3u Mathematics Marathon V 1.1 (1 Viewer)

[Riviet]

Nah, I haven't got any at the moment, I'll let Mountain Dew think of one. =D

 

[Mountain.Dew]

okay, okay...heres a trial paper one from my school

Question: prove by mathematical induction for n>= 0 that E(n) = 9ⁿ⁺² - 4ⁿ
is a multiple of 5

 

[Pianpupodoel]

okay, okay...heres a trial paper one from my school

Question: prove by mathematical induction for n>= 0 that E(n) = 9ⁿ⁺² - 4ⁿ
is a multiple of 5

Spoiler

Holds true for n=0 ( 81-1= 5*16)
Assume true for n=k, that is,
9^k+2 - 4^k = 5m, for some integer m

9^k+2 - 5m = 4^k (1)

Prove true for n=k+1:
9^k+3 - 4^k+1=
9^k+3 - 4*4^k=

Sub (1) in​

9^k+3 - 4*(9^k+2 - 5m)=
9*9^k+2 - 4*9^k+2 + 20m=
(9 - 4)*9^k+2 + 20m=
5*9^k+2 + 20m=
5*(9^k+2 + 4m)

which is divisible by 5.
Since it holds for n=0, and if it is true for n=k, it is also true for n=k+1, the statement above is true for all integers n=>0.

For fear of getting relegated to the extracurricular forum again, prove by mathematical induction that x⁴ - x² is always divisible by 12 (for x>1).

 

[Slidey]

I think I got it. If I did, it's well within the scope of the 3u syllabus.

 

[Sober]

I had to use induction twice, perhaps there is a simpler way?

Spoiler

- Let x=2:
2^4 - 2^2 = 12 = 12m (where m=1 ∴ true for x=2)

- Assume true for x=k:
k^4 - k^2 = 12m (where m is an integer)

- Prove true for x=k+1:
LHS = (k+1)^4 - (k+1)^2
= k^4 + 4k^3 + 5k^2 + 2k
= (k^4 - k^2) + 4k^3 + 6k^2 + 2k
= 12m + (4k^3 + 6k^2 + 2k)

Now all we have to do is show that (4k^3 + 6k^2 + 2k) is divisible by 12:

- Let k = 2:
4*2^3 + 6*2^2 + 2*2 = 32 + 24 + 4 = 60 (which is divisble by 12, ∴ true for k=2)

- Assume true for k=j:
4j^3 + 6j^2 + 2j = 12m (where m is an integer)

- Prove true for k=j+1:
LHS = 4(j+1)^3 + 6(j+1)^2 + 2(j+1)
= 4j^3 + 18j^2 + 26j + 12
= 12m + 12j^2 + 24j + 12 (from the induction hypothesis)
= 12m + 12(j^2 + 2j + 1)

∴ by the theory of mathematical induction... yadda yadda.. *twice*

 

[Mountain.Dew]

i'll save everyone from posting up the next question:

Question:
You are given that 4sin²[(pi)/10] + 2sin[(pi)/10] = 1

(i) deduce that sin[(pi)/10] = (sqrt 5 - 1)/4
(ii) use result in (i) to prove that: sin[pi/5]*cos[pi/10] = sqrt 5 / 4

 

[Riviet]

i'll save everyone from posting up the next question:
Question:
You are given that 4sin²[(pi)/10] + 2sin[(pi)/10] = 1
(i) deduce that sin[(pi)/10] = (sqrt 5 - 1)/4
(ii) use result in (i) to prove that: sin[pi/5]*cos[pi/10] = sqrt 5 / 4

Spoiler

(i) Let u=sin(pi/10)
then 4u²+2u-1=0
by the quadratic formula,
u=[-2+sqrt(20)]/8
sin(pi/10)=sqrt4 - 1 QED

(ii) LHS=sin(pi/5)cos(pi/10)
=2sin(pi/10)cos²(pi/10
=2sin(pi/10)[1-sin²(pi/10)]
=1/2.(sqrt5-1).(1-[6-2sqrt5]/16) [from part (i)]
=[16sqrt5-16-(8sqrt5-16)]/64
=8sqrt5 / 32
=sqrt5 / 4 QED

Next Question:

Prove tan 2θ = (sin 4θ - cos 4θ + 1) / (sin 4θ + cos 4θ + 1).

