VCE Specialist 3/4 Question Thread!

[deStudent]

For these couple problems, I'm having trouble understanding the logic of the answers provided http://m.imgur.com/a/Z17e5

For Q6g), so the answer said x is in quadrant 4. I didn't know how to find out if it's in quadrant 2 or 4. Are we supposed to assume that the angle is less than 90 degrees so it can't be in quadrant 2?

(i): I got the same answer as the book but their working suggested that it might be an error. It said "arctan(0.7) E [-pi/2,0]. But then gave the answer as positive 7/(root149) when we're calculating sinx?

Q7b) what I wrote is what the answer had pretty much, but how does that even prove the statement given?

Thanks

[Syndicate]

For these couple problems, I'm having trouble understanding the logic of the answers provided http://m.imgur.com/a/Z17e5

For Q6g), so the answer said x is in quadrant 4. I didn't know how to find out if it's in quadrant 2 or 4. Are we supposed to assume that the angle is less than 90 degrees so it can't be in quadrant 2?

(i): I got the same answer as the book but their working suggested that it might be an error. It said "arctan(0.7) E [-pi/2,0]. But then gave the answer as positive 7/(root149) when we're calculating sinx?
Q7b) what I wrote is what the answer had pretty much, but how does that even prove the statement given?

Thanks

Q6g)


6i) I don't really know what you mean... But here's the working out



7b) I guess you can check your answer with my working out.
Let \( sin^{-1}(\frac{3}{5}) \) = A, and \( sin^{-1}(\frac{5}{13})\) = B
\( sin^{-1}(\frac{16}{65}) \) = A-B
\(\therefore \space \frac{16}{65} \) = sin(A-B)
RHS= sinAcosB - cosAsinB

(where sinA = 3/5, sinV = 5/13,  cosA = 4/5 and cosB = 12/13)

=> \( \frac{3}{5} \times \frac{12}{13} - \frac{4}{5} \times \frac{5}{13} \)
=> \( \frac{36}{65} - \frac{20}{65} \space = \space \frac{16}{65} \) = LHS

[Shadowxo]

For these couple problems, I'm having trouble understanding the logic of the answers provided http://m.imgur.com/a/Z17e5

For Q6g), so the answer said x is in quadrant 4. I didn't know how to find out if it's in quadrant 2 or 4. Are we supposed to assume that the angle is less than 90 degrees so it can't be in quadrant 2?

(i): I got the same answer as the book but their working suggested that it might be an error. It said "arctan(0.7) E [-pi/2,0]. But then gave the answer as positive 7/(root149) when we're calculating sinx?

Q7b) what I wrote is what the answer had pretty much, but how does that even prove the statement given?

Thanks

Hi, for the questions in 6, you have to remember the domains for inverse functions are restricted. For sin inverse, domain is restricted to [-pi/2,pi/2]. Cos inverse, domain is [0,pi]. Tan inverse, domain is [-pi/2,pi/2]. In 6g, tan inverse is negative, meaning the angle must be in the 4th quadrant (as it must be either 1st or 4th, and if it were in the first it would be positive). For 6i, same sort of thing, as tan inverse is positive meaning the angle must be in the 1st quadrant, resulting in a positive sin value. If the book said it was in the 4th, it was likely an error.

For 7b, I'd define sin^-1(3/5)=a, so sin(a)= 3/5, and define sin^-1(5/13)=b so sin(b) = 5/13, then find cos(a) etc
solve sin (a-b) = 16/65
sin^-1(16/65) = a-b
sin^-1(16/65) = sin^-1(3/5) - sin^-1(5/13)
Therefore you've proven it.

Syndicate beat me to it but hope this helps anyway

[deStudent]

Thanks Syndicate and Shadowxo

Sorry another question...inverse trig is kicking my ass lol.

http://m.imgur.com/J04a8nP I'm pretty lost on these questions.

