3u Mathematics Marathon V 1.0 (2 Viewers)

[Slidey]

This thread is not intended as a place to ask for help. Please make a new thread for that.

Rules:

1) No spamming. Please try to stick to strictly answering and questioning. If you have an alternate way of doing a problem, you can post that, too.

2) If you answer a question, try to find a new question for the next person to try. Hard or uncommon textbook questions and and questions from past papers are good places to look.

3) Please surround your answer in spoiler tags: (spoiler) solution (/spoiler) replacing the round brackets with these square brackets: [

Have fun.

Q 1)

Find the cartesian equation of the curve whose parametric equations are:
x=2cos@
y=3sin@

 

[rama_v]

Solution to Question 1

Spoiler

x = 2cos@
x² = 4cos²@
x² = 4(1-sin²@)
x² = 4 - 4sin²@
4sin²@ = 4 - x²
sin²@ = (4 - x²) / 4

y = 3sin@
y² = 9sin²@
y² = 9(4-x²)/4
y² = (36 - 9x²)/4

Question 2
The acceleration of a particle is given by the equation a = -4 / (x+1)²

If the particle begins at the origin with velocity 6 m/s , find an expression for the velocty, v, of the particle.

 

[>monkey<]

Solution to Q1
make cos@ the subject: cos@=x/2
make sin@ the subject: sin@=y/3

then (sin@)^2 + (cos@)^2 = 1

so x^2/4 + y^2/9 = 1

 

[word.]

Spoiler

a = -4/(x+1)²

d/dx(1/2*v²) = -4/(x+1)²
1/2*v² = 4/(x+1) + C
at x = 0, v = 6
18 = 4 + C, C = 14
1/2*v² = 4/(x+1) + 14
v = +- Sqrt{8/(x+1) + 28}
but at x = 0, v = 6
so v = Sqrt{8/(x+1) + 28}

Question 3
Find the cubic equation whose roots are each one less than the roots of the cubic equation 24x³ - 2x² - 5x + 1 = 0.

 

[Slidey]

I can solve that using 4u, but... hmm.

Spoiler

a+b+y=1/12
ab+ay+by=-5/24
aby=-1/24

(a-1)+(b-1)+(y-1)=1/12-3=-35/12
(a-1)(b-1)+(a-1)(y-1)+(b-1)(y-1)=ab-a-b+1+ay-a-y+1+by-y-b+1=-5/24-2*(1/12)+3=21/8
(a-1)(b-1)(y-1)=(ab-a-b+1)(y-1)=aby-ab-ay+a-by+b+y-1=-1/24+5/24+1/12-1=-3/4

.'. the equation is:

x^3+(35/12)x^2+(21/8)x+3/4=0
Or:
24x^3+70x^2+63x+18=0

Question 4) Differentiate x.cos^-1(x) - sqrt(1-x^2) and hence or otherwise integrate cos^-1(x) from 0 to 1, w.r.t. x.

 

[haboozin]

Question 3
Find the cubic equation whose roots are each one less than the roots of the cubic equation 24x³ - 2x² - 5x + 1 = 0.

this is 4unit?

Spoiler

24(x + 1)^3 -2(x + 1)^2 -5(x + 1) + 1 = 0
24(x^3 + 3x^2 + 3x + 1) - 2(x^2 + 2x + 1) -5(x + 1) + 1 = 0
24x^3 + 72x^2 + 72x + 24 -2x^2 -4x - 2 -5x -5 + 1 = 0
24x^3 + 70x^2 + 63x + 18 = 0

Question 4) Differentiate x.cos^-1(x) - sqrt(1-x^2) and hence or otherwise integrate cos^-1(x) from 0 to 1, w.r.t. x.

