2u Mathematics Marathon v1.0 (1 Viewer)

[icycloud]

Question 45:
Differentiate y=log_2 x

Spoiler

y = log_2 x
= (1/ln(2)) * ln(x)

Let u = ln(x)
du = dx / x

y = (1/ln(2)) u
dy = (1/ln(2)) du
= (1/ln(2)) / x dx

Thus, dy/dx = 1 / (x ln(2))

Question 46:
By expressing sec(x) and tan(x) in terms of sin(x) and cos(x), show that sec²(x) - tan²(x) = 1.

 

[chinkyeye]

jesus........... u guys are all so smart..... i'm ranked number 1 in my school for maths but i don't knoe half da shit is going on hahahahaha

 

[MarsBarz]

Answer to question 46:
By expressing sec(x) and tan(x) in terms of sin(x) and cos(x), show that sec²(x) - tan²(x) = 1.

Spoiler

Prove that sec²(x) - tan²(x) = 1
LHS = sec²(x) - tan²(x)
LHS = (1/cos²x) - (sin²x/cos²x)
LHS = (1-sin²x)/cos²x
LHS = cos²x/cos²x
LHS = 1
.:. LHS = RHS

Question 47:
Given log₇2=0.36 and log₇5=0.83 find log₇14
Note: Nasty trap, this is a part of a longer question, you may therefore disregard some of the given information.

 

[icycloud]

Question 47:
Given log₇2=0.36 and log₇5=0.83 find log₇14

Spoiler

log₇14 = log₇(7 * 2)
= log₇7 + log₇2
= 1 + 0.36
= 1.36 #

Hehe, did I fall for the trap??? :S

Question 48:
Consider the geometric series: 1 - tan²x + tan⁴x - ....
For what values of x in the interval -pi/2 < x < pi/2 does the limiting sum of the series exist?

 

[word.]

Spoiler

A limiting sum exists when |r| < 1
tan²x < 1
-1 < tanx < 1
-pi/4 < x < pi/4

Question 49
An urn contains W white and B black marbles. If the probability of selecting 2 white marbles at random is 1/3 while the probability of selecting 3 white marbles at random is 1/6, find the number of white marbles in the urn.

(Source: JR 1997 2U Trial)

 

[MarsBarz]

Hehe, did I fall for the trap??? :S

No you got it . I know, it was an easy question, just that it's so simple some people get stuck ahahah.

Question 49
An urn contains W white and B black marbles. If the probability of selecting 2 white marbles at random is 1/3 while the probability of selecting 3 white marbles at random is 1/6, find the number of white marbles in the urn.

(Source: JR 1997 2U Trial)

Spoiler

At least 3 white marbles.

 

[icycloud]

Question 49
An urn contains W white and B black marbles. If the probability of selecting 2 white marbles at random is 1/3 while the probability of selecting 3 white marbles at random is 1/6, find the number of white marbles in the urn.

(Source: JR 1997 2U Trial)

Spoiler

W is the number of white marbles
B is the number of black marbles

P(W and W) = (W/(W+B)) * (W-1)/(W+B-1) = 1/3 ----- (1)
P(W and W and W) = P(W and W) * (W-2)/(W+B-2) = 1/6

Thus, (W-2)/(W+B-2) = 1/6 * 3/1 = 1/2
2W-4 = W+B-2
B = W-2 ---- (2)

But, in (1),

W(W-1) / (W+B)(W+B-1) = 1/3
3W(W-1) = (W+B)(W+B-1)
3W^2 - 3W = W^2 + WB - W + WB + B^2 - B
2W^2 - 2W - 2BW + B - B^2 = 0 ----- (3)

Substituting (2) into (3), we have:

2W^2 - 2W - 2W(W-2) + W-2 - (W-2)^2 = 0
2W^2 - 2W - 2W^2 + 4W + W-2 - W^2 + 4W - 4 = 0

Thus, we have:
-W^2 + 7W -6 = 0
W^2 - 7W + 6 = 0
(W-6)(W-1) = 0

W = 1, 6

However, logically W cannot be 1 because then it is impossible to draw 2 or 3 marbles from the urn. Thus, it leaves one logical answer, W = 6

Therefore, the number of white marbles in the urn is 6. #

Question 50
Differentiate (x+1)(x+2)(x+3) without expansion.

 

[Riviet]

Answer Q50:

Spoiler

d/dx(x+1)(x+2)(x+3)=(x+1)(x+2)+(x+3).d/dx(x+1)(x+2)
=(x+1)(x+2)+(x+3)[x+1+x+2]
=(x+1)(x+2)+(x+3)(2x+3)

Q51: Solve x²+5x-3=0 by completing the square.

 

[klaw]

Q41:

Spoiler

P(Jeremy wins the game)=P(he wins on his first throw)+P(Wins the game after throwing a 4 or 10 on his first throw)+P(he wins the game after throwing a 5 or 9 on his first throw+P(he wins the game after throwing a 6 or 8 on his first throw
=2/9+1/18+4/45+25/198=244/495<0.5
.: In the long run the odds for winning are in the casino's favour

 

[icycloud]

Q51: Solve x^2+5x-3=0 by completing the square.

