3u Mathematics Marathon V 1.1 (1 Viewer)

[pLuvia]

Spoiler

We need A(x,y) and let P(10,8)
10=[(4*2-1*x)/(4-1)]
30=8-x
-22=x
8=[4*3-1*y)/(4-1)]
24=12-y
-12=y
A(-22,-12)

Next question:
If a particle is projected up vertically with an initial velocity of 30m/s, from the origin. What is the speed of the particle as it lands back onto the horizontal ground

 

[Riviet]

Next question:
If a particle is projected up vertically with an initial velocity of 30m/s, from the origin. What is the speed of the particle as it lands back onto the horizontal ground

Spoiler

Take upwards to be the positive direction.
..
y =-g
.
y=-gt+c

when:
.
y=30 and t=0,
c=30
.
y=-gt+30

y=-gt²/2 + 30t + C

When t=0 and y=0,
c=0
.'. y=-gt²/2 + 30t

when:
.
y=0

gt-30=0
t=30/g

.'. y=-g(30/g)²/2 + 30(30/g)
=450/g

Starting at the max height:
..
y = -g
.
y = -gt + c

when:
.
y=0 and t=0,
c=0
.
y=-gt

y=-gt²/2 + C

when t=0 and y=450/g, taking initial position as the max height reached,
c= 450/g

.'. y=-gt²/2 + 450/g

when y=0,
gt²=450/g
t=(450/g²)^1/2
=(rt450)/g
.
y=-g(rt450)/g
=-rt450

.'. particle hits the ground at rt450 m/s.

 

[Riviet]

Next Question:
By completing the square and rationalising of the numerator with a substitution of u=x-7, evaluate:
8
∫(14x-24-x²)^1/2/{14x-24-x²} dx
7

 

[SoulSearcher]

You made it less obvious than it was before, but I'll do it anyway

Spoiler

14x-24-x² = (x-2)(12-x)
Therefore
(14x-24-x²)^1/2/(14x-24-x²)
= (14x-24-x²)^1/2/{(14x-24-x²)^1/2*(14x-24-x²)^1/2}
= 1/(14x-24-x²)^1/2
= 1/{(x-2)(12-x)}^1/2

u = x-7
du/dx = 1
Therefore du = dx
x = u+7, therefore
(x-2)(12-x) = (5+u)(5-u) = 25 - u²
x = 8, u = 1
x = 7, u = 0
Therefore int. [(14x-24-x²)^1/2/(14x-24-x²)] (8->7) dx
= int. [1/{(x-2)(12-x)}^1/2] (8->7) dx
= int. [1/(25 - u²)^1/2] (1->0) du
= [ sin^-1x/5 ] (1->0)
= sin^-1(1/5) - sin^-1(0/5)
= sin^-1(1/5) units²

 

[Riviet]

By completing the square, I meant:

Spoiler

25-(x-7)²

But your method works too of course.

 

[SoulSearcher]

Yeah I saw that method as well, however it worked both ways I'll put a question up tomorrow

Next Question:

Let f(x) = x³ + 3bx² + 3cx + d
a) show that y = f(x) has two distinct turning points if and only if b² > c
b) If b² > c, show that the vertical distance between the turning points is 4(b²-c)^3/2

A hint for b) as given by the textbook:

Spoiler

use the sum and product of the roots of the derived function

 

[followme]

Spoiler

a)

f(x)=x^3+3bx^2+3cx+d
f'(x)= 3x^2+6bx+3c
let f'(x)=0
ie 3x^2+6bx+3c=0
for 2 roots, delta>0
36b^2- 4*3*3c>0
36(b^2-c)>0
so b^2>c

b)
m=alpha, n=beta

3x^2+6bx+3c=0
m+n=-2b
mn=c


vertical distance = |f([FONT=宋体]m[/FONT])-f([FONT=宋体]n[/FONT])|


so m^3+3bm^2+3cm+d-(n^3+3bn^2+3cn+d)
[FONT=宋体]= [/FONT]m^3+3bm^2+3cm+d - n^3- 3bn^2- 3cn- d
[FONT=宋体]= [/FONT]m^3 - n^3 +3bm^2 - 3bn^2 +3cm- 3cn
= (m^3 - n^3) + 3B (m^2-n^2) + 3c(m-n)

------------------------------------------------
m^3 - n^3=(m-n)(m^2+n^2+mn)

m^2-n^2 = (m+n)(m-n)
------------------------------------------------

(m-n)^2=(m+n)^2-4mn
= 4b^2-4c
= 4 (b^2-c)
so (m-n)= 2 root (b^2-c)

(m^2+n^2+mn)= (m+n)^2-mn
= 4b^2-c

so |f([FONT=宋体]m[/FONT])-f([FONT=宋体]n[/FONT])|=| (2 root(b^2-c)) (4b^2-c) + 3b (-2b) (2 root(b^2-c)) + 3c (2 root(b^2-c)) |

= |(2 root(b^2-c)) (4b^2-c-6b^2+3c)|
= |(2 root(b^2-c)) -2(b^2-c)|
= 4 (b^2-c)^3/2 as required

please show that :
nC1+nC3+nC5+... = 2^(n-1)

 

[Sober]

nC1+nC3+nC5+... = 2^(n-1)

Consider the binomial expansion of 2^n = (1+1)^n whilst acknowldgeing this relationship:

nCn = nC0
nC(n-1) = nC1
nC(n-2) = nC2
etc...