I've managed to do some of them but I think I've just been lucky. Using Q8a as an example, this is my method:

To find the domain: you consider the range of the function within the brackets -> see which values are shared by the domain of the function outside of the brackets.

Solve for x/adjust range of values that can be taken. This will be the domain.

Range can be found by subbing the max/min values of the domain in to y.

Would this be correct?

[Shadowxo]

Thanks Syndicate and Shadowxo

Sorry another question...inverse trig is kicking my ass lol.

http://m.imgur.com/J04a8nP I'm pretty lost on these questions.

I've managed to do some of them but I think I've just been lucky. Using Q8a as an example, this is my method:

To find the domain: you consider the range of the function within the brackets -> see which values are shared by the domain of the function outside of the brackets.

Solve for x/adjust range of values that can be taken. This will be the domain.

Range can be found by subbing the max/min values of the domain in to y.

Would this be correct?

Yep that's how I'd do it (inverse trig annoyed me too )

[deStudent]

Yep that's how I'd do it (inverse trig annoyed me too )

Thanks, looking back at this I'm still confused.

For 8a) the domain is [0, pi]. Using my method, range of cosx = [-1,1], domain of arcsin(x) = [-1,1]. Wouldn't this mean nothing is changed. So cosx E [-1,1]?

What am I doing wrong?

Edit: nvm I think what I wrote is correct, unless I'm wrong in thinking I'm correct lol.

[Syndicate]

Thanks, looking back at this I'm still confused.

For 8a) the domain is [0, pi]. Using my method, range of cosx = [-1,1], domain of arcsin(x) = [-1,1]. Wouldn't this mean nothing is changed. So cosx E [-1,1]?

What am I doing wrong?

Edit: nvm I think what I wrote is correct, unless I'm wrong in thinking I'm correct lol.

Range of cos(x) is [-1,1], and is a subset of the domain of sin^-1(x), which means you will use all of the principal domain of cos(x).

dom( sin^-1(cosx)) = dom(cosx) = [0, pi]

ran( sin^-1(cosx)) = ran(sin^-1x)= [-pi/2, pi/2]

[nice!]

Could someone help me with chaper 3c question 8 in general ?? I'm writing out the domain and range for both the inner and outer functions but don't understand how they are getting the answers. (esp q8d!)

[Syndicate]

Could someone help me with chaper 3c question 8 in general ?? I'm writing out the domain and range for both the inner and outer functions but don't understand how they are getting the answers. (esp q8d!)

Hi, I know these trigonometric composite function questions are really time-consuming and annoying.
Here's a way I would do it myself:

First of all, write the domain and range of both functions (I'll do question 8d as an example).

dom sin(x) = [-pi/2, pi/2]
ran sin(x) = [-1,1]

dom -cos^-1(x) = [-1,1]
ran cos^-1(x) = [-pi,0]

ran cos^-1(x) must be a subset of dom sin(x)
[-pi, 0] must be a subset of [-pi/2, pi/2], but it's not... so we have to restrict ran cos^-1(x) to [-pi/2, 0]

if cos^-1(x) has a range of [-pi/2, 0], then it will have a domain of [0,1] (You can work these out by using basic algebra)
if sin(x) has a domain of [-pi/2,0], then it will have a range of [-1,0]
dom sin(cos^-1x) = dom cos^-1(x) = [0,1]
ran sin(cos^-1x) = ran sin(x) = [-1,0]

I am really sorry if this doesn't make sense, I am probably not qualified enough to explain this quite well

Edit: fixing some of my grammatical and spelling errors

[nice!]

