MedVision ad

HSC 2015 MX1 Marathon (archive) (2 Viewers)

Status
Not open for further replies.

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,385
Gender
Male
HSC
2006
Post any questions within the scope and level of Mathematics Extension 1. Once a question is posted, it needs to be answered before the next question is raised.

I encourage all current students in particular to participate in this marathon.

To start off:

 
Last edited:

jyu

Member
Joined
Nov 14, 2005
Messages
623
Gender
Male
HSC
2006
Re: HSC 2015 3U Marathon

h depends on d?
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

i) Using tan(a-b) = tana - tanb / 1+tanatanb,
show that 1+tannθtan(n+1)θ = cotθ(tan(n+1)θ-tannθ)

ii) Prove by mathematical induction that
tanθtan2θ + tan2θtan3θ + ... + tannθtan(n+1)θ = -(n+1) + cotθtan(n+1)θ for all integers n>=1
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

i) Using tan(a-b) = tana - tanb / 1+tanatanb,
show that 1+tannθtan(n+1)θ = cotθ(tan(n+1)θ-tannθ)

ii) Prove by mathematical induction that
tanθtan2θ + tan2θtan3θ + ... + tannθtan(n+1)θ = -(n+1) + cotθtan(n+1)θ for all integers n>=1
Files too large and I don't feel like typing my working. Refer to links:
i) http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI07012015_zpsdc70650f.jpg
ii) http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI07012015_0001_zps84b4191b.jpg
http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI07012015_0002_zpsa8ac050b.jpg

Next question:
Using the substitution u=ln(x), or otherwise, find Capture.PNG

Hint for those who haven't done logs (highlight to show):
d/dx ln(x) = 1/x
 
Last edited:

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

Files too large and I don't feel like typing my working. Refer to links:
i) http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI07012015_zpsdc70650f.jpg
ii) http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI07012015_0001_zps84b4191b.jpg
http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI07012015_0002_zpsa8ac050b.jpg

Next question:
Using the substitution u=ln(x), or otherwise, find View attachment 31575

Hint for those who haven't done logs (highlight to show):
d/dx ln(x) = 1/x
Nice first post!
http://imgur.com/bNK5TeN (I think this is how you do it)

Next question:
Differentiate tanx using first priciples
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

Nice first post!
http://imgur.com/bNK5TeN (I think this is how you do it)

Next question:
Differentiate tanx using first priciples
Thanks! Yep that's how to do it
http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI08012015_zpsa3c8b8da.jpg

Next question:
(i) Explain why d/dx c = 0 (for some constant c)
(ii) Prove by mathematical induction and derivative laws that the nth derivative of x^n is n!
(iii) Show that for any polynomial P(x)=a+bx+cx^2+...+zx^n, the (n+1)th derivative is always equal to 0
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

Thanks! Yep that's how to do it
http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI08012015_zpsa3c8b8da.jpg

Next question:
(i) Explain why d/dx c = 0 (for some constant c)
(ii) Prove by mathematical induction and derivative laws that the nth derivative of x^n is n!
(iii) Show that for any polynomial P(x)=a+bx+cx^2+...+zx^n, the (n+1)th derivative is always equal to 0
I like that induction question. Couldn't get it today, i'll try it tomorrow.
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

Thanks! Yep that's how to do it
http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI08012015_zpsa3c8b8da.jpg

Next question:
(i) Explain why d/dx c = 0 (for some constant c)
(ii) Prove by mathematical induction and derivative laws that the nth derivative of x^n is n!
(iii) Show that for any polynomial P(x)=a+bx+cx^2+...+zx^n, the (n+1)th derivative is always equal to 0
i) i) If c is a constant then y=c represents a horizontal line. Thus,the gradient of the tangent for any point P on that line will be 0;i.e. d/dx c= 0
ii) http://imgur.com/G0HdtqQ (may or may not be right)
iii) (working on it)
 
Last edited:

colinrocks95

New Member
Joined
Mar 23, 2012
Messages
5
Gender
Male
HSC
2012
Re: HSC 2015 3U Marathon