Spoiler

u = x v= cos^-1(x)
u' = 1 v'= -1/sqrt(1-x^2)

cos^-1(x) -x/sqrt(1- x^2) + x/(1 - x^2)^1/2 = cos^-1(x)

so integral of cos^-1(x) = x.cos^-1(x) - sqrt(1-x^2) + c

Question 4B
A ball is thrown from a point P with velocity V. at an inclination of a degrees to the horisontal and reaches a point Q after t seconds.
Show that if PQ is inclined at @ to the horizontal (where a > @), then the direction of motion of the ball, when at Q, is inclined to the horizontal at an acute angle of tan^-1[2tan@ - tana]

you may use the the results of motion:
x = Vtcosa
y = vtsina - 1/2gt²

 

[Slidey]

Please direct any comments about any of these questions here: http://community.boredofstudies.org/showthread.php?t=89440

 

[klaw]

Please direct any comments about any of these questions here: http://community.boredofstudies.org/showthread.php?t=89440

No Thread specified. If you followed a valid link, please notify the webmaster

Doesn't work

 

[thunderdax]

Question 4B
A ball is thrown from a point P with velocity V. at an inclination of a degrees to the horisontal and reaches a point Q after t seconds.
Show that if PQ is inclined at @ to the horizontal (where a > @), then the direction of motion of the ball, when at Q, is inclined to the horizontal at an acute angle of tan^-1[2tan@ - tana]

you may use the the results of motion:
x = Vtcosa
y = vtsina - 1/2gt²

k, i'll tackle this question since its this thread is getting pretty dead:

Spoiler

PQ is a line through O with acute angle @ with x-axis
So, eqn is y=tanx
Sub x=vtcosa, y=vtsina - 1/2gt²
vtsina-1/2gt²=vtcosatan@
t=0 or 1/2gt=vsina-vcosatan@ (1)
t=2v(vsina-vcosatan@)/g
Now, Let horiz. velocity be a, vert. velocity be b.
So, a=vcosa, b=vsina-gt
Sub in (1):
b=vsina-2vsina+2vcosatan@
b=v(2cosatan@-sina)
So, b/a=v(2cosatan@-sina)/vcosa
b/a=2tan@-sina/cosa
b/a=2tan@-tana
Therefore, the angle between the velocity and the horizontal is
tan^-1(2tan@-tana)

K, here's a question from my trial:

Question 5
a) Show that sin (sin^-1x-cos^-1x)=2x²-1
b) Hence solve sin^-1x-cos^-1x=sin^-1(5x-4)

 

[yabby]

Q5 a)

No point of a spoiler, didnt even get the first part!..someone correct me please!

LTP sin(sin^-1x-cos^-1x) = 2x^2-1

let sin^-1x = @
and cos^-1x = B
So i dont have to type too much
LHS = sin(@-B)
= sin@cosB - sinBcos@
= x^2 -sinBcos@

sinB = sin(90 - @)
= sin90cos@ - cos90sin@
= -x

cos@ = cos(90-B)
= cos90cosB - sin90sinB
= x

sub values in
x^2 - (x).(-x)
= 2x^2
where does the minus one come from?

 

[icycloud]

Question 5
a) Show that sin (sin^-1x-cos^-1x)=2x²-1
b) Hence solve sin^-1x-cos^-1x=sin^-1(5x-4)

Spoiler

(a) Let a = arcsin(x), b = arccos(x)
sin(a-b) = sin(a)cos(b)-sin(b)cos(a)

But, x = sin(a)=cos(b)
Sqrt(1-x^2)=sin(b)=cos(a) {using a right-angled triangle}

Thus, LHS = sin (arcsin(x)-arccos(x))
= sin(sin(a-b) = x^2 - (Sqrt(1-x^2))^2
= x^2 - 1 + x^2
= 2x^2 - 1
= RHS #

(b) arcsin(x) - arccos(x) = arcsin(5x-4)
sin(arcsin(x) - arccos(x)) = 5x-4

Hence, using part (a),

2x^2 - 1 = 5x - 4
2x^2 - 5x + 3 = 0
(2x - 3) (2x2 - 2) / 2 = 0
(2x - 3) (x - 1) = 0

Thus, x = 3/2, 1
But x cannot be 3/2 (domain of arc- functions)

Therefore, x = 1 #

Question 6
Find ∫ dx / Sqrt[x(1-x)] with the substitution z = x - 1/2.

 

[100percent]

Spoiler

z=x-1/2
dz=dx

∫dz/sq rt (z+1/2)(-z+1/2)
∫dz/sq rt (1/4-z²)
arcsin (z/0.25)
arcsin (4z)
arcsin (4x-2)+c

a game of poker with 52 cards, find the probability of getting 4 aces.