 

[Mountain.Dew]

Next Question:

Prove tan 2θ = (sin 4θ - cos 4θ + 1) / (sin 4θ + cos 4θ + 1).

more of my 2 cents.

Spoiler

RHS = (sin 4θ - cos 4θ + 1) / (sin 4θ + cos 4θ + 1)
= (2sin2θcos2θ - [1 - 2sin^2(2θ)] + 1)(2sin2θcos2θ + [2cos^2(2θ) - 1] + 1)
=(2sin2θcos2θ + 2sin^2(2θ))(2sin2θcos2θ + 2cos^2(2θ))
= (sin2θcos2θ + sin^2(2θ))(sin2θcos2θ + cos^2(2θ))
=sin2θ[cos2θ + sin2θ]/cos2θ[sin2θ + cos2θ] = sin2θ / cos2θ = tan2θ = LHS

QED...quite easily done

next question:

from Sydney Grammar 3U Trial Paper for 2003
Question 6(a)

In the diagram above (not give here ) the chord AB subtends an angle of θ radians at the centre of the circle with radius r, length OA = OB = r

(I) show that the ratio of the areas of the two segments is:

area of major segment / area of minor segment = [2pi - θ + sinθ] / [θ - sinθ]

(ii) now suppose that

area of major segment/ area of minor segment = [pi-1] / 1

(a' ) prove that θ - 2 - sinθ = 0
(b' ) show that the equation θ - 2 - sinθ = 0 has a root between θ = 2 and θ = 3
(c') taking θ = 2.5 as 1st approximation, use newton's method to find 2nd approximation to the root. answers given in 2 dp
(d' ) determine whether the 2nd approximation of θ yields a smaller value of |θ - 2 - sinθ| than the 1st approxmation

my addition: (e') do part (c' ) using the method of halving intervals.

 

[SoulSearcher]

next question:

from Sydney Grammar 3U Trial Paper for 2003
Question 6(a)

In the diagram above (not give here ) the chord AB subtends an angle of θ radians at the centre of the circle with radius r, length OA = OB = r

(I) show that the ratio of the areas of the two segments is:

area of major segment / area of minor segment = [2pi - θ + sinθ] / [θ - sinθ]

(ii) now suppose that

area of major segment/ area of minor segment = [pi-1] / 1

(a' ) prove that θ - 2 - sinθ = 0
(b' ) show that the equation θ - 2 - sinθ = 0 has a root between θ = 2 and θ = 3
(c') taking θ = 2.5 as 1st approximation, use newton's method to find 2nd approximation to the root. answers given in 2 dp
(d' ) determine whether the 2nd approximation of θ yields a smaller value of |θ - 2 - sinθ| than the 1st approxmation

my addition: (e') do part (c' ) using the method of halving intervals.

Spoiler

(i) the area of the minor segment is equal to 1/2 r² ( θ - sin θ )
the area of the major segment is equal to the area of the major segment plus the area of the triangle OAB, which is equal to
[ 1/2 r² ( 2pi - θ ) ] + [ 1/2 r² sin θ ] = 1/2 r² [ 2pi - θ + sin θ ]
therefore area of major segment / area of minor segment = 1/2 r² [ 2pi - θ + sin θ ] / 1/2 r² ( θ - sin θ )
= ( 2pi - θ + sin θ ) / ( θ - sin θ )