Thanks! Yep that's how to do it
http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI08012015_zpsa3c8b8da.jpg

Next question:
(i) Explain why d/dx c = 0 (for some constant c)
(ii) Prove by mathematical induction and derivative laws that the nth derivative of x^n is n!
(iii) Show that for any polynomial P(x)=a+bx+cx^2+...+zx^n, the (n+1)th derivative is always equal to 0
For (iii), we can work out each derivative as follows:
P'(x) = b + 2cx + 3dx^2 + ... + znx^(n-1)
P"(x) = 2c + 6dx + ... + zn(n-1)x^(n-2)
...
P^n(x) = zn(n-1)...(n-[n-1])x^(n-n) = zn!, where n! = n(n-1)(n-2)...1

Since P^n(x) is a constant, P^(n+1)(x), i.e. the (n+1)th derivative, is zero.

--

Next question:
By expanding sin(A+B), show that sin(3x) = 3sin(x) - 4sin^3(x). Hence or otherwise find the smallest positive value of u that satisfies the equation 12u^3 - 9u + 2 = 0. Give your answer to two decimal places.

By the way, you should check out Sci School's HSC programs. I know the presenter and he's a genius + really good at explanations.
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

For (iii), we can work out each derivative as follows:
P'(x) = b + 2cx + 3dx^2 + ... + znx^(n-1)
P"(x) = 2c + 6dx + ... + zn(n-1)x^(n-2)
...
P^n(x) = zn(n-1)...(n-[n-1])x^(n-n) = zn!, where n! = n(n-1)(n-2)...1

Since P^n(x) is a constant, P^(n+1)(x), i.e. the (n+1)th derivative, is zero.

--

Next question:
By expanding sin(A+B), show that sin(3x) = 3sin(x) - 4sin^3(x). Hence or otherwise find the smallest positive value of u that satisfies the equation 12u^3 - 9u + 2 = 0. Give your answer to two decimal places.

By the way, you should check out Sci School's HSC programs. I know the presenter and he's a genius + really good at explanations.
http://imgur.com/XDVhTNU

But still, how can you tell that it's the smallest possible number? o.o
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

Next question:
(i) Show that the equation of the tangent at P(2ap,ap^2) on the parabola x^2=4ay is y=px-ap^2
(ii) Show that the point of intersection of the tangents at P and Q(2aq,aq^2) on said parabola is (a(p+q),apq)

Boring question, but I'm a bit out of ideas.
 
Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
Re: HSC 2015 3U Marathon

Next question:
(i) Show that the equation of the tangent at P(2ap,ap^2) on the parabola x^2=4ay is y=px-ap^2
(ii) Show that the point of intersection of the tangents at P and Q(2aq,aq^2) on said parabola is (a(p+q),apq)

Boring question, but I'm a bit out of ideas.
Stock standard question hahaha
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

Next question:
(i) Show that the equation of the tangent at P(2ap,ap^2) on the parabola x^2=4ay is y=px-ap^2
(ii) Show that the point of intersection of the tangents at P and Q(2aq,aq^2) on said parabola is (a(p+q),apq)

Boring question, but I'm a bit out of ideas.
lol

i) y = x^2/4a
y' = x/2a
f'(2ap) = p
y -ap^2 = p(x-2ap)
y = px -ap^2

ii) Similarly, at Q, y = qx-aq^2 -------1
px-ap^2 = qx-aq^2
x(p-q) = a(p+q)(p-q)
therefore, x = a(p+q)---------2
2 -> 1
y = q(ap+aq)-aq^2
= apq

Therefore, point of intersection = (a(p+q),apq)
 
Last edited:

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

Next question!

Express (30 x 28 x 26 x ... x 2) / (15 x 13 x 11 x ... x 1)
in factorial notation

(i'm running out of ideas too)
 

Kaido

be.
Joined
Jul 7, 2014
Messages
798
Gender
Male
HSC
2015
Re: HSC 2015 3U Marathon

rofl, pre sure u dont learn that in yr 12 ^

i reckon its 30!!/15!! ?
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top