3u Mathematics Marathon V 1.0 (2 Viewers)

[Estel]

1/6 n(n+1)(n+2) + 1/2(n+1)(n+2) by assumption.
= 1/6(n+1)(n+2)(n+3), as reqd by factorising.

Q14
The roots of x^2-bx+c=0 are X and Y.
Find |X^4-Y^4|

 

[icycloud]

Q14
The roots of x^2-bx+c=0 are X and Y.
Find |X^4-Y^4|

Spoiler

x^2 - bx + c = 0

Let the roots be X and Y,

X + Y = b
XY = c

X^2 + Y^2 = (X+Y)^2 - 2XY
= b^2 - 2c

(X-Y)^2 = X^2 - 2XY + Y^2
= b^2 - 4c

X-Y = +/- Sqrt(b^2 - 4c)

Thus, |X^4 - Y^4| = |(X^2 - Y^2)(X^2 + Y^2)|
= |(X+Y)(X-Y)(b^2 - 2c)|
= |b(b^2-2c) * +/- (b^2-4c)^(1/2)|
= b(b^2-2c) * Sqrt(b^2-4c) #

Question 15
A fair, six-sided die is thrown seven times. What is the probability that a '6' occurs on exactly 2 of the 7 throws?

 

[haboozin]

Question 15
A fair, six-sided die is thrown seven times. What is the probability that a '6' occurs on exactly 2 of the 7 throws?

Spoiler

binomial

7c2(1/6)^2(5/6)^5

Q16
Fred has three uniform tetrahedra (triangular pyramids). Each of these tetrahedra has one face black, one face white, one face red and one face green. Whne tossed onto a table, three faces of each tetrahedron can bee seen. If the probability of any coloured face not being seen is equally likely, what is the probability that:
i, no black face can be seen?
ii, exactly 2 black faces can be seen
iii, at least 2 red faces can be seen
iv, 3 white faces and only 1 green face can be seen.

 

[Riviet]

oh shit, are these questions hard or is it just me?

Some are hard, some are alright.

 

[word.]

Spoiler

i. P(no black) = (3/4)³
ii. P(2 black) = ³C₂(3/4)(1/4)²
iii. P(at least 2 red) = 1 - P(no red) - P(1 red) = 1 - (3/4)³ - ³C₁(1/4)(3/4)²
iv. P(3 white and 1 green seen) = P(2 green on bottom and 1 red or back on bottom) = ³C₂(2/4)(1/4)²

Question 17
Consider the circle x² + y² - 6x + 8y = 0. Find the value(s) of k for which the line x + 2y = k is a tangent to the circle.

 

[damo676767]

Question 17
Consider the circle x² + y² - 6x + 8y = 0. Find the value(s) of k for which the line x + 2y = k is a tangent to the circle.

Spoiler

2x + 2y dy/dx - 6 + 8 dy-dx = 0

dy/dx = (6 - 2x)/(2y + 8)

for x + 2y = k to be a tgt dy/dx = -1/2

-1/2 = (6 - 2x)/(2y + 8)
4x - 12 = 2y + 8
2x - 6 = y + 8
2x - y = 14

y = 2x - 14

x² + (2x - 14)² - 6x + 8(2x - 14) = 0
x² + 4x² - 56x + 196 - 6x + 16x - 112 = 0
5x² -46x + 84 = 0

x = (46 +- root(46² - 4 * 84 * 5))/10
=(46 +- 2root109)/10

2 posible points

([23 + root109]/5, [46 + 2 root 109]/5 - 14)
([23 - root109]/5, [46 - 2 root 109]/5 - 14)

2 pos values for k

k = [23 + root109]/5 + [92 + 4 root 109]/5 - 28
= (115 + 5 root109)/5 - 28
= root 109 - 5

k = [23 - root109]/5 + [92 - 4 root 109]/5 - 28
= (115 - 5 root109)/5 - 28
= -5root109 - 5

Question 18
show that (x-1)(x-2) is a factor of
p(x) = xⁿ(2^m - 1) + x^m(1 - 2ⁿ) + (2ⁿ - 2^m)
where m and n are positive intergeres

 

[香港!]

Question 18
show that (x-1)(x-2) is a factor of
p(x) = xⁿ(2^m - 1) + x^m(1 - 2ⁿ) + (2ⁿ - 2^m)
where m and n are positive intergeres

Spoiler

p(x)=x^n(2^m-1)+x^m(1-2^n)+(2^n-2^m)

p(1)=2^m-1+1-2^n+2^n-2^m=0
p(2)=2^n(2^m-1)+2^m(1-2^n)+2^n-2^m
=2^n2^m-2^m2^n-2^n+2^n+2^m-2^m=0
.: 1 and 2 are both roots
.: (x-1)(x-2) is a factor of p(x)

Question 19
At an air show, a Harrier Jump Jet leaves the ground 200 metres from an observer and rises vertically at the rate of 25 m/sec. At what rate is the observer's angle of elevation of the aircraft changing when the jet is 500 metres above the ground?