(ii) (a') since area of major segment/ area of minor segment = [pi-1] / 1, then ( 2pi - θ + sin θ ) / ( θ - sin θ ) = [pi-1] / 1
2pi - θ + sin θ = ( θ - sin θ )( pi - 1 )
2pi - θ + sin θ = piθ - θ - pi * sin θ + sin θ
2pi = piθ - pi * sin θ
piθ - 2pi - pi * sin θ = 0
pi ( θ - 2 - sin θ ) = 0
but as pi cannot equal 0, then θ - 2 - sin θ = 0

Note: θ is measured in radians
(b') to show a root beteen two points, it must be shown that the points lie on opposite sides of the θ-axis. Let θ - 2 - sin θ = f (θ)
when θ = 2, f (2) = 2 - 2 - sin 2 = -0.91 (to 2 dp)
when θ = 3, f (3) = 3 - 2 - sin 3 = 0.86 (to 2 dp)
compare f(2) and f(3). Since f(2) and f(3) have opposite signs, f (θ) has a root between θ = 2 and θ = 3

(c') let the first value of approximation equal θ₁ = 2.5
f (θ) = θ - 2 - sin θ
f (2.5) = -0.098
f '(θ) = 1 - cos θ
f '(2.5) = 1.801
since θ₂ = θ₁ - f (θ₁) / f '(θ₁)
then θ₂ = 2.5 - ( -0.098 / 1.801 )
θ₂ = 2.5 - (-0.05)
θ₂ = 2.55 to 2 dp

(d') substitute the values of θ₁ and θ₂ into |θ - 2 - sinθ|
θ₁ = 2.5
|2.5 - 2 - sin 2.5| = | -0.0985 |
= 0.0985
θ₂ = 2.55
|2.55 - 2 - sin 2.55| = | 0.000872 |
= 0.000872
as θ₂ is the second approximation, the second approximation of θ does yield a smaller value than the first approximation.

(e') let θ₁ = 2 and θ₂ = 3
therefore the value of θ₃ = ( 2 + 3 ) / 2 = 2.5
substitute the value of θ₃ into θ - 2 - sinθ
2.5 - 2 - sin 2.5 = -0.0985
as the value of f ( θ₃ ) < 0, then the value of θ must be between θ₃ and θ₂
therefore the value of θ₄ = ( 2.5 + 3 ) / 2 = 2.75
therefore an approximate value for the root of θ - 2 - sinθ is 2.75

 

[Riviet]

SoulSearcher, don't forget to post up the next question after answering the last one.

 

[SoulSearcher]

Alright then.

Next question:
Calculate the value of n if the coefficients of the second, third and fourth terms in the expansion of (1+x)ⁿ are sucessive terms in an arithmetic progression.

 

[Mountain.Dew]

Alright then.

Next question:
Calculate the value of n if the coefficients of the second, third and fourth terms in the expansion of (1+x)ⁿ are sucessive terms in an arithmetic progression.

okay, here goes:

Spoiler

(1+x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ⁿC₃x³ + .... + ⁿC_nxⁿ

now, the 2nd, 3rd and 4th coefficients are: ⁿC₁, ⁿC₂ and ⁿC₃

therefore, if these are in arithmetric progression, THEN:
ⁿC₃ - ⁿC₂ = ⁿC₂ - ⁿC₁

so we get:

n! / (n-1)! 3! = [2 * n! / (n-2)! 2!] - [n!/(n-1)! 1!]

now, simplify --> get denominator to equal (n-1)! 3!

(n-1)(n-2) n! / (n-1)! 3! = [2*3*(n-1)n!/(n-1)! 3!] - [3!n! / (n-1)! 3!]

times all sides by (n-1)! 3!, divide by n!...

