VCE Specialist 3/4 Question Thread!

[Sanguinne]

a bit of a simplistic problem.

lets say we had -2cosx=0 with no domain
             step 1)     cosx=0

after this, id assume we would use general solution to get x= (2npi) +- (pi/2) ,where n is Z
however the answer states x = pi/2 + npi, n is Z

are these answers equivalent?

[revcose]

You get exactly the same set of answers with both ways, the npi +pi/2 solution is just neater.

[Sanguinne]

You get exactly the same set of answers with both ways, the npi +pi/2 solution is just neater.

So im guessin you would get the marks either way.

thanks

[b^3]

Just a basic partial fractions question

Express into partial fractions

= 

Thus, 2x-5 = a (4-x) + b
 
Let x = 0 to eliminate "b"

-5 = 4a
a = -5/4

Let x = 4 to eliminate "a"
Thus, b = 3

HOWEVER, the answers are different.
It seems that "a" actually equals 2 and "b" equals "3" and this occurred because "a" was eliminated first (ie. They let x = 4 first and obtained b= 3 THEN let x = 0)
Can anyone explain why order matters when eliminating (why must I eliminate "a" by letting x =4 first) when trying to get partial fractions?

When you let you're not eliminating because isn't multiplied by , it's just a constant term with what you have left, and the equation would be . So it's easier to eliminate first and then substitute the value back in with any value of other than to find .

[Phy124]

I think it's easier to just expand it into the form .

Then you know that (as the x coefficient must be the same) and (as the constant term must be the same).

[Cort]

Just a basic partial fractions question

Express into partial fractions

= 

Thus, 2x-5 = a (4-x) + b
 
Let x = 0 to eliminate "b"

-5 = 4a
a = -5/4

Let x = 4 to eliminate "a"
Thus, b = 3

HOWEVER, the answers are different.
It seems that "a" actually equals 2 and "b" equals "3" and this occurred because "a" was eliminated first (ie. They let x = 4 first and obtained b= 3 THEN let x = 0)
Can anyone explain why order matters when eliminating (why must I eliminate "a" by letting x =4 first) when trying to get partial fractions?

I think that in this case, I think it's necessary to entirely eliminate the value, rather than having it floating it around. Because
2x-5 = a (4-x) + b
with x = 0 means that
-5= 4a + b. Hence, b still exists. It's only AFTER that you find the letter by itself do you use  x=0 to find the other.
e.g. Eliminate a via x =4. This allows you to find b by itself.
3 = b. Hence b=3. To FIND a, sub b=3 and x=0 to find a. Hence:
-5 =4a+3
-8 = 4a.
a = 2

[M_BONG]

Oh crap.

Thanks guys! I realised when I let x = 0 I accidentally removed "b" as well so I got 4a = 5;  a = 5/4 (thus the initially inexplicable mistake/error).

[Cort]

The question below had asked to find the range of f(x). However, after reading through the solutions below, I'm confused in some points:
1. Why does the solution just substitute the values as sin^-1 = pi/2, and not pi/2 into the full equation? That is,
instead of just sin^-1 = pi/2, instead (I thought) that it would be like
cos(sin^-1(2*pi/2)). When do you actually not substitute and instead swap it over to the value itself?

In another note, this works just like a composite function, right? That's why the range of sin^-1 was taken into the account of the dom of cos?

Thanks,
Cort.

[b^3]

They've found the value of rather than using at the extreme points (that is when you minimise and maximise ). This means that you're plugging into , and the values associated with it rather than into , they've already done the plugging in of values into .

For the second question, yeah like a composite function. "The range of the second () must be equal to or a subset of the domain of the first ().

[Cort]

They've found the value of rather than using at the extreme points (that is when you minimise and maximise ). This means that you're plugging into , and the values associated with it rather than into , they've already done the plugging in of values into .

For the second question, yeah like a composite function. "The range of the second () must be equal to or a subset of the domain of the first ().