Spoiler

x^2+5x=3
x^2+5x+(5/2)^2 = 3 + (5/2)^2
(x+5/2)^2 = 3 + 25/4
=37/4

Thus, x+5/2 = +/- Sqrt(37/4)

x = {+Sqrt(37) - 5} / 2 or
x = {-Sqrt(37) - 5} / 2

Question 52:
Use Simpson's rule with 3 function values to find an approximation to the area under the curve y = 1/x between x = a and x = 3a, where a is positive.

Using the result above, show that ln(3) is approximately equal to 10/9.

 

[word.]

Q52:

Spoiler

by Simpsons rule, h = a,
A = a/3[(1/a + 1/(3a)) + 4/(2a)] = 10/9

evaluating the definite integral dx/x from a to 3a
= ln(3a) - ln(a)
= ln3

so ln3 approx. = 10/9

Question 53
The quadratic equation x² + Lx + M = 0 has one root which is twice the other. Prove that 2L² = 9M.

 

[Nerd Queen]

Spoiler

let roots = a and 2a (one root is twice the other)

(a +2a) = -L ________ 1

(a *2a) = M _________ 2

now solve simultaneously:

from 1:
3a = -L
a = -L/3

sub 'a' into 2:
2a² = M
2(-L/3)² = M
2(L²/9) = M

2L² = 9M

wow i really hope those power n spoiler things worked ive never used them b4

k next questio can be:

for what values of k has the equation x²-2kx+8k -15 = 0 real and distinct roots?

 

[gosh]

Spoiler

x²-2kx+8k -15 = 0
a=1, b=2k, c=8k-15

descriminant= b² - 4ac
= (2k)² - 4(1)(8k-15)
= 4k² - (32k - 60)
= 4k² - 32k + 60

for distinct and real roots: descriminant > 0
b² - 4ac > 0
4k² - 32k + 60 >0
(4k -20)(k-3) > 0
.: distinct and real roots when
k<3 and k>5

Express 3x² +4x +5 in the form A(x-1)² + B(x-1) +C

 

[skyrockets1530]

How do u do that? i'm probably mistaken but couldnt u just:

Spoiler

3x^2 + 4x +5
let x = (x-1)
therefore: 3(x-1)^2 + 4(x-1) + 5

 

[rama_v]

Spoiler

3x² + 4x + 5 = A(x-1)² + B(x-1) + C

note that the = sign should actually be a 'congruent' sign, i dunno exactly what you call it

put x = 1
3(1) + 4(1) + 5 = A(0) + B (0) +C
3 + 4 + 5 = C
.: C = 12
3x² + 4x + 5 = A(x-1)² + B(x-1) + 12
RHS = A (x² - 2x + 1) + B(x-1) + 12
= Ax² - 2Ax + A + Bx - B + 12
= Ax² - x(2A - B) + A + 12
therefore, equating coefficient of x²:
A = 3
equating coefficent of x
2A - B = -4
2(3) - B = -4
6 - B = -4
B = 10
C = 12

.: 3x² + 4x + 5 = 3(x-1)² +10(x-1) + 12

Next question:
A max invests $900 at the beginning of eachyear in a superannuiation fund. If interest paid is 12% p.a. on his investment, how much would he expect to receive if he retired after he had contributed to the fund for 40 years?

 

[chin music]

Spoiler

Alright so first investment grows to 900*(1.12)^40
Second is 900*(1.12)^39
Last is 900(1.12)
Therefore it makes a GP with a = 900(1.12) and r = 1.12
Therefore this equals a(r^n-1)/(r-1) which equals 900(1.12)((1.12)^40-1)/.12
that equals $773228.15. Someone correct me if im wrong

If someone already hasnt answered this as i worked it out then next question is find the equation of the normal to the curve y=x^5+4x^3-3x^2+2x at x=3.

 

[jake2.0]

If someone already hasnt answered this as i worked it out then next question is find the equation of the normal to the curve y=x^5+4x^3-3x^2+2x at x=3.

Spoiler

y'=5x^4+12x^2-6x+2
at x=3 y'=492
so gradient of norm.=-1/497 (m₁m₂=-1)
at x=3 y=330
so eq. of norm: y-330=(-1/497)(x-3)

find the point of inflection of the curve y=xe^x

 

[skyrockets1530]

Spoiler

y=xe^x
y' = e^x(1+x)
y'' = e^x(2+x)
let y'' = 0 to find point of inflexion
0 = e^x(2+x)
therefore x = -2
point of inflexion at (-2,-2e^-2)

next question
find the equation of the curve that is always concave upwards with a stationary point at -1,2 and y-intercept = 3

 

[chinkyeye]

hey i need help do not understand this trig identity.. HSC PAST PPER 2004 Q9a i
"consider the GP 1 - tan^2"theta" + tan^4"theta"...when is da limitngsum exists, find its value in simplest form... i done all da steps, buti don' understand last line of working..their last line is 1/sec^2"theta" and then jumps to da anwser of cos^2"theta" wtf pelase explain

 

[skyrockets1530]

hey i need help do not understand this trig identity.. HSC PAST PPER 2004 Q9a i
"consider the GP 1 - tan^2"theta" + tan^4"theta"...when is da limitngsum exists, find its value in simplest form... i done all da steps, buti don' understand last line of working..their last line is 1/sec^2"theta" and then jumps to da anwser of cos^2"theta" wtf pelase explain

sec^2theta = 1 / cos^2theta
thus 1 / (1 / cos^2theta) = cos^2theta