2022 CSSA MATH EXT 2 Thoughts (2 Viewers)

Flise · Aug 8, 2022

How’d y’all find the exam?

Flise · Aug 8, 2022

Couldn't solve 16aii) and Q16b)

Q16a)
$Prove \int_0^1 x^p (1-x)^q dx \ = \frac{q}{p+1}\ \int_0^1 x^{p+1}(1-x)^{q-1}dx$
$Hence, prove \int_0^1 x^p (1-x)^q dx \ = \frac{p!q!}{(p+q+1)!}$

Q16b) Consider the concavity of y=(x)^1/3
Prove (a-b)^1/3 + (a+b)^1/3 < 2(a)^1/3

Any ideas?

Lith_30 · Aug 8, 2022

for question 16) a) use IBP
$I=\int_0^1{x^p(1-x)^qdx$

let $u=(1-x)^q\implies{}du=-q(1-x)^{q-1}dx$

let $dv=x^p\ dx\implies{}v=\frac{x^{p+1}}{p+1}$

hence the integral can be written as $\left[\frac{q(1-x)^{q-1}x^{p+1}}{p+1}\right]^1_0+\frac{q}{p+1}\int_0^1{x^{p+1}(1-x)^{q-1}dx=\frac{q}{p+1}\int_0^1{x^{p+1}(1-x)^{q-1}dx$

then by continually using IBP to the integral

$\\\frac{q}{p-1}\int_0^1{x^{p+1}(1-x)^{q-1}dx\\\\=\frac{q(q-1)}{(p+1)(p+2)}\int_0^1{x^{p+2}(1-x)^{q-2}dx=...$

you get that $I=\frac{q(q-1)(q-2)...(q-(q-1))}{(p+1)(p+2)...(p+q)}\int_0^1{x^{p+q}(1-x)^{q-q}dx$

multiply top and bottom by p!

$\\I=\frac{q!p!}{1\times2\times...(p+q)}\left[\frac{x^{p+q+1}}{p+q+1}\right]^1_0\\\\I=\frac{p!q!}{(p+q+1)!}$ as required

notme123 · Aug 8, 2022

Flise said:
Couldn't solve 16aii) and Q16b)

Q16a)
$Prove \int_0^1 x^p (1-x)^q dx \ = \frac{q}{p+1}\ \int_0^1 x^{p+1}(1-x)^{q-1}dx$
$Hence, prove \int_0^1 x^p (1-x)^q dx \ = \frac{p!q!}{(p+q+1)!}$

Q16b) Consider the concavity of y=(x)^1/3
Prove (a-b)^1/3 + (a+b)^1/3 < 2(a)^1/3

Any ideas?

16 b: Prove
$(a-b)^\frac{1}{3} + (a+b)^\frac{1}{3} < 2a^{\frac{1}{3}}$
Looking at the graph $y=x^{\frac{1}{3}},$ we can deduce
$y'=\frac{1}{3} x^{-\frac{2}{3}}$ and
$y''=-\frac{2}{9} x^{-\frac{5}{3}}<0$
for all x in the positive reals, meaning it is concave down always for $x>0$ .
Using a concave down function and letting $0<b\leq a$ for all positive real a, you want to show graphically that the y-coordinate of the midpoint between the points $(a-b,(a-b)^{\frac{1}{3}}), (a+b,(a+b)^{\frac{1}{3}})$ is less than the y-coordinate of $(a,a^{\frac{1}{3}})$ , or $\frac{(a-b)^{\frac{1}{3}}+(a+b)^{\frac{1}{3}}}{2} < a^{\frac{1}{3}}$ , which you multiply by 2 on both sides to arrive at the statement to be proved.

Also, 16a is basically identical to one of the questions in my tutorial sheet for uni and is categorized as a hard one as well. CSSA did the same thing last year with polynomials and proving factor theorem.

