VCE Methods Question Thread!

[Moe112]

Hi everybody, can I please get some help with this composite function question:

Q9, f:{x:x≤3}→R,f(x)=3-x and g:R→R,g(x)=x^2-1
b) Define a restriction g* of g such that f o g*is defined and find f o g*

I have a feeling that the answer is painfully obvious but just cant see it... Thanks in Advance

[aznxD]

The composite function f(g(x)) is only defined if the range of g(x) is a subset of the domain f(x).
Rang is [-1,infinity)
Domf is (-infinity,3)
So the max range of g* has to be [-1,3]
From there, work out the domain of g required to obtain this range.

[Moe112]

The composite function f(g(x)) is only defined if the range of g(x) is a subset of the domain f(x).
Rang is [-1,infinity)
Domf is (-infinity,3)
So the max range of g* has to be [-1,3]
From there, work out the domain of g required to obtain this range.

So g:R→R,g(x)=x^2-1 becomes g:[0,2]→R,g(x)=x^2-1 and then f o g* = 4-x^2?

[butene]

since the rang has to be [-1,3], the new domain is [-2,2] for fog to exist
draw the graph and ull get it

[Moe112]

since the rang has to be [-1,3], the new domain is [-2,2] for fog to exist
draw the graph and ull get it

Many Thanks I finally get it, when working it out I forgot (0,-1) is the turning point so naturally I forgot all the basics including the Square Root of 4 is equal to plus or minus 2.  Bah it's been too long since I sat down and done some maths 

[Bee49]

I've been struggling with this question for too long, help please!

Find the value of m for which the simultaneous equations
3x + my = 5 and (m + 2)x + 5y = m have:

a) an infinite number of solutions
b) no solutions

I understand that for (a) the equations are the same and for (b) the gradients are the same.

[monkeywantsabanana]

I've been struggling with this question for too long, help please!

Find the value of m for which the simultaneous equations
3x + my = 5 and (m + 2)x + 5y = m have:

a) an infinite number of solutions
b) no solutions

I understand that for (a) the equations are the same and for (b) the gradients are the same.

3x+my=5
(m+2)x+5y=m

For no unique solution, determinant = 0
15-m(m+2)=0
15-m^2-2m=0
-m^2-2m+15=0
-(m-3)(m+5)=0
m=-5 or m= 3

sub m=-5
we have:
3x-5y=5
-3x+5y=-5

These two equations ARE THE SAME thus infinite number of solutions

sub m= 3
3x+3y=5
5x+5y=5

These two equations HAVE THE SAME GRADIENT thus no solutions

Hope this helps.

[butene]

Adding to the post above, not only do they have the same gradient but it's also because they have different y-intercepts
this tells us that both lines will never intersect so there'd be no solutions

[Panicmode]

I've been struggling with this question for too long, help please!

Find the value of m for which the simultaneous equations
3x + my = 5 and (m + 2)x + 5y = m have:

a) an infinite number of solutions
b) no solutions

I understand that for (a) the equations are the same and for (b) the gradients are the same.

The easiest method of solving this using the matrix method.

Set up a matrix as if you were going to solve these two equations simultaneously, then work from there. I have my working attached.

If you have more questions feel free to PM me.

Hope I helped. 

[Bee49]

I've been struggling with this question for too long, help please!

Find the value of m for which the simultaneous equations
3x + my = 5 and (m + 2)x + 5y = m have:

a) an infinite number of solutions
b) no solutions

I understand that for (a) the equations are the same and for (b) the gradients are the same.

3x+my=5
(m+2)x+5y=m

For no unique solution, determinant = 0
15-m(m+2)=0
15-m^2-2m=0
-m^2-2m+15=0
-(m-3)(m+5)=0
m=-5 or m= 3

sub m=-5
we have:
3x-5y=5
-3x+5y=-5

These two equations ARE THE SAME thus infinite number of solutions

sub m= 3
3x+3y=5
5x+5y=5

These two equations HAVE THE SAME GRADIENT thus no solutions

Hope this helps.

Thank you! Brilliant, I understand completely now

[Bee49]

Thank you butene and panicmode, very helpful I love this forum.

[lbeste12]

Been raking through my brain to see the solution, just not getting it.

Let a be a positive number, let f:[2,∞)→R, f(x)=a – x and let g(-∞,1] →R, g(x) = x² + a. Find all values of a for which f◦g and g◦f.

Its in the Essentials book, chapter 1F question 13.

[gumby97]

Hi guys

Find the value of a for which (1-2a)x^2 + 5ax + (a-1)(a-8) is divisible by (x-2) but not by (x-1)

help would be much appreciated, never come across a question like this before

[aznxD]

For f◦g to exist: rang has to be a subset of domf
For g◦f to exist: ranf has to be a subset of domg

Domf = [2,∞)
Ranf = (-∞,a-2]

Domg = (-∞,1]
Rang = [a,∞)

For f◦g: (I work these out by drawing the respective values on a number line)
a2
For g◦f:
a-21
a3

Therefore:

[butene]

Hi guys

Find the value of a for which (1-2a)x^2 + 5ax + (a-1)(a-8) is divisible by (x-2) but not by (x-1)

help would be much appreciated, never come across a question like this before

since it's divisible by x-2
solve for P(2) = 0
should get a = 3 or 4 but becareful, it said that this equation cannot be divisable by x-1 so
solve for P(1) =/ 0
(a-3)^2=/0, a =/3
so since a=/3
a=4 which is the other option