VCE Specialist 3/4 Question Thread!

[Stick]

I'd do this by drawing out a right angled triangle and applying SOH CAH TOA and Pythagoras' theorem.

If you get stuck:

Spoiler

(the "opposite" side of your triangle is now and the "adjacent" side is 1)



(used Pythagoras' theorem to find the hypotenuse)





Answer is A.

EDIT: beaten by b^3

[ahat]

I miss this thread so much, it's only been like 2 months since our last exams, but I've already forgotten half of my spec knowledge! D:
I want to go back to school ♥
But all of you guys who continue to help even during this time, are amazing. I truly respect you all

[M_BONG]

Can anyone please clear up my misconception here:

a) Solve for x:
Divide both sides by gives . Therefore there is no solution for x.

The answer is wrong as there are solutions for x.

Can anyone please tell me why it is wrong to divide both sides by  in this case?


b) Also, how would you guys sketch using coordinate geometry method? Ie. finding nature of asymptotes, vertical and horizontal asymp etc. (in other words, not using second derivative), if that makes sense.


Thanks heaps!

[nhmn0301]

Can anyone please clear up my misconception here:

a) Solve for x:
Divide both sides by gives . Therefore there is no solution for x.

The answer is wrong as there are solutions for x.

Can anyone please tell me why it is wrong to divide both sides by  in this case?


b) Also, how would you guys sketch using coordinate geometry method? Ie. finding nature of asymptotes, vertical and horizontal asymp etc. (in other words, not using second derivative), if that makes sense.


Thanks heaps!

a) Whenever you get something like ab=0. there are always 2 possibilities: a= 0 or b= 0. So in this case, since x can be divined therefore,  e^x must not = 0. then the only possibility that we have is (4x + 2 + x^2 )=0. From there, you can easily find the value of x by DOPS ( there should be 2 solutions).
In brief, there is nothing wrong with dividing (4x + 2 + x^2) if the e^x can = 0 ( but the truth is it can't). So just in this scenario, you have to look at when x can be defined and when it cannot.
b) It's pretty hard to illustrate how to sketch that graph from here but what I can briefly tell you is this graph involves "adding" 2 separate graph to obtain a new graph using addition of ordinates. ( ordinates = y-value).
I find a video on youtube which can be quite helpful if you have a look at:
http://www.youtube.com/watch?v=h4R-IiZ91tQ
I hope that you can find how to deal with this graph from here
Hope this helps, let me know if there is any errors !

[psyxwar]

Can anyone please clear up my misconception here:

a) Solve for x:
Divide both sides by gives . Therefore there is no solution for x.

The answer is wrong as there are solutions for x.

Can anyone please tell me why it is wrong to divide both sides by  in this case?


b) Also, how would you guys sketch using coordinate geometry method? Ie. finding nature of asymptotes, vertical and horizontal asymp etc. (in other words, not using second derivative), if that makes sense.


Thanks heaps!

(a) imagine 5x=0. You can't do 5=0/x, therefore 5=0. In this case it is obvious x is 0 (null factor law), and we can't divide by zero. In the question you gave, e^x can't equal zero (you can't raise any base to a power so that it equals 0), meaning the other part must equate to zero, and that's what you divided by.

(b) I don't see why you'd need the second derivative for this, apart from finding where the inflection point is?

[lzxnl]

The second derivative is useful for classifying the type of stationary point you have.
In this case, find the x values where dy/dx=0. Then, sub these x values into the second derivative. A positive second derivative means a local minimum and a negative one means a local maximum.