(n-2)(n-1) = 6(n-1) - 6

n² - 3n + 2 = 6n - 6 - 6

n² - 9n + 14 = 0

(n - 7)(n-2) = 0 --> n = 2, 7

n=2 doesnt work (since we evaluated for the 4th term --> (1+x)² doesnt have a 4th term

therefore, n = 7

 

[Mountain.Dew]

okay, heres a projectile motion question. --> hasnt been a lot of these questions on these maths forums.
Barker College - Maths Ext 1 2003 Trials

Q 7(b)

A particle is projected under gravity with speed u ms^-1 and at an angle of π/4 from a point O on horizontal ground. it strikes the ground at P, where OP = R

(i) taking the x and y axis through O, show that the equation of the trajectory is given by y = x - g x²/u²

(ii) hence, or otherwise, show that R = u²/g

(iii) a ball is fired from O with velocity 30ms^-1 at an angle π/4 to the horizontal. find the speed of the ball when it has travelled a horizontal distance of 15m from its starting point.

 

[Mountain.Dew]

time to revive an old thread.

bump. please have a look at the previous post for the question to get the ball rolling!

 

[Riviet]

I don't think many schools have done this topic, it's usually done a little later this year.

 

[icycloud]

Mountain.Dew's Q:
(someone might like the check my answers)

Spoiler

(i) x'' = 0
x' = ucos(pi/4) = u/Sqrt[2]
x = ut/Sqrt[2]

y'' = -g
y' = -gt + usin(pi/4) = -gt + u/Sqrt[2]
y = -1/2 gt² + ut/Sqrt[2]

We have t = xSqrt[2]/u
t² = 2x²/u²

Thus, y = -1/2 g(2x²/u²) + x
= -gx²/u² + x
= x - gx²/u² #

(ii) x = R when y = 0, where x is not 0

Thus, 0 = x(1-gx/u²)

x not equal to 0, thus gx/u² = 1

x = u²/g #

(iii) We have u = 30m/s
When x = 15, t = 15/30 Sqrt[2] = Sqrt[2]/2

At t = Sqrt[2]/2, x' = 30/Sqrt[2] and y' = -g(Sqrt[2]/2) + 30/Sqrt[2]
If we take g ~ 10m/s², we get y' = 20/Sqrt[2]

V² = (x')² + (y')²
= 900/2 + 400/2
= 650

V = 25.50 m/s (2dp) #

Next Q
Describe in terms of range, domain and equation the graph of Sec[ArcSec[x]].

 

[pLuvia]

*Bump*
Describe in terms of range, domain and equation the graph of Sec[ArcSec[x]].

Spoiler

Not sure about this question, only just did inverse trig

y=Sec[ArcSec[x]]=x
y=x
D: all real x
R: all real y

Next Question
There are 4 identical maths books, and 4 identical physic books on a shelf
(i) How many different ways can these be arranged?
(ii) If 5 are to be used how many different arrangements are possible?

 

[Sober]

*Bump*
Describe in terms of range, domain and equation the graph of Sec[ArcSec[x]].

Spoiler

Not sure about this question, only just did inverse trig

y=Sec[ArcSec[x]]=x
y=x
D: all real x
R: all real y

Next Question
There are 4 identical maths books, and 4 identical physic books on a shelf
(i) How many different ways can these be arranged?
(ii) If 5 are to be used how many different arrangements are possible?

Concerning the previous question: y = Sec[ArcSec[x]]:

-1 ≮ x ≮1
-1 ≮ y ≮1

Next question:

Spoiler

8 books are arranged 8! ways
Divide by 4! twice as the orders of the same kind is irrelivent
8!/4! = 8x7x6x5/2 = 840

And I do not quite understand what (ii) is asking for but I will assume 5 of each book instead of 4 in which case: 10! / (2x5!) = 15120

 

[Riviet]

Part (ii) means how many ways you can select 5 books from the 8 on the shelf.
Answer:

Spoiler

There are 4 cases:

4P and 1M~ 5! / 4! = 20 ways
3P and 2M~ 5! / (3! x 2!) = 10 ways
2P and 3M~ 5! / (2! x 3!) = 10 ways
1P and 4M~ 5! / 4! = 20 ways.

.'. total number of ways = 60.

 

[pLuvia]

(i) part is actually

Spoiler

8 book, can be arranged in 8! ways, since there are identical ones, then 8!/(4!x4!)=70