Hi, I know this trigonometric composite function questions are really time-consuming and annoying.
Here a way I would do it myself:

First of all, write the domain and range of both functions (I'll do question 8d as an example).

dom sin(x) = [-pi/2, pi/2]
ran sin(x) = [-1,1]

dom -cos^-1(x) = [-1,1]
ran cos^-1(x) = [-pi,0]

ran cos^-1(x) must be a subset of dom sin(x)
[-pi, 0] must be a subset of [-pi/2, pi/2], but it's not... so we have to restrict ran cos^-1(x) to [-pi/2, 0]

if cos^-1(x) has a range of [-pi/2, 0], then it will have a domain of [0,1] (You can work these out by using basic algebra)
if sin(x) has a domain of [-pi/2,0], then it will have a range of [-1,0]
dom sin(cos^-1x) = dom cos^-1(x) = [0,1]
ran sin(cos^-1x) = ran sin(x) = [-1,0]

I am really sorry if this doesn't make sense, I am probably not qualified enough to explain this quite well

omg i get it now !! thank u so much , it was very well explained )

[deStudent]

http://m.imgur.com/J04a8nP

h) I'm abit lost here. My domain was: all R, except -pi/2 and pi/2, the answer says it's all R. So my range was (-1,1) which was also what the answer had. I'm not sure what I've done wrong?

Also, for these questions are we allowed to find the range by directly subbing in the max and min values of the domain? Since these are 1-1 functions, I assumed it would be possible. I think this didn't work for all of these problems though.

[Syndicate]

http://m.imgur.com/J04a8nP

h) I'm abit lost here. My domain was: all R, except -pi/2 and pi/2, the answer says it's all R. So my range was (-1,1) which was also what the answer had. I'm not sure what I've done wrong?

Also, for these questions are we allowed to find the range by directly subbing in the max and min values of the domain? Since these are 1-1 functions, I assumed it would be possible. I think this didn't work for all of these problems though.

Yes, it is possible for you to calculate the range/domain by substituting values (can you please tell me which question you are talking about?)

dom sin(x) = [-pi/2, pi/2]
ran sin(x) = [-1,1]

dom tan^-1 = R
ran tan^-1 = (-pi/2, pi/2)

ran tan^-1(x) must be subset of dom sin(x) for the function to exist, which it already does.

Restricted sin(x) (which has a domain of (-pi/2, pi/2) will have a range of (-1,1).
Whereas, tan^-1(x) is not restricted and will have a domain of R.

dom sin(tan^-1x) = dom tan^-1(x) = R
ran sin(tan^-1x) = ran sin(x) = (-1,1)

[deStudent]

Yes, it is possible for you to calculate the range/domain by substituting values (can you please tell me which question you are talking about?)

dom sin(x) = [-pi/2, pi/2]
ran sin(x) = [-1,1]

dom tan^-1 = R
ran tan^-1 = (-pi/2, pi/2)

ran tan^-1(x) must be subset of dom sin(x) for the function to exist, which it already does.

Restricted sin(x) (which has a domain of (-pi/2, pi/2) will have a range of (-1,1).
Whereas, tan^-1(x) is not restricted and will have a domain of R.

dom sin(tan^-1x) = dom tan^-1(x) = R
ran sin(tan^-1x) = ran sin(x) = (-1,1)

I need help on question (h).

[itsangelan]

Hi Guys, if anyone could please help me solve this, it would be greatly appreciated!

The polynomial P(z) = 2z^3 + az^2 + bz + 5, where a and b are real numbers, has 2-i as one of its zeroes.

Find a quadratic factor of P(z) and hence calculate the real constants a and b.
(I've already found the quadratic factor, just have trouble finding a and b)

Mod edit: Your thread has been merged with the main Q&A thread for specialist maths.

[Shadowxo]

z-2+i = 0, so  z-2-i = 0 also
so quadratic factor is (z-2)²+1 = z²-4z+4+1 = z²-4z+5
Third linear factor is (cz+d) (as all constants are real, ends in +5 there's no i component in P(z))
(z²-4z+5)(cz+d) = P(z)
cz*z²=2z³ so c = 2
d*5 = 5 so d = 1
So third factor is (2z+1)
so P(z)=
(z²-4z+5)(2z+1)
-8z²+z² = az²
a = -7
-4z + 10z = bz
b = 10 - 4 = 6