[survivor]

Hi could someone please help me factorise this into linear factors over C:

z^3 - 4z^2 - 4z - 5

thanks

[nhmn0301]

Hi could someone please help me factorise this into linear factors over C:

z^3 - 4z^2 - 4z - 5

thanks

Sub z=5 into the equation, you should get f(5) = 0. Therefore, using factor theorem, you will get (z-5) is a factor. From this stage, you can use long division or you can try "guessing the number" ( sorry, don't know a proper name for it). What I did is (z-5) (z^2 + bz +1 ) should be the final answer ( try to think why I can get the +1 in the quadratic). Then just expand the whole equation above out, equating coefficient, you shall get the b=1 and the final answer is (z-5) (z^2 +z +1).
Hope this helps

[survivor]

Sub z=5 into the equation, you should get f(5) = 0. Therefore, using factor theorem, you will get (z-5) is a factor. From this stage, you can use long division or you can try "guessing the number" ( sorry, don't know a proper name for it). What I did is (z-5) (z^2 + bz +1 ) should be the final answer ( try to think why I can get the +1 in the quadratic). Then just expand the whole equation above out, equating coefficient, you shall get the b=1 and the final answer is (z-5) (z^2 +z +1).
Hope this helps

Thanks!

[Thorium]

A question has risen in my mind which has baffled me for weeks now. x_x

I know that cosec is the reciprocal of sin, and I also know that arcsin is the inverse of sin.
If I am not wrong, inverse of sin and reciprocal of sin are both the same, 1/sin (or sin^(-1)), but for some reason, there's different terminologies used for each of them. If cosec=arcsin=1/sin, then why are they plotted differently. But if they are not equal, then it doesn't make sense again since they are both 1/sin.

Or is it that a reciprocal is not equivalent to inverse?

Anyone's help will be appreciated.

[brightsky]

okay there is a difference between sin^(-1)(x) and 1/sin(x), although the laws of algebra seems to tell us that the two are equal. mathematicians use sin^(-1)(x) or arcsin(x) to signify the inverse function of sin(x) over the domain [-pi/2,pi/2]. recall the definition of an inverse function. let's say we have a function f such that when we input the number a, the number b is outputted. the inverse function of f, denoted by f^(-1), is a function such that when we input the number b, the number a is outputted. the inverse function of f is a function that 'undoes' f. now, this is radically different from the reciprocal function of f, which is merely 1/f. mathematicians use cosec(x) to signify the reciprocal function of sin(x) over the domain R, i.e. 1/sin(x).

hope this makes sense.

[Thorium]

okay there is a difference between sin^(-1)(x) and 1/sin(x), although the laws of algebra seems to tell us that the two are equal. mathematicians use sin^(-1)(x) or arcsin(x) to signify the inverse function of sin(x) over the domain [-pi/2,pi/2]. recall the definition of an inverse function. let's say we have a function f such that when we input the number a, the number b is outputted. the inverse function of f, denoted by f^(-1), is a function such that when we input the number b, the number a is outputted. the inverse function of f is a function that 'undoes' f. now, this is radically different from the reciprocal function of f, which is merely 1/f. mathematicians use cosec(x) to signify the reciprocal function of sin(x) over the domain R, i.e. 1/sin(x).

hope this makes sense.

thanks, that made sense, that was taught in yr 11 methods, but apparently I have alzheimer lol.
So if we use the definition of an inverse function:
f(x)=y=sinx
f^(-1) (x)=x=siny for yϵ[-π/2,π/2]
y=sin^(-1)x
y=1/sinx

Now in this case the derived function is both inverse and reciprocal, but it is graphed different accordingly with the two terminologies. So should I graph it as an arcsin or an cosec?

[nhmn0301]

Hi guys, I need help on a question.
Solve for z ( I've already done part a, just putting in up since part b links to part a)
a) z^3 = -8
b) Using your result from a), hence solve z^3 + 6z^2 + 12z + 8 = -8  ( I just need help on part b )
Thanks heaps

[brightsky]

(z+2)^3 = -8

so it's all the solutions from a) minus 2.

[nhmn0301]

(z+2)^3 = -8

so it's all the solutions from a) minus 2.

Oh thanks silly me didn't realise z^3 + 6z^2 + 12z + 8 = (z + 2)^3
Thanks brightsky.