VCE Methods Question Thread!

[ahat]

The graph of is transformed to , where k is a positive real number.

i) Find the value of k such that the equation    has only one solution in the domain and that the solution is at .

ii) Find the values of k such that the equation   has only one solution in the domain .

iii) Find the values of k such that the equation   has exactly three solutions in the domain .

Help on this question would be appreciated. My teacher used trial/error, but surely there's a better method?

Thanks

Solutions

Spoiler

[Alwin]

The graph of is transformed to , where k is a positive real number.

i) Find the value of k such that the equation    has only one solution in the domain and that the solution is at .

ii) Find the values of k such that the equation   has only one solution in the domain .

iii) Find the values of k such that the equation   has exactly three solutions in the domain .

Help on this question would be appreciated. My teacher used trial/error, but surely there's a better method?

Thanks

Solutions

Spoiler

First up, you should note that: but for the sake of simplicity I will use the form you gave your question in, as I understand that not all methods students are comfortable with double angle formulas.

Preamble, but PLEASE read as it is important

Now, lets look at what the question is asking (It's important to read the entire question before starting it!!!): for each part. Maybe this is important? Let's take a closer look:


Question seem a bit easier already?

Now if we solve for say, 6 solutions -1≤ n ≤ 1, we get:

a. Only one solution at and

working for part a

We could go through each solution and sub in to find a value of k:

subbing in

Going from left to right, these values would be:

But, if we look closely only if we let the forth solution will there be one solution in the domain.
In other words,

Therefore

b. Only one solution

working for part b

This one is a bit more tricky, because we have inequalities. We want:

Therefore, combining the inequalities we get

c.  Three solutions in the domain
This one is even more tricky, because we have to consider which three will be in the domain. But as I have gone though parts a and b already and want to get to other methods, I'm hoping you understand my logic and can work this one out yourself. I believe in you ^^


Other methods - Graphical.


The green is the area in the restricted domain.
The points of intersection are:

For part a, you can clearly see that we need to 'move' the intersection of to .
Consider  : It's a dilation by factor 1/k from the y axis is it not?
So, then since we want x=2pi -> x=6pi, it's by a factor of 3.
Therefore 1/k=3 -> k =1/3

For part b, the solution can be anywhere but the original intersections at x=-2pi and x= 6pi cannot be in the green zone.
Thus, after the dilation:

When solved, this gives: 1/3 ≤ x < 2

For part c, again I'll let you tackle this yourself and try using a graphical approach.


Good luck and hope it helps + make sense

[Sanguinne]

is there a difference between these two general solutions for sine

x=2npi + sin-1(a)      x=(2n+1)pi - sin-1(a)

[Damoz.G]

is there a difference between these two general solutions for sine

x=2npi + sin-1(a)      x=(2n+1)pi - sin-1(a)

Is that sine inverse? If so, then there is a difference between the two.

The best way is to enter your answer on your CAS and make it equal to the solution's answer. If it comes up with "true", then it is correct, if not, then it isn't correct. I'm just guessing that you got a different answer, and wondering if its the same answer in the solutions but in a different format, right?

[zvezda]

is there a difference between these two general solutions for sine

x=2npi + sin-1(a)      x=(2n+1)pi - sin-1(a)

Yes there is.
Say for angle "a", sin is positive. sin^-1(a) equals pi-sin^-1(a), correct? Sin^-1(a) is in the first quadrant and pi-sin^-1(a) is in the second, assuming sin^-1(a) is between 0 and pi/2. Now, for both of these angles, add 2npi where n is an integer, which just gives you the solutions for an indefinite number of periods around the unit circle. Rearrange these two new expressions and you will find that you get the two that you have there

[ahat]

I believe in you ^^

Alwin, I apologise, but I think I'm set to disappoint. You must forgive me in advance for the oncoming barrage of blasphemous questions, especially in your godly mathematical dominion, but I fear, a peasantly denizen of mathematics such as myself could never quite match your prowess.

