1


Q3 - I got the answer as AUB, maybe you made a mistake while doing the reverse distributive law too quickly

 

[HeroicPandas]

Also help me out with 2010 2B Q2 plz

hm....

ii) f(g(x)) is an exponential function, so clearly for every y in R, there exists at most one x in R such that f(g(x)) = y (this is the idea in your mind)
To show it in the test, just show that if x,y in R and f(x) = f(y), then x = y!! (this actually means that if you begin with x and y, and it maps to the same y-value, then x should be equal to y --> Peter Brown says this in his video of solutions to 2008 T1 V1A)

f(x) = f(y)

e^(2x) = e^(2y) (you can log both sides and simplify)
2x = 2y
x = y

Extension to this question: Is f(g(x)) an onto function? Is g(f(x)) an onto function? Give reasons for your answers

 

[Drewk]

Q3 - I got the answer as AUB, maybe you made a mistake while doing the reverse distributive law too quickly

Hmm so for Q 3 i'm not sure how you mean i did the reverse distributive law wrong... here is my working
On 1st line we get (AUB^c)U(AUB) - De morgans law, second line we use the reverse distributive law
The distributive law says: "A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)"
So we note the A's common in both brackets on the RHS in the above proof and apply to Q taking A...(...)
Then the sign between the brackets in the law is placed in next to the A so then we have A∪..(...)
The the remaining sign is in between brackets hence for our Question A∪(..∪..)
Finally placing the B and B^c in A∪(B∪B^c) am i still wrong ??

Also i found out that double angle formula will likely be tested

 

[Thief]

If anyone will be revising today, there's solutions on the mathsoc website.

Also i found out that double angle formula will likely be tested

It's confirmed true by lecturer.

 

[HeroicPandas]

Drewk, read the question again

Your (AUB^c)U(AUB) should be (A intersect B^c) U (A U B)

AND say if you began from (AUB^c)U(AUB) - you can't use the distributive law

 

[Thief]

Drewk, read the question again

Your (AUB^c)U(AUB) should be (A intersect B^c) U (A U B)

AND say if you began from (AUB^c)U(AUB) - you can't use the distributive law

What degree are you doing Heroic?