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July 24, 2025, 01:32:35 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2549630 times)  Share 

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Jaswinder

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Re: Specialist 3/4 Question Thread!
« Reply #1200 on: January 27, 2013, 06:11:34 pm »
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smartt! thanks Jenny_2108

duhherro

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Re: Specialist 3/4 Question Thread!
« Reply #1201 on: January 28, 2013, 03:37:57 pm »
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Could anyone help me find the asymtotes with the equation...

4(x-1)^2 /3   - (y-1)^2 /3 = 1

I've tried to use the y-k = plus/minus b(x-h)/a

But not sure how to get by with a 4/3 :/

Stick

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Re: Specialist 3/4 Question Thread!
« Reply #1202 on: January 28, 2013, 03:41:43 pm »
+1
Hint: Make the denominator of the first expression 3/4. :)
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duhherro

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Re: Specialist 3/4 Question Thread!
« Reply #1203 on: January 28, 2013, 03:45:16 pm »
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Hint: Make the denominator of the first expression 3/4. :)

Thanks for that! What a nice trick to simply make the denominators in your favour :P

b^3

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Re: Specialist 3/4 Question Thread!
« Reply #1204 on: January 28, 2013, 03:51:17 pm »
+1
Hint: The following two expressions are the same


Spoiler

So our asymptotes will be


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Mr. Study

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Re: Specialist 3/4 Question Thread!
« Reply #1205 on: January 28, 2013, 03:51:55 pm »
+2
Just a trick I found, last year, when finding asymptotes for hyperbolas.

Spoiler








or

or

I find this 'method' easier than remembering y-k = plus/minus b(x-h)/a, as I don't jumble h,a,k,b or w/e.

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shaiga95

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Re: Specialist 3/4 Question Thread!
« Reply #1206 on: January 28, 2013, 11:30:25 pm »
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Reduce to following to partial fractions I keep getting it wrong
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brightsky

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Re: Specialist 3/4 Question Thread!
« Reply #1207 on: January 28, 2013, 11:38:47 pm »
+1
long divide first. (in general, if the degree of the numerator is not less than that of the denominator, then long divide before performing partial decomp.)
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Re: Specialist 3/4 Question Thread!
« Reply #1208 on: January 29, 2013, 12:11:22 am »
+3
The working if you really are still stuck, but try to do it on your own first :)
Spoiler

« Last Edit: January 29, 2013, 06:02:11 pm by b^3 »
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shaiga95

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Re: Specialist 3/4 Question Thread!
« Reply #1209 on: January 29, 2013, 12:15:24 am »
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thanks i cross multiplied A and B silly me
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Re: Specialist 3/4 Question Thread!
« Reply #1210 on: January 29, 2013, 02:11:45 pm »
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Hi guys,
I came accross this question, I thought I had the answer but I didn't:
For pi/2<A<pi and 0<B<pi/2  sinA=t and cosB=t, cos(B+A)=?????

i got 0 but the answer says -2t(1-t^2)^(1/2)
How did they get this?
Many thanks.

Limista

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Re: Specialist 3/4 Question Thread!
« Reply #1211 on: January 29, 2013, 02:29:23 pm »
+1
Hi guys,
I came accross this question, I thought I had the answer but I didn't:
For pi/2<A<pi and 0<B<pi/2  sinA=t and cosB=t, cos(B+A)=?????

i got 0 but the answer says -2t(1-t^2)^(1/2)
How did they get this?
Many thanks.
pi/2<A<pi is in the second quadrant.
0<B<pi/2 is in the first quadrant.

cos (B + A) = cosBcosA - sinBsinA
We need to work out what cosA and sinB is equal to.

cosA = -(1-t2)^(1/2)......why?
sinA = t, so opposite side of triangle = t, hypotenuse of triangle = 1, and adjacent side of triangle = (1/t2)^(1/2). Because A is in the 2nd quadrant and cosine is negative in the 2nd quadrant, cosA = -(1-t2)^(1/2)......

sinB = (1-t2)^(1/2)....why?
follow the same logic as above, except this time sinB will remain positive, because B is in the 1st quadrant and sine is positive in the 1st quadrant.

Substituting these values into the cos (B + A) formula....

t x (-(1-t2)^(1/2)) - (1-t2)^(1/2) x t
= -2t(1-t2)^(1/2)
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bucklr

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Re: Specialist 3/4 Question Thread!
« Reply #1212 on: January 29, 2013, 02:36:55 pm »
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Thanks... LOL i thought they were the same triangle. I must stop assuming everything.

Thanks again.

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Re: Specialist 3/4 Question Thread!
« Reply #1213 on: January 30, 2013, 01:13:05 pm »
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I have some problems with this question.
Could someone help me :D
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Re: Specialist 3/4 Question Thread!
« Reply #1214 on: January 30, 2013, 01:39:30 pm »
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OK, so ABD forms an isosceles triangle if you draw a chord between B and D. After that, you'll see angle ADB is also x, and then by utilising the tangent/chord angle in the opposite segment rule, you'll also see angle BCD is x.

For part b, you'll need to use the cross over of two secants rule so that you get y^2=a(a+b).
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