First up, you should note that:

Thanks to Spec, I understand this, and thus, I tried my hand at finding a general solution as you did:


 


From observation:







First of all, have I done this correctly. Second of all,
Could you please tell me what it should be, i.e. what is n an element of? I'm pretty sure I've always used J in Methods and Z in Spec? (maybe)? What do they all mean Alwin?! And in what cases do we use a specific number system (e.g. you used N - natural numbers)? ლ(ಠ益ಠლ)

Secondly

Now if we solve for say, 6 solutions -1≤ n ≤ 1, we get:

Why do we use -1≤ n ≤ 1? And if you got 6 solutions, this means that you would have to have used fractions, so is the number system N not integers?  If, say, I had used my general solution would I still get  these results?

But, if we look closely only if we let the forth solution will there be one solution in the domain.
In other words,

I need some help for this part. I'm struggling to understand why we reject solutions being outside of the domain. For example, say if K was 10, then the solution:
  substitute K = 10 would still be in this domain. Conversely, say if K = then , the solution we used, would be 8 and therefore, outside of the domain?  So, are we just considering the numerator to see if these values of x will be within the specified domain, because k is merely a constant? Essentially, how did you reject the solutions?
**Shoot - I just realised. Did you just go along and substitute each k-value to see if there was 1/2/3 etc solutions? Isn't tha time consuming?***

I think if I can grasp this, I'll easily understand the other parts.

Much gratitude Alwin, be thankful you are forsaken of the tumultuous confusion experienced by your simpletoon peers.

[clıppy]

I get a bit lost on questions like these when there is a dilation parallel to the x applied to the function, how do you work with these?

Let f be a differentiable function for all real x, where f(x) <= 0 for all x E [0,a]

If then is equal to

A = -1/2
B = -1
C = -2
D = -4
E = -8

The answer is C

[RKTR]

I get a bit lost on questions like these when there is a dilation parallel to the x applied to the function, how do you work with these?

Let f be a differentiable function for all real x, where f(x) <= 0 for all x E [0,a]

If then is equal to

A = -1/2
B = -1
C = -2
D = -4
E = -8

             = 2 x
                                                                      = 2 x
                                                                      = 2 x -2
                                                                      = -4

[clıppy]

             = 2 x
                                                                      = 2 x
                                                                      = 2 x -2
                                                                      = -4

Sorry RKTR, i forgot to mention that the answer was C (-2)

Updated main post

[RKTR]

oops .. i got the wrong answer

[SocialRhubarb]

Let's say that we have some function, called  ,  such that the derivative of    is  .

What would be the derivative of  ? It would actually be  ,  because we have to apply the chain rule.

So working backwards, what is the integral of  ? Given that the derivative of    is  ,  the antiderivative of is , and hence, the antiderivative of   is  .

You'll realise that RKTR's working is just missing this step, and if you add on the half, his answer is correct.

[ahat]

I get a bit lost on questions like these when there is a dilation parallel to the x applied to the function, how do you work with these?

Let f be a differentiable function for all real x, where f(x) <= 0 for all x E [0,a]

If then is equal to

A specialist method, if it doesn't make sense, just ask:

First integral:

Let
When
When ,

So this integral is transformed to:


Second integral:
 

Let ,
When
When

So this integral is transformed to:


Look familiar?


Therefore,
Substituting the first integral:

[darklight]

A car is 250 km due east of a motorbike and travelling 60km/hr west. Motorbike is travelling north at a rate of 80km/hr.
Find the time at which the vehicles are closest to each other (and this distance).

[SocialRhubarb]

Okay, so to do this problem, we'll first need to establish an expression for the distance between the car and the bike.

So we can treat the distance between the two as the hypotenuse of a right angled triangle, with one side the north-south component of the distance, and the other side being the east-west component of our distance, meaning we can use Pythagoras' theorem in this instance.



Now, the north-west component is quite simple to express, since they both start at the origin, but the motorbike is moving north at a speed of 80 km/h, we can say that the northerly component is equal to 80t, where t is time.

The east-west component component starts at 250 km and gets smaller at a rate of 60 km/h, so we can express the east-west component of that distance as 250-60t.

Putting these values back into our expression for distance:



At this point, it is a simple case of differentiating this expression, finding the value of t for which D(t) is a minimum, and the value of D(t) at this point.

[Stevensmay]

Following on.


Let for a minimum and solve.


To find our distance just substitute t into the distance function we found earlier.
Distance is 200km.