 

[damo676767]

Question 19
At an air show, a Harrier Jump Jet leaves the ground 200 metres from an observer and rises vertically at the rate of 25 m/sec. At what rate is the observer's angle of elevation of the aircraft changing when the jet is 500 metres above the ground?

Spoiler

d@/dt

h = 500tan@
dh/d@ = 500sec²@

dh/dt = 25

d@/dt = d@/dh * dh/dt
= 25/500sec²@
= cos²@/20

at h = 500
@ = 45

d@/dt = cos²45/20
= 1/20 root2 m/s

Question 20

show that

xⁿ(1 + x)ⁿ(1 + 1/x)ⁿ = (1+x)²ⁿ

hence prove that

1 + ⁿC₁²+ ⁿC₂²+ ... + ⁿC_n² = + ²ⁿC_n

 

[haboozin]

omg how ironic... I JUST DID THIS QUESTION

Question 20

show that

xn(1 + x)n(1 + 1/x)n = (1+x)2n

hence prove that

1 + nC12+ nC22+ ... + nCn2 = + 2nCn

Spoiler

LHS
(x(1 + 1/x)^n(1 + x)^n
(x + x/x)^n(1 + x)^n
(x + 1)^n(1 + x)^n
(1 + x)^n + n
= (1 + x)^2n
= RHS


Coefficient of x^n of RHS
= (2n n)

Coeff of x^n LHS
(n 0)(n n) + (n 1)(n n – 1) + ….+ (n n)(n 0)

Symmetrical property (n n) = (n 0) , (n n -1) = (n 1) and so on

So (n 0)^2 + (n 1)^2 + (n 2)^2 + …+ (n n)^2
Now (n 0) = 1
So 1^2 + (n 1)^2 + (n 2)^2 + … + (n n)^2
= 1 + (n 1)^2 + (n 2)^2 + …+ (n n)^2
#

Q21
<b>FOR 3unit People ONLY</b>
no 4unit ppl solve it please.

 

[damo676767]

was that question in the 4 unit exam?

i rember it but i dont know from where

 

[word.]

Spoiler

i. Comparing areas:
da/2 = bc/2
a = sqrt(b² + c²)
d*sqrt(b² + c²) = bc
squaring both sides: d²(b² + c²) = b²c²

ii. let A = alpha, B = beta, C = gamma
AC = h/tanB
AB = h/tanA
AP = h/tanC

from (i):
AC = b
AB = c
BC = a
AP = d

so b = h/tanB
c = h/tanA
d = h/tanC

d²(b² + c²) = b²c²
so (h²/tan²C)[(h²/tan²B) + (h²/tan²A)] = h⁴/(tan²A*tan²B)

[(tan²A*tan²B)/(tan²C*h²)][(h²/tan²B) + (h²/tan²A)] = 1

tan²A/tan²C + tan²B/tan²C = 1

hence tan²A + tan²B = tan²C

Question 22
Given the polynomial P(x) = 2x³ - 9x² + kx + 6
i. Find the value of k if (x - 3) is a factor of P(x).
ii. Hence, or otherwise, determine all the roots of the equation P(x) = 0.

 

[damo676767]

Question 22
Given the polynomial P(x) = 2x³ - 9x² + kx + 6
i. Find the value of k if (x - 3) is a factor of P(x).
ii. Hence, or otherwise, determine all the roots of the equation P(x) = 0.

Spoiler

p(3) = 0
= 2 *27 - 9*9 + 3k +6
3k = 21
k = 7


(x - 3)/(2x³ - 9x² + 7x + 6)

= 2x² - 3x - 2

p (x) = (x - 3)(2x² - 3x - 2)
= (x - 3)(x - 2)(2x + 1)
x = 3,2,-1/2

q 23
proove by indution
1 + 2 + 4 + ... +2^{n - 1} = 2ⁿ-1

 

[Bellow]

Spoiler

Let n=1
LHS=1 RHS=2^1-1=1
true for n=1
Assume the statement is tru for n=k
i.e 1+2+4+.....+2^(k-1)=2^k-1
Prove the statement for n=k+1
i.e Prove that 1+2+4+.....+2^(k-1)+2^k=2^(k+1)-1

LHS=1+2+4+.....+2^(k-1)+2^k
=2^k-1 + 2^k
=2.2^k-1
=2^(k+1)-1
=RHS
true for n=k+1
therefore if the statement is true for.....................hence it is true for all n positive integers.

