2009 HSC, Extension 1 Mathematics Question 1 (12 marks)(a) Factorise 8x^3 +27 (2 marks)
Whew! This question is lovely, because in terms of factorising a cubic, this is as easy as it gets. It's of the form (a^3+b^3), where a=2x and b=3. This is really easy if you can remember the formula:
(a^3+b^3)=(a+b)(a^2-ab+b^2)
If you don't remember the formula, don't fret! You can look at this problem and kind of just figure it out by inspection and a little trial and error.
We're dealing with cubics, and we've got 8, x^3, and 27.
So 2 is the cube root of 8, x is the cube root of x^3, and 3 is the cube root of 27. Even if you're just guessing after this, you've got a pretty good idea of where to start.
(2x+3)
The reason I would 'trial' with the (+) is simply because the original polynomial has a (+). If it doesn't work out, try again with a minus sign. As I mentioned before, this is just if you don't remember the formula and you're stuck but still want to spend the time to work it out. I wouldn't do trial and error in an exam unless I was pretty sure I knew I was heading in the right direction. It's not worth spending lots of time on something if you're going in totally blind.
Anyway, so you have (2x+3) as one potential factor. Now let's see what's left over when we pull that out.
(2x+3)(4x^2-6x+9)
8x^3 +12x-12x+18x-18x +27
YAAAAAY!!!
(b) Let f(x)=ln(x-3). What is the domain of f(x)? (1 mark)
Because this is one mark, it means you should be able to do this by inspection based on your knowledge of the domain of the natural log function, or it should take just one or two simple steps. This should not take you longer than a minute.
What do we know about the function 'ln'??
Well, ln0 is undefined. Basically the ln(x) function is saying 'e' to the power of what number equals x? So 'e' to the power of what number equals zero? In this case, that number is undefined. Equally, e to the power of what number equals a negative number? Also undefined.
So now we can very easily answer this question. What are the values for x that satisfy the condition that (x-3) must be greater than zero?
(x-3)>0
Add 3 to both sides to have x as the subject and you get...
Domain: x>3
(c) Find lim ((sin(2x))/x) (1 point)
x->0
That looks confusing trying to type it out, but you can see below. Once again, this is a 1 point question so should take you a minute or so. Just a couple of steps.
Okay, there are a couple of things to note here.
First of all, it's a good idea to just know this:
If you know this, then you should know how to do this question. It is based on l'Hopital's rule.
This working should be sufficient for a 1 point question in the HSC. You may want to just say something about "from l'Hopital's rule..." or whatever but really for 1 point I can't imagine they'd need a proof as long as you know it.
Here is an explanation of how to use l'Hopital's rule:
(d) Solve the inequality (x+3)/(2x) > 1 (3 points)
Okay, so since this is three points we probably need a few steps, but solving inequalities should just be a mechanical process and if you have to stress about this or you're spending too much time thinking about it rather than just going through the motions of solving it, I suggest you just do a whole heap of these in preparation for your exam because it should be easy marks. Anything mechanical like this you want to just be able to churn out with no stress or effort. There will be longer harder questions later that deserve that stress and effort, but not this one.
Okay, so we want to find the critical point. We know that x can't equal zero because the term '2x' is in the denominator and this would be undefined.
So change the inequality into an equality for the time being to find the critical point.
(e) Differentiate x(cos^2(x)) (2 marks)
Once again, a mechanical question. Worth 2 marks, so I guess 1 mark for knowing what the product rule is and another mark for actually getting the correct answer.
The product rule is what you use to differentiate a function that is two functions multiplied together.
In this case, one function is x and the other is cos^2(x).
The product rule is...
f(x) = uv, where u and v are functions
then,
f'(x)= uv'+vu'
Pretty straight forward. You just need to memorise it.
(f) I have no idea how to do integration symbols and whatnot keyboard, so here's part (f) written out. It's worth 3 marks.
So there you go, question 1 of the 2009 extension 1 mathematics exam.
There's nothing in there that should cause any problems if you practice. Don't waste time writing out pretty notes about stuff or doing flashcards or whatever. Just do actual past papers! Do question 1 of 2009, 2008, 2007, 2006, 2005... You will start to see the same types of questions over and over again and this is how you really truly internalise it. Don't sit around memorising what l'Hopital's rule looks like if you have no idea how or when to apply it.
Anyway, with Question 1 it's often all mechanical, going through the motions type maths.
If you have any questions, please let me know I'm happy to help!
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I kinda like the first way of writing it. It's just easier to say in your head and have it stick so you can just have it up in your memory for your exam. The second way with f(x) and g(x) is good when you are first learning this method, perhaps, because it helps visualise two separate functions and what's happening with these functions. It's really up to you which one you like, but I will be using the 'u' and 'v' notation rather than f(x) and g(x).
So if you integrate the first part of the product rule, it's like the integral is undoing the d/dx, because it's like what's the integral of the derivative of u*v? The integration and differentiation undo each other.
Any questions? Issues? Disputes? Confused? Just a bit of a Nigel and need a friend? Comment below!
