VCE Methods Question Thread!

[paper-back]

Thanks monsieur kebab and alchemy

[Zealous]

Hey guys can someone help me out with this?

let f(x)=1/x  -3 and g(x) =-ax. The values of a for which the graphs of y=f(x) and y=g(x) have two unique intersection points are?

a) a<9/4
b) a>9/4
c) a>4/9
d) a<4/9
e) -infinity <a <0 or 0<a <9/4

How do i go about doing this? Thanks

Equate the two functions to find interesection points:



Multiply both sides by x to remove the x on the denominator:



Solving this equation for x will give us the intersection points of f(x) and g(x). We can now use the discriminant which has to be greater than 0 for this equation to have 2 unique solutions (or two unique intersection points).





Now it seems the answer is D, but there is one more thing you need to consider.

g(x)=-ax is a linear function, and if a=0 then the graph will be a horizontal line, which could only intersect once with the graph of f(x), so we need to exclude 0 as a possible value of a.

Therefore:

[IndefatigableLover]

also, i'm starting to learn change of bases for log at school now.. is there an easy way to understand/learn the rules?

Haha yeah depending on your textbook it might explain it or not (Maths Quest doesn't whilst Essentials does) but essentially Change of Base for logs is using the formula:

where 'b' is just any number that's not negative (so long as the top and bottom are the same then it'll work [personally I like to just use the natural log]). It's the same thing as what paper-back has written:

Do you mean as in:
?

Prime example of this question being used was in the VCAA 2011 Exam 2 MCQ 22 where they asked a question where you needed to utilise the change of base rule.
If you use Essentials then Exercise 5F Questions 7-8 are quite similar to the VCAA question on Change of Base actually..

Examiner's report doesn't say much about the question but 45% only got that question right in that year so it's a good question to ask since it isn't really focussed on in Methods I guess (well not in class for me anyway)..

[paper-back]

How would you do this question:
The equation 3sin(x)-1=b where b is a positive real number, has one solution in the interval (0,2pi). The value of b is?
Thanks

[Rishi97]

Evaluate the following
tan(240^o)

[Orb]

Evaluate the following
tan(240^o)

tan (240) is the same as tan (180+60)

Tan is positive in the third quadrant (CAST rule), therefore tan(240) = root 3

[Tasmania Jones]

Evaluate the following
tan(240^o)

tan (240) is the equivalent of . If you look at the unit circle, it is in the 3rd quadrant, where tan is positive.
is
so is

If you don't know the exact values of tan, it can be found using

EDIT: beaten and fixing

[Tasmania Jones]

How would you do this question:
The equation 3sin(x)-1=b where b is a positive real number, has one solution in the interval (0,2pi). The value of b is?
Thanks

I'm not 100% sure but because there is only 1 solution, then it must be at either of the turning points of the sin graph.
If the solution to sin(x) was anything other than -1 or 1, then it would have 2 solutions.
rearranging the equation->   where it equals -1 or 1
because b is positive, then b can only equal
*not sure if I'm correct though

EDIT: So the graph of 3sin(x)-1 with domain (0,2) has a positive turning point with y-intercept 2, and this is the only part of the graph with a single positive y-intercept within the domain.
Therefore b=2

[Rishi97]

tan (240) is the same as tan (180+60)

Tan is positive in the third quadrant (CAST rule), therefore tan(240) = root 3

The answer is 0
Why?

[Tasmania Jones]

The answer is 0
Why?

Is it a typo in the answers? How can it equal 0?

[Rishi97]

Is it a typo in the answers? How can it equal 0?

I don't think so.
Because the previous answers were all 1,-1,0,undefined etc...
eg tan270 = undefined

[Stevensmay]

From the question you have given, tan(240 degrees) definitely =

[Rishi97]

ok so there must be a typo.
Thanks

[paper-back]

Thanks Tasmania Jones,
 The answer is b=2 which is a max on the graph,

Rishi, is that on maths quest? I remember my friend pointing that question out, I'm guessing its a typo too

[Tasmania Jones]

Thanks Tasmania Jones,
 The answer is b=2 which is a max on the graph,

Rishi, is that on maths quest? I remember my friend pointing that question out, I'm guessing its a typo too

So the graph of 3sin(x)-1 with domain (0,2) has a positive turning point with y-intercept 2, and this is the only part of the graph with a single positive y-intercept within the domain.
Therefore b=2