VCE Specialist 3/4 Question Thread!

[Mattjbr2]

Hey guys! What's wrong with my logic here?
\[y= \frac{2x-4}{x^2}\\ \frac{\partial y}{\partial x}=\ \frac{-2}{x^2} + \frac{8}{x^3}\\ if\ \frac{\partial y}{\partial x}=0\ then \frac{-2}{x^2} + \frac{8}{x^3}=0\\ \therefore \frac{2}{x^2} = \frac{8}{x^3}\\ \therefore 2x^3 = 8x^2\\ \therefore 2x^3-8x^2=0\\ \therefore 2x^2(x-4)=0\\ \therefore 2x^2=0\ or\ x-4=0\\ \therefore x=0\ or\ x=4\]

My CAS calculator and book say x=4 only, and not zero. Why? Where's my mistake? I know you can cancel x out early, but I saw on one of the exam reports that you shouldn't do that, because you'd be cancelling out potential solutions. x always equaled zero in questions similar to this, why not now? In line 4 I multiplied both sides by x² and x³ to obtain line 5, apparently my CAS says this is not allowed, as the solutions to line 4 and line 5 are different, why isn't it allowed?

[Buddster]

Hey guys! What's wrong with my logic here?
\[y= \frac{2x-4}{x^2}\\ \frac{\partial y}{\partial x}=\ \frac{-2}{x^2} + \frac{8}{x^3}\\ if\ \frac{\partial y}{\partial x}=0\ then \frac{-2}{x^2} + \frac{8}{x^3}=0\\ \therefore \frac{2}{x^2} = \frac{8}{x^3}\\ \therefore 2x^3 = 8x^2\\ \therefore 2x^3-8x^2=0\\ \therefore 2x^2(x-4)=0\\ \therefore 2x^2=0\ or\ x-4=0\\ \therefore x=0\ or\ x=4\]

My CAS calculator and book say x=4 only, and not zero. Why? Where's my mistake? I know you can cancel x out early, but I saw on one of the exam reports that you shouldn't do that, because you'd be cancelling out potential solutions. x always equaled zero in questions similar to this, why not now?

x cannot equal 0 as the original function is undefined at that point, and therefore dy/dx is also undefined

[Mattjbr2]

Ahh I see, thanks!

[lzxnl]

The actual problem is when you multiply both sides by x. If x is zero, you've essentially multiplied both sides by zero. It's exactly like the fallacy going:
a = b
a^2 = b^2
(a-b)(a+b) = 0
a + b = 0
a = -b
a = -b = 0 for all a, b

Here, you can't divide by a-b if a = b.

[j.wang]

For question 6b in the first attachment, my answers were 58.67 degrees or 121.33 degrees (rounded off to 2 decimal places). But the correct answer is only 121.33 degrees because the "acute angle is inconsistent with the given angle 35 degrees". What does this mean?

For question 10 in the second attachment, how come there's 2 answers? I thought there wasn't any "ambiguity" with the cosine rule (2 sides and 1 non-included angle), like there can be with the sine rule

[Syndicate]

For question 6b in the first attachment, my answers were 58.67 degrees or 121.33 degrees (rounded off to 2 decimal places). But the correct answer is only 121.33 degrees because the "acute angle is inconsistent with the given angle 35 degrees". What does this mean?

For question 10 in the second attachment, how come there's 2 answers? I thought there wasn't any "ambiguity" with the cosine rule (2 sides and 1 non-included angle), like there can be with the sine rule

6) Angle BAC is 23.67 degrees. Which means angle ACB must be 121.33 degrees.

10) Since you end up solving for the length of BC as a quadratic, you will conclude with two values. To prove that they both work, you can calculate the magnitude of angle ABC (since it is an non-inclusive angle + you have two sides, you should get two values).

Angle ABC = 38.68, 141.32

Since you have two possible angles for ABC, you will also have two different lengths of BC.

[Gogo14]

How do you evaluate questions that have an inverse trig inside a trig?
E.g. Sin(tan^-1(-theta))

[Buddster]

How do you evaluate questions that have an inverse trig inside a trig?
E.g. Sin(tan^-1(-theta))

Draw a triangle and label all the information you can. I can help more if you have an example question with values

[Gogo14]

Draw a triangle and label all the information you can. I can help more if you have an example question with values

Q2-3d
Also, dont understand Q 3-2

[RuiAce]

Q2-3d
Also, dont understand Q 3-2

[Gogo14]

Sorry got a bit confused with your explanation
How did you get 2 Arguments? I lost you at that point

[RuiAce]

Sorry got a bit confused with your explanation
How did you get 2 Arguments? I lost you at that point

I didn't get 2 arguments.

I merely split up the cases.

(The only thing was I assumed that theta was between -π and π, which may be wrong.)

[Syndicate]

Sorry got a bit confused with your explanation
How did you get 2 Arguments? I lost you at that point

-cis(x) is a rotated 180 from cis(x). [This is just a scenario I created for this situation] So if your cis(x) (same as a+bi) was in the first quadrant, then -cis(x) (-a-bi) will be in the third quadrant. Depending on where Arg(cis(x)) was originally located, Arg(-cis(x) will be -pi + x. Which must right as it follows the principal argument convention (this works only and only if the x is a principal argument as well) .


Q2-3d)

we already know that tan(\theta) = opposite/ adjacent

[RuiAce]

-cis(x) is a rotated 180 from cis(x). [This is just a scenario I created for this situation] So if your cis(x) (same as a+bi) was in the first quadrant, then -cis(x) (-a-bi) will be in the third quadrant. Depending on where Arg(cis(x)) was originally located, Arg(-cis(x) will be -pi + x. Which must right as it follows the principal argument convention (this works only and only if the x is a principal argument as well) .


Q2-3d)

we already know that tan(\theta) = opposite/ adjacent

Problem was in the principal argument falling out of range for negative values.


For positive theta (less than pi) yeah that's right.

[Gogo14]

Ah I see, Thnx

-cis(x) is a rotated 180 from cis(x). [This is just a scenario I created for this situation] So if your cis(x) (same as a+bi) was in the first quadrant, then -cis(x) (-a-bi) will be in the third quadrant. Depending on where Arg(cis(x)) was originally located, Arg(-cis(x) will be -pi + x. Which must right as it follows the principal argument convention (this works only and only if the x is a principal argument as well) .


Q2-3d)

we already know that tan(\theta) = opposite/ adjacent

Did you stop mid sentence? If you draw a triangle of tan^-1(-2/3) doesnt that leave you with theta being in either quadrant 2 or 4? but the ans only gives 1 soln