Flise · Aug 8, 2022

16ci) There was a part one which was to find v in terms of t given resistance is kv^2, mass of skydiver is 'm', and take up as positive (weirdly)
Basically integrating a = 1/m (kv^2 - mg), note: assume at rest when t=0

They give you the required result:
$v=\sqrt{\frac{mg}{k}}(\frac{e^{-t\sqrt{\frac{kg}{m}}}-e^{t\sqrt{\frac{kg}{m}}}}{e^{-t\sqrt{\frac{kg}{m}}}+e^{t\sqrt{\frac{kg}{m}}}}\)$

Part two gives you values and wants you to find the time when the skydiver is 1500m from the ground
Dropped from an altitude of 5000m, mass = 100kg, k=0.25, g=10

I integrated the equation to get x in terms of t but was stuck thereafter.

d1zzyohs · Aug 8, 2022

Couldn't get out 16 a) and b); besides that everything was fine...
That's okay I guess. 16 c) for my school was replaced since we haven't finished mechanics.
Definitely was way harder than last years - last year's question 16 was an absolute snore.

Flise · Aug 8, 2022

Also what did yall get for the distance travelled for Q15b) the fricken vector-helix question combined with mechanics. My unwise brain saw it was 1 mark and I just did the velocity (4m/s) times 10 (since they wanted after 10s)! Surely it wasn't that simple...

d1zzyohs · Aug 8, 2022

Flise said:
Also what did yall get for the distance travelled for Q15b) the fricken vector-helix question combined with mechanics. My unwise brain saw it was 1 mark and I just did the velocity (4m/s) times 10 (since they wanted after 10s)! Surely it wasn't that simple...

I too got that. It was just length; not distance; and it was also a 1 marker, so yeah, it was that simple. Was probably trying to catch out some people trying to over-complicate it. That question was fun.

I really enjoyed the Question 14; reduction formulae question. Was a satisfying result.
How were you supposed to do the first Complex Numbers question of Question 15? I don't know if I did it right.

notme123 · Aug 8, 2022

Flise said:
16ci) There was a part one which was to find v in terms of t given resistance is kv^2, mass of skydiver is 'm', and take up as positive (weirdly)
Basically integrating a = 1/m (kv^2 - mg), note: assume at rest when t=0

They give you the required result:
$v=\sqrt{\frac{mg}{k}}(\frac{e^{-t\sqrt{\frac{kg}{m}}}-e^{t\sqrt{\frac{kg}{m}}}}{e^{-t\sqrt{\frac{kg}{m}}}+e^{t\sqrt{\frac{kg}{m}}}}\)$

Part two gives you values and wants you to find the time when the skydiver is 1500m from the ground
Dropped from an altitude of 5000m, mass = 10kg, k=0.25, g=10

I integrated the equation to get x in terms of t but was stuck thereafter.

subbing in the given values you get
$v=20 \left(\frac{e^{-0.5t}-e^{0.5t}}{e^{-0.5t}+e^{0.5t}} \right)$
Integrating from t=0 ->T
$x-5000=\int_0^{T} 20 \left(\frac{e^{-0.5t}-e^{0.5t}}{e^{-0.5t}+e^{0.5t}} \right) dt = -20 \int_0^{T} \frac{-e^{-0.5t}+e^{0.5t}}{e^{-0.5t}+e^{0.5t}} dt = -\frac{20}{0.5} \int_0^{T} \frac{-0.5 e^{-0.5t}+0.5 e^{0.5t}}{e^{-0.5t}+e^{0.5t}} dt = -40 \ln(e^{-0.5t}+e^{0.5t}) \Biggr|_0^{T} = 40 \ln(2) - 40 \ln(e^{-0.5T}+e^{0.5T})$
$x=5000+40\ln(2)- 40 \ln(e^{-0.5T}+e^{0.5T})$
I feel like somethings wrong

jklol · Aug 8, 2022

d1zzyohs said:
I too got that. It was just length; not distance; and it was also a 1 marker, so yeah, it was that simple. Was probably trying to catch out some people trying to over-complicate it. That question was fun.