Question 24.
In a flock of 1000 chickens, the number P infected with a disease at time t years is given by
P=1000/(1+ce^-1000t) where c is a constant
i.Show that eventually, all the chickens will be infected.
ii.Suppose that when time t=0, exactly one chicken was infected. After how many days will 500 chickens be infected?
iii.Show that dP/dt=P(1000-P)

 

[haboozin]

Question 24.

Spoiler

i.
lim as t --> infinity
e^-1000t --> 0
so P --> 1000/1 = 1000
so all will be infected

ii.
P = 1
t = 0
1 = 1000/( 1 + c)
c = 999

P = 500
t = ?

1 + 999e^-1000t = 2

e^-1000t = 1/999
-1000t = ln(1/999)
t = - ln(1/999)/1000
t = 0.0069... x 365 (yrs to days)
= 2.4...
= 2.5 days

iii,
p = 1000(1 + ce^-1000t)^-1
P = (1000(1 + ce^-1000t)^-1)^2. e^-1000t

but ce^-100t = 1000/P - 1 (just rearranged)
=P^2 (1000/p - 1)
= p (1000 - P)

<b>Q25</b>

The probability that n car accidents occur at a given set of traffic lights during a year is

P_n = {e^-2.6x2.6ⁿ}/n!
By considering values of n for which P_{n+ 1}/P_n >= 1, determine the most likely number of accidents at this intersection in a given one year period

 

[jake2.0]

for greatest term isnt it P_n/P_n+1 >= 1 ??

 

[KFunk]

Spoiler

<b>Q25</b>
Wasn't this one on a 4U paper once?

P_{n+ 1} &ge; P_n

{e^-2.6x2.6ⁿ⁺¹}/(n+1)! &ge; {e^-2.6x2.6ⁿ}/n!

2.6/(n+1) &ge; 1

2.6 &ge; n+1

n &le; 1.6 ===> n=1 , &there4; P₂/P₁ >1, P₂ > P₁

If n &ge; 2 then P_n+1/P_n <1 so P₂ > P₃ > ...

Hence the most likely number of accidents is 2.

Q.26 Given n points in a plane, prove by Induction that the maximum number of lines joining any two points is (1/2)n(n-1), for n &ge; 2

 

[damo676767]

Q.26 Given n points in a plane, prove by Induction that the maximum number of lines joining any two points is (1/2)n(n-1), for n &ge; 2

this would me much eaiser proved without induction, but any way

Spoiler

rtp true for n=2
2 points there isonly 1 line
form = (1/2)*2*(2-1)
=1

true for n = 2


assume yadda yadda yadda


amount of lines = (1/2)k(k-1) + k
= k ((1/2)(k-1)+1)
= k/2 (k - 1 + 2)
= (1/2)(k+1)k
= form


yadda yadda yadda

Q 27

a projectile is launced of a 50m cliff and hits the ground 200 from the foot of the cliff. it was fired with a velocity 40m/s and angle @

find the 2 posible values of @

 

[100percent]

Q 27

a projectile is launced of a 50m cliff and hits the ground 200 from the foot of the cliff. it was fired with a velocity 40m/s and angle @

find the 2 posible values of @

Spoiler

x=vtcos@
t=5/cos@
y=vtsin@-0.5gt²+50
0=200tan@-25g/2-[25/2]gtan²@+50
gtan²@-16tan@+(g-4)=0
tan@={8±sq rt [64+4g-g]}/g
@=arctan ({8±sq rt [64+4g-g]}/g)

Q28
show that a particle with displacement x=atan(nt) is not moving in SHM

 

[Stefano]

Q28
show that a particle with displacement x=atan(nt) is not moving in SHM

Spoiler

x=atan(nt)
x'=ansec²(nt)
x''=[2an²sin(nt)]/[cos²(nt)]
=/ -n²x

<b>Q29</b> Show that:
C₀ + 2C₁ + 3C₂ + ... + (n+1)C_n = (n+2)2^n-1

 

[damo676767]

Q28
show that a particle with displacement x=atan(nt) is not moving in SHM

Spoiler

when nt = pi/2 , x becomes undefind, therefore it is not moving in simole harmonic motion

Question 29

i am x meteres away from a diving board, the angle (as i see it) between then 1 m and the 4m board is @, they are vertically alinged

show that @ = tan^-1(4/x_ - tan^-1(1/x)

show @ is a min when x=2

deduce that the min @ = tan^-1 (3/4)