Mathematics Marathon HSC 09 (3 Viewers)

[Aquawhite]

if your cos(C)

and im the reciprocal of you

what am i

You're sec(C)!

 

[00iCon]

2-6+8-8+32-10+...

find the sum of the first 20 terms of this series

10 terms of arithmetic:
S10=10/2(2(-6)+9(-2))
=5(-12-18)
=-150
10 terms of GP:
S10=[2(4^10-1)]/[1-4]
=2*1048575/3
=699050

.'. 699050-150=698900

now, class! integrate: [1-sqroot(x)]/[1-x]

 

[untouchablecuz]

hold up guys

this relationship is moving way too fast

i wanna get to know you guys first

 

[00iCon]

hold up guys

this relationship is moving way too fast

i wanna get to know you guys first

the world works like this: do maths or have a social life.
just answer the damn question!

 

[eldore44]

integrate: [1-sqroot(x)]/[1-x]

[1-sqroot(x)]/[1-x] = [1-sqroot(x)]/[1-sqroot(x)][1+sqroot(x)]

=1/[1+sqroot(x)]
u=sqroot(x)
du/dx=1/sqroot(x)
du.sqroot(x)=dx
du.u=dx
du.u/(1+u)
=1 - 1/(1+u)
integral of 1 - 1/(1+u) = u - log(1+u) + c
=sqroot(x) - log(1+sqroot(x)) + c

.. i think.

Differentiate tan^-1(x^4) with respect to x

EDIT: there should be some 2's floating around from the derivative of sqroot(x)

 

[untouchablecuz]

sec(C)?

ahahaha i c what u did there!

integrate: [1-sqroot(x)]/[1-x]

i dont think its at 2 unit level

all i can think of is substituting x=u²

 

[star of the sea]

its not 2unit its integration using substitution its a part of 3unit....can som1 doing it using a 2unit method coz im stuck

 

[00iCon]

[1-sqroot(x)]/[1-x] = [1-sqroot(x)]/[1-sqroot(x)][1+sqroot(x)]

=1/[1+sqroot(x)]
u=sqroot(x)
du/dx=1/sqroot(x)
du.sqroot(x)=dx
du.u=dx
du.u/(1+u)
=1 - 1/(1+u)
integral of 1 - 1/(1+u) = u - log(1+u) + c
=sqroot(x) - log(1+sqroot(x)) + c

.. i think.

Differentiate tan^-1(x^4) with respect to x

EDIT: there should be some 2's floating around from the derivative of sqroot(x)

Methinks perfect square in the numerator is an easier method.
Actually, your first step is wrong!
btw, that was a trial ext2 Q7 (c) worth 2 marks.

dont ask 4 the method but:
d[tan^-1(x^4)]/dx
=[4x^3]/[1+x^8]

i reiterate:
integrate: [1-sqroot(x)]/[1-x]
GL

 

[star of the sea]

(4x^3)sec^2(x^4)...i think

solve the following equation for x
e^2x+3e^x-10=0

 

[AlexJB]

e^2x+3e^x-10=0
Let u = e^x
u^2 + 3u - 10 = 0
(u-2)(u+5) = 0
u = 2 or -5
e^x = 2 or e^x = -5
x = ln(2) or x = ln(-5) [not solution]
Therefore x = ln2

Differentiate y = cos(ln(x) + 1)

 

[star of the sea]

can som1 plzz integrate: [1-sqroot(x)]/[1-x] using a 2unit method....i kinda gave up on 3unit afta the trials

 

[star of the sea]

y'=-sin(lnx+1).1/x
=-sin(lnx+1)/x

the first three term of an arithmetic series are -1+4+9...
i) find the 60th term
ii) hence or otherwise find the sum of the first 60 terms of the series

 

[eldore44]

[1-sqroot(x)]/[1-x] = [1-sqroot(x)]/[1-sqroot(x)][1+sqroot(x)]

=1/[1+sqroot(x)]
u=sqroot(x)
du/dx=1/2sqroot(x)
du.2sqroot(x)=dx
du.2u=dx
du.2u/(1+u)
=2 - 2/(1+u)
integral of 2 - 2/(1+u) = 2u - 2log(1+u) + c
=2sqroot(x) - 2log(1+sqroot(x)) + c

That looks fine. Where is the mistake?

Also, I dont know if this could even be a 2 unit question.

