VCE Methods Question Thread!

[knightrider]

For methods 3/4 are you actually required to know the binomial theorem.

Can it be assessed in VCAA exams. 

[keltingmeith]

For methods 3/4 are you actually required to know the binomial theorem.

Can it be assessed in VCAA exams.

You won't be asked about the binomial theorem directly, but you should be able to expand all expressions of form , even if just out of necessity. For example, the only way to compute is to expand the brackets.

[catherine_mimi]

You take comfort in knowing that VCAA will never ask this on a tech-free exam, then you turn on your calculator and use that to graph it. Where did you pull this question from?

Thanks for letting me know. The question is from chapter 7 in the Cambridge methods textbook.
It asks me to find the range of the function by graphing it.
Even if I graph it using CAS, I'm unsure of finding the horizontal asymptote, and from that, finding the range.
The answer is (1, infinite)
Please help

[psyxwar]

Thanks for letting me know. The question is from chapter 7 in the Cambridge methods textbook.
It asks me to find the range of the function by graphing it.
Even if I graph it using CAS, I'm unsure of finding the horizontal asymptote, and from that, finding the range.
The answer is (1, infinite)
Please help

We can treat e^(2/sqrt(x)) as the composite function f(g(x)), where f(x)=e^x and g(x)=2/sqrt(x)

The domain of g(x) is (0,infinity) because sqrt(x) is the denominator (meaning it can't be negative or 0). Hence, the range of the graph is also (0, infinity), because sqrt(x) is always positive, implying that 2/sqrt(x) is also always positive.

The range of g(x) inputted into f(x) gives us the range of f(g(x)) (because its a composite function and effectively all it means is that instead of inputting x into the function f, you are inputting g(x)).

Thus, to find the range of f(g(x)) we sketch f(x) for x e (0,infinity), which gives us (1, infinity).

Hope that makes sense!

[SE_JM]

Hello!
I just come across a weird question. I was wondering if someone could help?

For which values of k does the equation kx²-2kx=5 have one solution for x?


Seemed like an easy question, and I solved it like this:

If it has one solution that means, b²-4ac = 0. then (-2k)²-4(k)(-5)=0

Thus,
4k²+20k=0
4k(k+5)=0,

Which made me realise k= -5 or k=0.

But the answer tells me the correct answer is k=-5 only

Why is this? I know if I substitute k=0 into the original equation, two solutions exist. But if I had done this correctly, I shouldn't have gotten k=0 as one of my answers...

Could someone tell me why?

[psyxwar]

Hello!
I just come across a weird question. I was wondering if someone could help?

For which values of k does the equation kx²-2kx=5 have one solution for x?


Seemed like an easy question, and I solved it like this:

If it has one solution that means, b²-4ac = 0. then (-2k)²-4(k)(-5)=0

Thus,
4k²+20k=0
4k(k+5)=0,

Which made me realise k= -5 or k=0.

But the answer tells me the correct answer is k=-5 only

Why is this? I know if I substitute k=0 into the original equation, two solutions exist. But if I had done this correctly, I shouldn't have gotten k=0 as one of my answers...

Could someone tell me why?

First of all, if you sub in k=0 you get no solutions. Not sure if you just mistyped but yeah.

There's nothing wrong with your working, you just need to make sure that you check the solutions by subbing them back into the equation. Remember that the discriminant comes from the quadratic formula. If k is 0, then your a value is 0, meaning the quadratic formula becomes something divided by 0 which is not defined.

[SE_JM]

Thank you. I'll sub it in the original equation from now on

Could I ask one more thing? I attached a picture and it's what I wanted to ask. It says Note: (-1,2) is a point of zero gradient.
They found this by first finding a mapping of the translational transformations only and they applied this transformation to the point (0,0) which is the point of zero gradient in the standard x³ graph.

I was just wondering if this method works for other graphs such as x⁵ and also ones with even powers like x⁶?
and if I use this method to find point of zero gradient, is this always going to yield a correct answer, or are there exceptions to this that does not work?
Thank you

[psyxwar]

Thank you. I'll sub it in the original equation from now on

Could I ask one more thing? I attached a picture and it's what I wanted to ask. It says Note: (-1,2) is a point of zero gradient.
They found this by first finding a mapping of the translational transformations only and they applied this transformation to the point (0,0) which is the point of zero gradient in the standard x³ graph.

I was just wondering if this method works for other graphs such as x⁵ and also ones with even powers like x⁶?
and if I use this method to find point of zero gradient, is this always going to yield a correct answer, or are there exceptions to this that does not work?
Thank you

Yeah it'll always work, but its not really useful since its application is pretty limited. Calculus finds stationary points pretty easily anyway

[SE_JM]

Thanks!

[keltingmeith]

Yeah it'll always work, but its not really useful since its application is pretty limited. Calculus finds stationary points pretty easily anyway

Just to expand on this, if all you're doing is translating a graph by the rule (x, y)-->(x+h, y+k), then the gradient at the point (x, y) is the same as the gradient at the point of the new graph at (x+h, y+k).

For example, let's say a graph has gradient -5 at the point (2, 1), and then changes by the rule (x, y)--->(x+6, y-7). Then, at the point (8, -6), the new graph has a gradient of -5.

[knightrider]

How would you do this question?

State the coefficient of  , ii and iii in this equation?

[Orb]

How would you do this question?

State the coefficient of  , ii and iii in this equation?

Edit: woops completely wrong hahaha

[keltingmeith]

How would you do this question?

State the coefficient of  , ii and iii in this equation?

I'd expand it:



So, x^2 and x^4 have no coefficient, but the coefficient of x^3 is -540.

[knightrider]

I'd expand it:



So, x^2 and x^4 have no coefficient, but the coefficient of x^3 is -540.

Thanks EulerFan 101 

[knightrider]

How would you do this question?

In the expansion of , the coefficient of the second term is −192. Find the value of n.