VCE Methods Question Thread!

[revcose]

I tried to take an approach which didn't rely on referring to the graph. I feel like the wording is dodgy though.

"Because the gradient o f ln(2x+5) is always decreasing, as values of x increase, the amount of x values required for f(x) to change from 0 to 1 or 1 to 0 increases. This means that between any two intercepts, fewer x values will be required to increase it to its stationary point (from f(x)=0 to f(x)=1) than back down to f(x)=0, thus the stationary point is to the left of ((x1+x2)/2), f((x1+x2)/2)."

[~T]

I tried to take an approach which didn't rely on referring to the graph. I feel like the wording is dodgy though.

"Because the gradient o f ln(2x+5) is always decreasing, as values of x increase, the amount of x values required for f(x) to change from 0 to 1 or 1 to 0 increases. This means that between any two intercepts, fewer x values will be required to increase it to its stationary point (from f(x)=0 to f(x)=1) than back down to f(x)=0, thus the stationary point is to the left of ((x1+x2)/2), f((x1+x2)/2)."

Okay, it's always a good idea to be able to visualise things without the graph.

I see where the answers are going and will try to clarify it... firstly, f(x) can be seen as a composite function of |sinx| and ln(2x+5)

When we think about any log function, it increases very rapidly to begin with, but this 'slows down' as the  values get larger (this is what the answers suggest by saying that the gradient is always decreasing)

Because the log function is the 'inside' function of the composite function we talked about earlier, it can be seen as a kind of 'input' to the function |sinx|

Now, if this input is rapidly increasing early on, and slowly increasing later on, then we are going to get a sin graph that is rapidly cycling through its period early on, and slowing down / being stretched out later on.

If we try to visualise this, we get the idea that for each 'up and down' motion of the graph, it will always increase to the peak 'faster' than it will decrease to zero. The result will be a kind of sin function slanted to the left, thus the peak will occur before the middle of two intercepts.

Does it make more sense?

[jono88]

When I solve on my CAS (casio) it gives 2 incorrect solutions and says "warning: more solutions may exist" any idea how to fix this?

Place domain restrictions. problem solved.

[rhinwarr]

The graph of cubic function g(x) has only one x-intercept (-2,0) and a y-intercept (0,2). g(x) has a quadratic factor ax^2+bx+c . Find the relationship between a and b.

So far I've subbed in the first two co-ordinates to get:
1) -8a+4b-2c+d=0
2) d=2
Therefore, -8a+4b-2c+2=0

I don't know what to do with the quadratic factor thing. If that is the quadratic factor, does it mean that the linear factor is (x+d)? But if you sub that into the cubic equation you just get the same thing as I've already got above.

[SocialRhubarb]

By quadratic factor, it means an irreducible quadratic. We can also tell this from the fact that there is only one x-intercept.

As a result, the quadratic is never equal to 0, which means that for the cubic function g(x) to equal 0, the linear factor must be 0. Since this occurs at x=-2, (x+2) must be a factor of g(x).



Subbing in the y-intercept gives us:







And finally since the quadratic is irreducible, there must be no solutions to .

Therefore,

.

I don't know if this is the specific relationship they wanted, or if there's more to the question, but it's a relationship at the very least.

[rhinwarr]

Yep that's right. Thanks

[BasicAcid]

By quadratic factor, it means an irreducible quadratic. We can also tell this from the fact that there is only one x-intercept.

As a result, the quadratic is never equal to 0, which means that for the cubic function g(x) to equal 0, the linear factor must be 0. Since this occurs at x=-2, (x+2) must be a factor of g(x).



Subbing in the y-intercept gives us:







And finally since the quadratic is irreducible, there must be no solutions to .

Therefore,

.

I don't know if this is the specific relationship they wanted, or if there's more to the question, but it's a relationship at the very least.

What do you mean by the quadratic is irreducible?

And rhinwarr, what exams are you doing? I haven't seen anything like these questions you've been asking on any of the ones I've been doing haha, these are really tricky!

[rhinwarr]

This question is from an itute worksheet.
http://www.itute.com/download-free-vce-maths-resources/free-maths-worksheets-for-year-5-12-students/
They're mostly just the 'basics' but the style of questions is kind of different from the actual exams.

[SocialRhubarb]

Irreducible quadratic means it cannot be made into the product of linear factors.

It just means that the equation ax^2+bx+c=0 has no solutions, which we can tell from the fact that the cubic only has one intercept.

[BasicAcid]

This question is from an itute worksheet.
http://www.itute.com/download-free-vce-maths-resources/free-maths-worksheets-for-year-5-12-students/
They're mostly just the 'basics' but the style of questions is kind of different from the actual exams.

Oh wow they're free as well, thanks for that haha.
I guess if you're able to confidently do all of these, you should be fine for the VCAA ones.


And ah thanks rhubarb, I didn't see that "only one x intercept" in the question.

[rhinwarr]

Solve for x.

[Jaswinder]

is the answer 1 and -ln(3)

[SocialRhubarb]

Could use quadratic formula as well.

[abcdqdxD]

Can someone explain the idea of integration by recognition and the steps needed to solve these types of problems?

Thanks

[Dayman]

Hey y'all I'm having difficulty with a question on the 2010 insight exam 2.

So f(x)=|sin(x)*(sin(x)-cos(x))| and the question asked to find a general equation that gives x values of the turning point yet when I put d/dx(f(x))=0 and solve it gives me solutions rather than a general equation please help.

Btw I  did it without the mod sign and still did not work. And I have it nspire cx