 

[Zak Ambrose]

e^2x+3e^x-10=0
Let u = e^x
u^2 + 3u - 10 = 0
(u-2)(u+5) = 0
u = 2 or -5
e^x = 2 or e^x = -5
x = ln(2) or x = ln(-5) [not solution]
Therefore x = ln2

Differentiate y = cos(ln(x) + 1)

y= cos (ln (x) + 1)
y'= -1/x (-sin (ln(x) + 1))
= [sin (ln (x) + 1)]/x


enough calculus

(c) A chessboard has 32 black squares and 32 white squares. Tanya chooses three different squares at random.
(i) What is the probability that Tanya chooses three white squares?
(ii) What is the probability that the three squares Tanya chooses are the same colour?
(iii) What is the probability that the three squares Tanya chooses are not the same colour?

 

[ninetypercent]

the first three term of an arithmetic series are -1+4+9...
i) find the 60th term
ii) hence or otherwise find the sum of the first 60 terms of the series

i) T = a+(n-1)d
= -1 + (60-1)5
= 294

ii) S = n/2[2a + (n-1)d]
= 60/2[2x-1 + (60-1)5]
= 30[-2 + 295]
= 8790

 

[00iCon]

[1-sqroot(x)]/[1-x] = [1-sqroot(x)]/[1-sqroot(x)][1+sqroot(x)]

=1/[1+sqroot(x)]
u=sqroot(x)
du/dx=1/2sqroot(x)
du.2sqroot(x)=dx
du.2u=dx
du.2u/(1+u)
=2 - 2/(1+u)
integral of 2 - 2/(1+u) = 2u - 2log(1+u) + c
=2sqroot(x) - 2log(1+sqroot(x)) + c

That looks fine. Where is the mistake?

Also, I dont know if this could even be a 2 unit question.

The first line!

=integral(1+sqroot(x))dx
=x+(2/3)(x^(3/2))+C

It's not that hard, im not sure of the extent of the 2unit syllabus

 

[untouchablecuz]

y= cos (ln (x) + 1)
y'= -1/x (-sin (ln(x) + 1))
= [sin (ln (x) + 1)]/x


enough calculus

(c) A chessboard has 32 black squares and 32 white squares. Tanya chooses three different squares at random.
(i) What is the probability that Tanya chooses three white squares?
(ii) What is the probability that the three squares Tanya chooses are the same colour?
(iii) What is the probability that the three squares Tanya chooses are not the same colour?

i) ³²C₃/⁶⁴C₃=5/42
ii) ³²C₃/⁶⁴C₃ + ³²C₃/⁶⁴C₃=5/21
iii) P(not the same colour)=1-P(same colours)=1-5/21=16/21

 

[untouchablecuz]

The first line!

=integral(1+sqroot(x))dx
=x+(2/3)(x^(3/2))+C

It's not that hard, im not sure of the extent of the 2unit syllabus

-.-

thats a diff integral

u gave us initially:

integral of (1-sqrt(x))/(1-x)

 

[AlexJB]

y= cos (ln (x) + 1)
y'= -1/x (-sin (ln(x) + 1))
= [sin (ln (x) + 1)]/x


enough calculus

(c) A chessboard has 32 black squares and 32 white squares. Tanya chooses three different squares at random.
(i) What is the probability that Tanya chooses three white squares?
(ii) What is the probability that the three squares Tanya chooses are the same colour?
(iii) What is the probability that the three squares Tanya chooses are not the same colour?

You missed a minus in your answer. Been a while since I did probability. I'm assuming by different she can't pick the same square twice?

i. 32P3/64P3 = 5/42
ii. 2 x 5/42 = 5/21
iii. 1- 5/21 = 16/21

Let me know if I'm wrong.

---

Using the substitution u = 2x + 1 or otherwise, integrate [4x]/[2x + 1], limits 1,0

 

[00iCon]

(c) A chessboard has 32 black squares and 32 white squares. Tanya chooses three different squares at random.
(i) What is the probability that Tanya chooses three white squares?
(ii) What is the probability that the three squares Tanya chooses are the same colour?
(iii) What is the probability that the three squares Tanya chooses are not the same colour?

i) 1/2 * 31/63 * 15/31
460/3906
230/1953

ii) 460/1953

iii)1493/1953

i have no 2 unit questions.