VCE Methods Question Thread!

[kinslayer]

Dilate by 1/2 from y-axis, that is, f(2x) to get:
f(2x) = 16x^4 + 2x
Translate one unit up, that is, f(2x) + 1, to get:
f(2x) + 1 = 16x^4 + 2x + 1 = g(x)

This is good but you forgot the reflection in the y-axis i.e. f(-x)

[knightrider]

When solving trig equations.

When you give your final answer and say x=bla,blah,etc

does your answer have to be in order of lowest x values to highest x values?

[kinslayer]

When solving trig equations.

When you give your final answer and say x=bla,blah,etc

does your answer have to be in order of lowest x values to highest x values?

No, it just has to include all of the possible solutions.

[Mershian]

This has really been bugging me.

Find the derivative of Sin^2(x)

So I've used the chain rule to get to: [2sin(x)] [cos(x)]

But what are the next steps? The answer is sin2x.

Any help would be much appreciated.

Thanks

[StupidProdigy]

This has really been bugging me.

Find the derivative of Sin^2(x)

So I've used the chain rule to get to: [2sin(x)] [cos(x)]

But what are the next steps? The answer is sin2x.

Any help would be much appreciated.
Thanks

Your answer is right. Its just been simplified using the double angle formula for sin(2x). Just type into google 'sin double angle formula' and you'll find out some more about it. Don't be too concerned about it though, its mainly specialist trig stuff, just know that it exists from a methods viewpoint

[Mershian]

Your answer is right. Its just been simplified using the double angle formula for sin(2x). Just type into google 'sin double angle formula' and you'll find out some more about it. Don't be too concerned about it though, its mainly specialist trig stuff, just know that it exists from a methods viewpoint

Alright thanks a lot, I'll look it up for tomorrow.

So if that pops up in a SAC or exam I can leave my answer in that form?

You don't understand how long I spent trying to comprehend that question haha 

[knightrider]

For the question (attached).

i got the answer k=3 or k=-3.

The book has an answer that k=3.

Who is right and why?

[knightrider]

How would you do this question?(attached)

[knightrider]

anyone?  Thanks 

How would you do this question?(attached)

For the question (attached).

i got the answer k=3 or k=-3.

The book has an answer that k=3.

Who is right and why?

[cosine]

How would you do this question?(attached)

And of course you can confirm k, by substituting it into your integral:

[kinslayer]

For the question (attached).

i got the answer k=3 or k=-3.

The book has an answer that k=3.

Who is right and why?

The function f(x) = 2/x has a vertical asymptote at x = 0. Even though which has the same value for , the integral has a vertical asymptote in its domain, which doesn't make sense.

In other words, the book is right. Soz

[knightrider]

And of course you can confirm k, by substituting it into your integral:

Thanks so much cosine 

[knightrider]

The function f(x) = 2/x has a vertical asymptote at x = 0. Even though which has the same value for , the integral has a vertical asymptote in its domain, which doesn't make sense.

In other words, the book is right. Soz

Thanks kinslayer 

wait but shouldnt it be .

also when you antiderive .

[knightrider]

Here is a question (attached)?
here is my working out.











The answer says  k=3 only.

But doesnt the absolute value mean that k can also equal -3 too.
The  CAS calculator also says k=3 or k=-3.

Who is right?

[HighTide]

Here is a question (attached)?
here is my working out.











The answer says  k=3 only.

But doesnt the absolute value mean that k can also equal -3 too.
The  CAS calculator also says k=3 or k=-3.

Who is right?

The absolute value method is correct. You can also think of it this way with a graphical perspective. The definite integral in this case is giving you the area under the curve with respect to the x axis. Hence the value of k must exist the whole way for the area to be calculated. As the graph has an asymptote at x=0, in this case, the graph cannot cross to x=-3. Therefore, you have to reject the solution k=x=-3, and thus the only solution is k=3.
As for your question regarding the calculator: The calculator does indeed provide two solutions. However, if you sub in k=-3 to the definite integral, you will see that it will come as undefined. This is because you can't calculate the area under the curve in the case that there is an undefined region in between x=a and x=b (in integral).
Hope this helps