4U Revising Game (4 Viewers)

ronnknee

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Don't you just feel like doing maths for fun, but when you go onto BOS there aren't any questions? Well you can here now. Maths is fun after all =)

Rules:
1. Attempt the question from the newest poster. If you can do it, post the solution up (this includes working out). Remember to leave another question for someone else to solve
2. If there's an alternative method to a question (and is significantly more efficient) that is previously done, quote the question and post the solution; don't leave another question though
3. Challenging questions are encouraged but make sure that you can do them yourself
4. Since it's still early in the year, I assume that most of the grade haven't touched mechanics at school yet, so try stay out of that topic
5. Questions must be within the HSC syllabus


Previously done questions:

Question 1 (Complex)
Express (3 + 2i)(5 + 4i) and (3 - 2i)(5 - 4i) in the form of a + ib. Hence find the prime factors of 72 + 222.


Question 2 (Harder 3U)
u1= 1
un = (2un-1^3 + 27)/3un-12 for n >= 2

Show: un>3 for n >= 2
un+1 < un for n >=2


Question 3 (Conics)
The point P{acosx, bsinx} is any point on the ellipse x2/a2 + y2/b2 = 1 with focus S.

The point M is the midpoint of the interval SP.

Show that as P moves on the ellipse, M lies on another ellipse whose centre is midway between the origin O and the focus S.


Question 4 (Integration)
Int {0 -> a} [(x + a) / (x2 + a2)] dx


Question 5 (Polynomial)
The equation x3 + qx + r = 0 has roots A, B and C. Find the expression for A5 + B5 + C5


Question 6 (Conics)
The point P(x0,y0) lies on the hyperbola x2/a2 - y2/b2 = 1

(i) Show that the acute angle @ between the asymptote satisfies tan@ = 2ab / (a2 - b2)

(ii) If M and N are the feet of the perpendiculars drawn from P to the asymptotes, show that MP.NP = a2b2 / (a2 + b2)

(iii) Hence find the area of triangle PMN


Question 7 (Integration)
Int {1 -> e} [ln x] dx


Question 8 (Integration)
Int {-1 -> 1} [tan x . ln (x2 + e)] dx


Question 9 (Integration)
Int [1 / (1 + sin x)] dx


Question 10 (Volume)
A solid figure has as its base, in the xy plane, the ellipse x2/16 + y2/4 = 1

Cross sections perpendicular to the x-axis are parabolas with latus rectums in the xy plane. Find the volume of the solid


Question 11 (Integration)
Int [1 / x.root(1 + x2)] dx


Question 13 (Complex)
Solve z2 = -4


Question 14 (Integration)
If In = Int [tann x] dx where n >= 0, show that In = [tann-1x / (n - 1)] - In-2


Question 15 (Curve sketching)
Graph |x+y| = 2 labeling significant points.


Question 16 (Harder 2U)
A triangle has sides 3, 4 & 5. A circle has been inscribed.

Find the area of the circle.


Question 17 (Polynomial)
P(x) is a monic polynomial of degree 4 with interger coefficients and constant term 4. One zero is root 2, another zero is rational and the sum of the zeros is positive. Factorise P(x) over real numbers.


Question 18 (Integration)
Int {0 -> pi} [sin x / (a + b.cos2 x)] dx


Question 19 (Volume)
The top face of a container is a rectangle of sides 3 m and 4 m respectively.
The bottom is a rectangle of sides 2 m and 3 m respectively.
(i.e. side of TOP 4 m // BOT 3 m and side of TOP 3 m // BOT 2 m)

If the height of the container is 1.5 m, find the volume of this container by slicing.