Thank you very much. But forgive me for this possibly confusing next question: Does that mean that finding the implied domain in this case would always be the 'result' in the range(or is it dom?) inside, for any type of graph or something? So if I was given another question where  the dom and range of the two functions must be restricted for the NEW dom; those values would be used in place of the function inside? Hence, that is why, sin^-1 (x) = o and pi/2?

Also, you state that x is used at extreme points to min/max the function. Is this also the case for all functions (as in the exercise has told us that finding x individually) would be substituted into the inner function to find the range of the outer function?

[b^3]

Thank you very much. But forgive me for this possibly confusing next question: Does that mean that finding the implied domain in this case would always be the 'result' in the range(or is it dom?) inside, for any type of graph or something? So if I was given another question where  the dom and range of the two functions must be restricted for the NEW dom; those values would be used in place of the function inside? Hence, that is why, sin^-1 (x) = o and pi/2?

Partially confused with what you're asking. It depends at how you look at it, if you're looking at it as (that is the is restricted by the other function), then your implied domain would be the range of the inside function, which is what they've done above. If you look at it as , then your implied domain will be the domain of the inner function, BUT this will be restricted such that the output of this inner function fits into the input of the outer function (range of the second is equal to or subset of the domain of the first). The former is probably a bit easier when dealing with trigs, as you don't have to find the values for the inner function, rather just the range of the outer function. When it comes down to trig you'll normally get nice values there. If you're asked to find the domain of , then you'd have to do it the long way.

Also, you state that x is used at extreme points to min/max the function. Is this also the case for all functions (as in the exercise has told us that finding x individually) would be substituted into the inner function to find the range of the outer function?

Well if you had the domain of the inner function, you'd plug the 's that give extreme points that into the inner function, then plug that output into the outer function to give you the range.
If you had the range of the inner function, then you'd work out which values in that range would give extremes in the outer function, then plug those in to give you the range of , again the latter is normally easier for trig functions since you don't have to work back the original values, rather maximise the range of the inner function, which is easy for trigs and work off that.


Again though, if you want the domain of the whole thing you'll have to do it the long way.

Hope that helps, this can be quite confusing. At the end look back at what you've got and what you're plugging into what to check if it makes sense, and that you're not forgetting/overlooking something.

[IndefatigableLover]

A bathtub takes A minutes to fill up and B minutes to drain out completely.
If A<B, how long does it take for the bathtub to fill up if the sink hole is left unplugged?

[scribble]

if V is the total volume of water the bathtub can hold, then the bathtub fills at the rate Vliters per Aminutes or V/A L/min
the bathtub empties at the rate of V/B L/min
we're filling and emptying the bathtub at the same time, so the rate of this is V/A - V/B L/min (inflow-outflow)
If we make C the volume of water in the bathtub, and t the time in minutes, we can form an equation
C=(V/A - V/B)*t
the bathtub is full when C=V
so that's when V=(V/A - V/B)*t
V=V(1/A-1/B)*t
1=(1/A-1/B)*t
1=(B-A)/AB * t
t=AB/(B-A) mins

[alchemy]

A bathtub takes A minutes to fill up and B minutes to drain out completely.
If A<B, how long does it take for the bathtub to fill up if the sink hole is left unplugged?

Assume the bathtub is 1L (or any other value, it doesn't matter).
It fills up in A mins and drains in B mins. So, in A mins it would've filled up 1/A and in B mins it would've drained 1/B.
Since A<B, the net gain is 1/A - 1/B.
Because the sink hole is unplugged, water is going to be gained and drained at the same time. We can represent the total volume to be filled divided by the net gain to find the time it takes to fill with the hole left unplugged.
time=1/(1/A-1/B) = 1/(B-A)/AB = AB/B-A

EDIT: Beaten by scribble.

[alchemy]

I've attached a question and my working for it. My answer is correct, but I wanted to know if my method is legit (to transpose the equation in the way I did). Thanks.