VCE Methods Question Thread!

[~T]

Let







We want

We have a point that is 'too large' and the gradient is positive at this point, so we need a point before x=3, if that makes sense...

Let



for



therefore

Our solution is then approximately

[clıppy]

The graphs of and intersect at two distinct points for
A - a <= -1
B - a < -1
C - a = -1
D - a > -1
E - a >= -1

Solving with the cas I've figured out it's either D or E but the answers say it's D. Why can't it be E?

[Alwin]

The graphs of and intersect at two distinct points for
A - a <= -1
B - a < -1
C - a = -1
D - a > -1
E - a >= -1

Solving with the cas I've figured out it's either D or E but the answers say it's D. Why can't it be E?

If it's equal to -1 then there will only be one point of intersection.

[clıppy]

If it's equal to -1 then there will only be one point of intersection.

(Image removed from quote.)

That made perfect sense as soon as you showed it. If I've already proved that it can't be C, then -1 can't be included on any of them.
Thanks Alwin

[clıppy]

New question about transformations applied to graphs.
Let's say I have y=tan(2x) and its restricted from (-pi/4, pi/4). I dilate it by a factor of 3 from the x-axis and reflect it in the x-axis. Would these translations affect the restricted domain?
Which transformations would affect a restricted domain?

[e^1]

New question about transformations applied to graphs.
Let's say I have y=tan(2x) and its restricted from (-pi/4, pi/4). I dilate it by a factor of 3 from the x-axis and reflect it in the x-axis. Would these translations affect the restricted domain?
Which transformations would affect a restricted domain?

   ().

If we dilate from the x-axis and reflect it in the x-axis, such transformations do not affect the restricted domain, as the range will still remain . Hence, the restricted domain is unaffected.


Check me if this is wrong.

[~T]

New question about transformations applied to graphs.
Let's say I have y=tan(2x) and its restricted from (-pi/4, pi/4). I dilate it by a factor of 3 from the x-axis and reflect it in the x-axis. Would these translations affect the restricted domain?
Which transformations would affect a restricted domain?

No they wouldn't. The only transformations that will affect the domain are those that affect the graph's behaviour or placement along the horizontal, so to speak. This is of course limited to dilations from the y axis and translations parallel to the x axis, as the graph will respectively be stretched/compressed horizontally or shifted left/right horizontally, and in some cases reflections in the y axis as a domain of [4,7] will become[-7,-4], but [-2,2] will remain [-2,2].

It makes sense if you visualise the graph. We have a tan graph over one period; this singular line will not be over a different domain if we merely stretch it upwards (dilate 3 from x axis) or reflect it. The only actions that will alter this domain (the values of x for which the graph is defined) is if it is moved across or stretched to be wider or narrower.

I hope that makes sense

[phillp]

1)sand is poured into a heap in the shape of a cpne with angel 120º at the vertex. if the height pf the heap is increasing at 3cm per second when the height is 7cm, then volume of the cone is increasing at?
2) The amount of rainfall (mm) on a tropical island over a week can be modelled by a function fx)=-0.3x(x-7)+3
[0,7] . the average rainfall over 7 days?
3)For the function y=1/2x^4 the approximation value of the area between the graph and the x axis using the right end point esitmate wtg equal intervals -2  and 2 is equal to?

Practice McQ from 2006 past exam CSE

[SocialRhubarb]

1. Try to find a relationship between height and radius, and hence find a relationship between height and volume.

Spoiler

If you draw a cross section of the cone, you can see that we can split the 120 degrees at the vertex into two 60 degree angles, and create two right angled triangles.

If we do this,

[phillp]

thanks

[Essendon2013]

Solve √2 sin (2x) = -1 for x ∈ R

I have done all the working out but not sure how many solutions to include because x ∈ R. Never encountered a question like this where x ∈ R, only where x has been restricted over a domain.

Help would be appreciated!

[b^3]

In this case you can give the general solution, since this will give all solutions for .

You can also use the ... e.t.c general solution for sin if you want to as well, but both that solution and the two above will give all solutions for .

EDIT: Should also add for the second angle you could use like shadows has done below, as this will still give the same set of solutions, as you can use any integer for .

[shadows]

Have you heard of giving your answers as a general solution?

First find your root values

Assume that the values were defined from 0 to pi.
then x = 5 pi /8  and 7 pi /8

Since this function is has a period of pi, then your other values would just be n times pi added or minus onto these values. , where n is your nth set of solutions after your first set.


Therefore, usually you add a parameter n. (although it could be anything really)

x= 5pi/ 8 + pi.n    or x=7 pi /8 + pi.n   where n is an element of Z     ( . indicates multiply)

you need to include the last part to Z, as Z means that n is a natural number. (It's only whole numbers) so  n can be -1 or 3 but can't be 2.3

I really hope that makes sense.

LOL beaten

[Essendon2013]

In this case you can give the general solution, since this will give all solutions for .

You can also use the ... e.t.c general solution for sin if you want to as well, but both that solution and the two above will give all solutions for .

Thank you so much! One question, how did you get -pi / 4 on the fourth line?

[shadows]

Since the function repeats itself every pi units
-pi/4  would just be 7pi/8  minus pi

this is correct because there is no given domain.

EDIT: OOPS DW WHAT I WROTE.

LOL that didn't even make sense and I didn't read it properly..

-pi/4 is the solution to

sin(x)= -1/(sqr(2))

b^3 then just took over the two ... so it is - pi/8

-pi/8 is 7pi/8 minus pi.

you can include this as the root of your general solution since the parameters just  includes all your other sets of solutions in the domain.