Question 20 (Integration)
Prove Int {-pi/4 -> pi/4} [ 1 / (1 + sinx)] dx = Int {0 -> pi/4} [2 sec2x] dx


Question 21 (Volume)
Find the volume obtained by rotating the region enclosed by the circle (x - b)2 + y2 = a2

Where b > a, about the y-axis


Question 22 (Integration)
Find f'(x), if the function f is defined on all reals greater or equal to 0, and:
f(x) = Int {0 -> x2} [cos (root t)] dt


Question 23 (Complex)
Show that the roots of z6 + z3 + 1 = 0 are among the roots of z9 - 1 = 0. Hence find the roots of z6 + z3 + 1 = 0 in modulus/argument form


Question 24 (Complex)
If | z - 6i | = 5

Prove that arg ( z - 4 - 3i / z + 3 - 2i ) = pi/4 or 3pi/4


Question 25 (Integration)
Int {-1 -> 1} [1 / (1 + e-x) dx


Question 26 (Integration)
Int [sec3x] dx


Question 27 (Integration)
Int [4 / (x2 - 2x - 1)] dx


Question 28 (Polynomial)
Show that the equation x101 + x51 + x - 1 = 0 has exactly one real root


Question 29 (Integration)
Int [1 / 2.sin(2x) + cos x] dx


Question 34 (Integration)
By using the substitution (1 + x) / (1 - x) = z2, evaluate:

Int {-1 -> 1} sqrt [(1 + x) / (1 - x)] dx


Question 35 (Harder 3U)
i) Prove: 1 - r2 + r4 - r6 + r8 - ... = 1 / (1 + r2), for | r | < 1
ii) Hence find a series for tan inverse
iii) Hence show that pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 ...


Question 36 (Integration)
Int [rn / (1 - r)] dr = ln |1 - r| - (r + r2 / 2 + r3 / 3 + ... + rn / n)


Question 38 (Integration)
i) Int {0 -> 1} (5 / (2t + 1)(2 - t) dt
ii) Int {0 -> pi/2} (1 / 3 sin x + 4 cos x) dx


Question 39 (Polynomial)
Find the gradient of the tangent to the curve x2 - 3xy + 2y2 = 3 at the point (5, 2)


Question 40 (3U)
If pqr represents a three digit number and p + q + r = 3A, where A is a positive integer; show that the number pqr is also divisible by 3


Question 41 (3U)
By considering the graph of the hyperbola y = 1/x, show that
lim (1 + 1/n)n = e
n->∞


Question 42 (Integration)
Find the reduction formula for ex cosn x


Question 43 (Complex)
Prove that the roots of

z3 + 3pz2 + 3qz + r = 0

form an equilateral triangle if and only if p2 = q



Currently unsolved:

Question 12 (Harder 3U)
How many arrangements of 4 letters can be made using the letters of the words ZAPOPAN PARAGON?


Question 30 (Conics)
The normal to the parabola (4x2)/3 - 4y2 = 1 at the point (3sqrt2/4,sqrt2/4) is also a tangent to the circle x2 + y2 = 1 at the point (x1,y1). Find x1 and y1


Question 31 (?)
The temperature throughout the day is plotted continuously with respect to time. what is the formula for the average temperature over time-a to time-b?


Question 32 (Integration)
By considering the definition of the derivative
, prove lim {x->0} ln(1 + x) / x = 1


Int | x | dx


Question 33 (Integration)
If f is a continuous function such that

Int {0->x} f(t) dt = xe2x - Int {0->x} e-t . f(t) dt

For all x, find an explicit formula for f(x)


Question 37 (Conics)
The point P(acosθ, bsinθ) is any point on the ellipse: x2/a2 + y2/b2 = 1, with focus S.

The point M is the midpoint of the interval SP

Show that as P moves on the ellipse, M lies on another ellipse whose centre is midway between the origin O and the focus S
 
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Undermyskin

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(3+2i)(5+4i) = 7 + 22i (1)
(3-2i)(5-4i) = 7 - 22i (2)

Since (1) x (2) = 7^2 - 22^2 i^2 = 7^2 + 22^2

The prime factors are the 4 brackets given. (?)