 

[icycloud]

Question 7
a game of poker with 52 cards, find the probability of getting 4 aces.

Oops sorry, didn't read that it was for a game of poker. The answer below is for a game where 13 cards are dealt for one hand.

Spoiler

Number of possible hands: 52C13
Number of hands with 4 aces: 48C9

P(4aces in one hand) = 48C9 / 52C13
= 11/4165
= 0.26% (2dp) #

Question 8
Prove that {sin(3x) / sin(x)} - {cos(3x) / cos(x)} = 2. (where sin(x) and cos(x) not equal to 0).

 

[rama_v]

Solution to Question 8
Prove that {sin(3x) / sin(x)} - {cos(3x) / cos(x)} = 2. (where sin(x) and cos(x) not equal to 0).[/QUOTE]

Spoiler

[sin(3x)cos(x) - cos(3x)sin(x)] / sinxcosx

sin(3x-x)/(1/2)sin2x
= sin2x/[(1/2)sin2x)]
= 1/(1/2)
= 2

Question 9:
Find the general solution to the equation:
Cot y = 15^o

 

[Slidey]

No Thread specified. If you followed a valid link, please notify the webmaster

Doesn't work

Oops. This is it: http://community.boredofstudies.org/showthread.php?t=89508

It just keeps this thread cleaner and makes it easier to see what is happening.

 

[word.]

Spoiler

coty = 15^o
tan(pi/2 - y) = pi/12
pi/2 - y = npi + pi/12
y = 5pi/12 - npi
y = 75 - 180n... no idea
but anyway...

Question 10
Find INT.{0 to 1} x*Sqrt(1 - x)dx using the substitution x = 1 - u²

 

[icycloud]

Question 10
Find INT.{0 to 1} x*Sqrt(1 - x)dx using the substitution x = 1 - u^2

Spoiler

I = ∫x*Sqrt(1-x) dx

Let x = 1-u^2 (We restrict the domain to u >= 0)
dx = -2u du

I = ∫(1-u^2) * u * -2u du
= -2∫u^2 - u^4 du
= -2[(u^3)/3 - (u^5)/5] + C

Substitute boundaries, x = 0 --> x = 1
1-u^2 = 0 --> 1 - u^2 = 1
u = +1 --> u = 0

I(+1,0) = -2 (0 - (1/3-1/5))
= -2 * (-2/15)
= 4/15 #

Question 11
Multiply sec(x) by [sec(x) + tan(x)] / [sec(x) + tan(x)] and hence, or otherwise, find ∫sec(x) dx.

 

[100percent]

Question 11
Multiply sec(x) by [sec(x) + tan(x)] / [sec(x) + tan(x)] and hence, or otherwise, find ∫sec(x) dx.

Spoiler

∫sec(x) dx=ln (secx+tanx)+c

Question 12
show that sinx cosx= 0.5sin2x
hence find the exact value of ∫sin²x cos²x dx with boundaries pi/8 and 0

 

[Jago]

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

 

[icycloud]

Question 12
show that sinx cosx= 0.5sin2x
hence find the exact value of ∫sin²x cos²x dx with boundaries pi/8 and 0

Spoiler

sin[2x] = 2sin[x]cos[x]
sin[2x]/2 = sin[x]cos[x]

Thus, sin[x]cos[x] = 1/2 sin[2x]

∫sin^2[x] cos^2[x] dx
= ∫(sin[x]cos[x])^2 dx
= ∫(1/2 sin[2x])^2 dx
= 1/4 ∫ sin^2[2x] dx
= 1/4 ∫ (1-cos[4x])/2 dx
= 1/8 ∫ 1-cos[4x] dx
= 1/8 (x - 1/4 sin[4x]) + C

Bounds, 0 --> pi/8

I = 1/8 (pi/8 - 1/4 sin[pi/2] - 0)
= 1/8 (pi/8 - 1/4)
= pi/64 - 1/32
= (pi-2)/64 #

Question 13
Prove using mathematical induction that:

1+3+6+...+1/2 n(n+1) = 1/6 n(n+1)(n+2)

for all integers n = 1,2,3,...