I really enjoyed the Question 14; reduction formulae question. Was a satisfying result.
How were you supposed to do the first Complex Numbers question of Question 15? I don't know if I did it right.

It was a relationship between the roots of $z^9+1=0$ i.e you factorise it and get $(z+1)(z^8-z^7+z^6-z^5+z^4-z^3+z^2-z)$ and then you let them equal 0 $z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1= 0$ and luckily each of the roots have corresponding conjugates and hence adding each of the roots the imaginary parts go away and you just get left with the $cos's$

Flise · Aug 8, 2022

d1zzyohs said:
I too got that. It was just length; not distance; and it was also a 1 marker, so yeah, it was that simple. Was probably trying to catch out some people trying to over-complicate it. That question was fun.

I really enjoyed the Question 14; reduction formulae question. Was a satisfying result.
How were you supposed to do the first Complex Numbers question of Question 15? I don't know if I did it right.

I'm so pissed I screwed up my conversion of -1 into exponential form.
I wrote z^9 = cis(kpi) instead of z^9 = cis(pi + 2kpi) I'm so dumbbb. All you then do is sum of roots, one of the roots comes out as ±1/2 and just make note of -cos5pi/9 = cos4pi/9 and -cos7pi/9 = cos2pi

Flise · Aug 8, 2022

jklol said:
It was a relationship between the roots of $z^9-1=0$ i.e you factorise it and get $(z-1)(z^8+z^7+z^6+z^5+z^4+z^3+z^2+z)$ and then you let them equal 0 $z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1= 0$ and luckily each of the roots have corresponding conjugates and hence adding each of the roots the imaginary parts go away and you just get left with the $cos's$

Btw the question was z^9+1 = 0

jklol · Aug 8, 2022

lmao my bad

Flise · Aug 8, 2022

notme123 said:
subbing in the given values you get
$v=20 \left(\frac{e^{-0.5t}-e^{0.5t}}{e^{-0.5t}+e^{0.5t}} \right)$
Integrating from t=0 ->T
$x-5000=\int_0^{T} 20 \left(\frac{e^{-0.5t}-e^{0.5t}}{e^{-0.5t}+e^{0.5t}} \right) dt = -20 \int_0^{T} \frac{-e^{-0.5t}+e^{0.5t}}{e^{-0.5t}+e^{0.5t}} dt = -\frac{20}{0.5} \int_0^{T} \frac{-0.5 e^{-0.5t}+0.5 e^{0.5t}}{e^{-0.5t}+e^{0.5t}} dt = -40 \ln(e^{-0.5t}+e^{0.5t}) \Biggr|_0^{T} = 40 \ln(2) - 40 \ln(e^{-0.5T}+e^{0.5T})$
$x=5000+40\ln(2)- 40 \ln(e^{-0.5T}+e^{0.5T})$
I feel like somethings wrong

My bad mass was 100kg ain't a baby skydiving

Flise · Aug 8, 2022

d1zzyohs said:
I too got that. It was just length; not distance; and it was also a 1 marker, so yeah, it was that simple. Was probably trying to catch out some people trying to over-complicate it. That question was fun.

I really enjoyed the Question 14; reduction formulae question. Was a satisfying result.
How were you supposed to do the first Complex Numbers question of Question 15? I don't know if I did it right.

I'm so pissed as well for this question I literally did a variation of it three nights ago sinx + sin2x + ... + sinnx but literally mindblanked for part ii)
I got part i) luckily it was just a geometric progression, but I couldn't simplify the result...

I divided each term by e^itheta/2 when I was suppose to do that only for the bottom, while the top required me factoring e^i(ntheta) which I couldn't see in the exam!!

2022 CSSA MATH EXT 2 Thoughts (2 Viewers)

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