Questions:
u(1)= 1
u(n) = (2u(n-1)^3 + 27)/3u(n-1)^2 for n >= 2

Show: u(n)>3 for n >= 2
u(n+1) < u(n) for n >=2

Note: u (n) is the 'n' th value of 'u'
 

conics2008

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Undermy skin, i havn't done harder 3unit.. ronnkee sorry the girl posted up the answers.. ill post an Conics question << My favourite topic xD

The point P{acosx,bsinx} is any point on the ellipse x^2/a^2 + y^2/b^2 = 1 with focus S.

The point M is the midpoint of the interval SP.

Show that as P moves on the ellipse, M lies on another ellipse whose centre is midway between the origin O and the focus S.

This is the question, I will give hints, you may assume focus S(ae,0) S'(-ae,0) and Origin (0,0)

Good Luck.. PS I already know how to do this
 
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ronnknee

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Undermyskin said:
Questions:
u(1)= 1
u(n) = (2u(n-1)^3 + 27)/3u(n-1)^2 for n >= 2

Show: u(n)>3 for n >= 2
u(n+1) < u(n) for n >=2

Note: u (n) is the 'n' th value of 'u'
I'm guessing it's a mathematical induction question since we have to show for n >= 2

When n = 2,
u(2)
= (2 + 27) / 3
= 29 / 3
> 3
Therefore it is true for n = 2

Assume it is true for n = k
ie. u(k)
= (2[u(k-1)]^3 + 27) / 3([u(k-1)]^2
> 3

Consider n = k + 1
u(k + 1)
= (2[u(k)]^3 + 27) / 3[u(k)]^2

The problem is, if I sub in u(k) > 3, I get something like 3 > 3
 

ronnknee

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The equation x^3 + qx + r = 0 has roots A, B and C. Find the expression for
A^5 + B^5 + C^5
 

undalay

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the point P(x0,y0) lies on the hyperbola x2 / a2 - y2 / b2 = 1

(i) Show that the acute angle @ between the asymptote satisfies

tan@ = 2ab / (a2 - b2)

(ii) If M and N are the feet of the perpendiculars drawn from P to the asymptotes, show that MP.NP = a2b2 / (a2 + b2)

(iii) hence find the area of triangle PMN
 
Last edited:

independantz

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ronnknee said:
The equation x^3 + qx + r = 0 has roots A, B and C. Find the expression for
A^5 + B^5 + C^5
Since A,b and C are the roots of the eq, they satisfie they eq.

therefore, A^3=-qA-r
similarly, B^3=-qB-r
similarly, C^3=-qC-r

A^5=-qA^3-rA^2
B^5=-qB^3-rB^2
C^5=-qC^3-rC^2

therfore, A^5+b^5+c^5=-q(A^3+B^3=C^3)-r(A^2+B^2+C^2)


then you keep subbing in until you get it completely in terms of r and q , however it takes to long to type so i cbf.
 

ronnknee

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undalay said:
the point P(x0,y0) lies on the hyperbola x2 / a2 - y2 / b2 = 1

(i) Show that the acute angle @ between the asymptote satisfies

tan@ = 2ab / (a2 - b2)

(ii) If M and N are the feet of the perpendiculars drawn from P to the asymptotes, show that MP.NP = a2b2 / (a2 + b2)

(iii) hence find the area of triangle PMN
conics2008 should be able to do this without a problem yea?
 

undalay

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ronnknee said:
conics2008 should be able to do this without a problem yea?
This is a slightly modified question from my half yearlyy.
 

namburger

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undalay said:
This is a slightly modified question from my half yearlyy.
Why make it harder than it already than it is. The question was originally prove that the area = ......
Without it, i would of gotten it wrong lool
 

namburger

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3unitz said:
while we wait for conics:

evaluate the following integrals

e
|
| ln(x) dx
|
1

1
|
| tanx . ln (x^2 + e) dx
|
-1
1.
u= ln (x)
du = dx/x

dv = dx
v = x

e
|
| ln(x) dx = [x ln x - Integrate(dx)]
|
1

= [xln x - x] from e to 1
= 1

2.
1
|
| tanx . ln (x^2 + e) dx
|
-1

=
1
|
|tanx . ln (x^2 + e) + tan-x . ln (x^2 + e) dx
|
0

=
1
|
| 0 dx
|
0

=0
 
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undalay

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|
| 1 / (1 + sinx) dx
|

|
| 1-sinx / (1 - sin^2x) dx
|

|
| 1-sinx / cos^2 x dx
|

|
| sec^2 x - integral sinx / cos^2 x dx
|

= tanx + c1 + integral cos^-2 d (cosx)
= tanx - sec x + c2
 

ronnknee

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3unitz said:
1
|
| tanx . ln (x^2 + e) dx
|
-1
The more efficient way is to declare that it's an odd integral and say it's 0, instead of actually integrating it

Let f(x) = tan x . ln (x^2 + e)
f(-x) = tan -x . ln ([-x]^2 + e)
= - tan x . ln (x^2 + e)
= -f(x)

Therefore f(x) is an odd function
Therefore, the integral = 0

Haha guys, remember to leave questions behind after completing one that's not done before
 
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namburger

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3unitz said:
odd function in the integral :)

try:

|
| 1 / (1 + sinx) dx
|
let t = tan (x/2)
dt = (1/2)(sec^2 (x/2) ) dx
= (1/2) ( 1+t^2) dx

Therefore, dx = 2dt/ (1 + t^2)


|
| 1 / (1 + sinx) dx
|

=
|
| 1/ [1 + (2t)/ (1+ t^2)] . 2dt/ (1 + t^2) since sin x = 2t/(1+t^2)
|

=
|
| 2 / (t+1)^2 upon expansion
|

= -2/ (t+1)
= -2/ ( tan x/2 + 1)
 

Mark576

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Using t=tan(x/2):
int. dx/(1+sinx)
= int. [2dt/(1+t2)]/[1+(2t/(1+t2))]
= int. 2dt/(t2+2t+1)
= 2[int. dt/((t+1)2)]
= -2/(t+1) + c
= -2/(tan(x/2) +1) + c

Alternatively:

int. dx/(1+sinx)
Multiplying the numerator and the denominator by [1-sinx]:
I = int. dx(1-sinx)/cos2x
= int. sec2 dx - int. tanxsecx dx
= tanx - secx + c
 
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ronnknee

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Edit:
Okay, my solution to the volumes question is wrong as I interpreted the question incorrectly
This solution would fit the question "Find the volume of the solid formed by rotating the ellipse x^2/16 + y^2/4 = 1 around the y axis using the method of slicing"



End edit

I think that's right if I'm interpreting the question correctly


Int[1 / {x.root(1 + x^2)}]
 
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namburger

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ronnknee said:
Int[1 / {x.root(1 + x^2)}]
let u^2 = 1 + x^2
2u du = 2x dx
dx = u du/ x

Int [1 / {x.root(1 + x^2)}] . [u du/x]
= du / x^2 upon expansion
= du / (u^2 -1)

du / (u^2 -1) = A/(u-1) + B/(u-1)
= 1/2 int [1/(u-1)] -1/2 int [1/(u+1)] by use of partial fractions
= 1/2 ln [sqrt (1+x^2)-1] -1/2 ln [sqrt (1+x^2)+1] + C

Undalay's conics question still needs doing
Complex no.: Sketch the locus of the point z:
arg [(z-5)/(z+1)] = pi/4. Find its cartesian equation
 
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P

pLuvia

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How many arrangements of 4 letters can be made using the letters of the words ZAPOPAN PARAGON?
 

ronnknee

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AAAA
PPP
OO
NN
R
G
Z

Well this is a good way to start, by first grouping the letters by the frequency and then determining the cases
It'll involve a lot of cases I'm guessing

I shall try this later tonight
 

WannaBang?

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Substitute a number for each letter so that the equation works.
 

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