If n is odd:

(1+1)^n = nC0 + nC1 + nC2 + nC3 + ... + nC(n-1) + nCn
(1+1)^n = nCn + nC1 + nC(n-2) + nC3 + ... + nC1 + nCn
(1+1)^n = 2(nC1 + nC3 + ... + nC(n-2) + nCn)

Therefore:
nC1 + nC3 + ... + nCn = 2^(n-1) // for odd n

If n is even:

Hmmm, this seems more difficult, perhaps somebody else can prove it for even n.

 

[Riviet]

Next Question:
In how many ways can 3 boys and 3 girls be seated in two rows of three chairs if each boy must be behind/in front of another girl?

 

[Sober]

Spoiler

If we seat the boys on the back row and the girls on the front row then n = 3!², but we can then reverse any of the three colums so multiply by 2³ and get 288.

Next Question

If y = (√[1 - x²])·sin^-1x, find dy/dx.
Express (1 - dy/dx)·(x^-1 - x) in terms of y.

 

[Riviet]

That's pretty intuitive stuff Sober. I did it a much longer way. XD

P.S Could you put your solution in spoilers? Thanks.

 

[followme]

Spoiler

If y = (√[1 - x²])·sin^-1x, find dy/dx.
Express (1 - dy/dx)·(x^-1 - x) in terms of y.


dy/dx=√(1 - x²)*1/√(1-x²) + sin^-1x * 1/2 (1-x²)^-1/2 * -2x
= 1- [ (xsin^-1x)/√(1 - x²) ]


(1 - dy/dx)*(x^-1 - x)= (1-1+[ (xsin^-1x)/√(1 - x²) ] )*(1/x - x)
= [ (xsin^-1x)/√(1 - x²) ] *(1/x - x)
= [ (xsin^-1x)/√(1 - x²) *1/x ]- [ (xsin^-1x)/√(1 - x²)*x ]
= [sin^-1x/√(1 - x²)] - [(x²sin^-1x)/√(1 - x²) ]
= [(sin^-1x)*(1-x²)] / √(1 - x²)
= √(1 - x²) *sin^-1x
= y


hopefully this is right...

 

[Sober]

Correct

 

[Riviet]

Next Question

a) Show that:

i)
d(secx.tanx) = 2sec³x - secx
dx

ii) 1+(secx.tanx)² = sec⁴x - sec²x + 1

b) Hence evaluate

pi/4
∫ 2sec³x - secx dx
0 sec⁴x - sec²x + 1

 

[STx]

Spoiler

a)

i) d(secx.tanx)/dx
= (secx).(sec²x)+(tanx)(secx.tanx)
= sec³x+secx.(sec²x-1)
= 2sec³x-secx #

ii) 1+(secx.tanx)²
= 1+ sec²x.tan²x
=1+ (sec²x).(sec²x-1)
=1+ sec⁴x-sec²x #

b)
pi/4
∫ 2sec³x - secxdx/sec⁴x - sec²x + 1
0

let u =secx.tanx, du=2sec³x-secx dx (shown in i)
when x=0, u=0, x=pi/4, u=root[2]


₀∫^root[2] du/1+u² (since 1+ sec⁴x-sec²x=1+(secx.tanx)²

= [tan^-1(u)] (0-->root[2])
= tan^-1(root[2])

 

[Riviet]

New question please!

 

[STx]

Next Question

 

[followme]

Spoiler

Ross: annual contribution: $3120 r= 1.0019231
sum= 3120(r^26)^9 + 3120(r^26)^8 + ... + 3120
= 3120 [ (r^26)^10 - 1]/ [ (r^26) - 1]
= $39467.44

Eve: 3120(r^26)^7 + 3120(r^26)^6 + ... + 3120
= 3120 [ (r^26)^8 - 1]/ [ (r^26) - 1]
=29923.67

29923.67(1.01^12)^2
= $37995.12

therefore Ross pays for the bday

i'll post a Q if my ans is rite. Got so confused while doing it...

 

[STx]

Spoiler

Ross: annual contribution: $3120 r= 1.0019231
sum= 3120(r^26)^9 + 3120(r^26)^8 + ... + 3120
= 3120 [ (r^26)^10 - 1]/ [ (r^26) - 1]
= $39467.44

Eve: 3120(r^26)^7 + 3120(r^26)^6 + ... + 3120
= 3120 [ (r^26)^8 - 1]/ [ (r^26) - 1]
=29923.67

29923.67(1.01^12)^2
= $37995.12

therefore Ross pays for the bday

i'll post a Q if my ans is rite. Got so confused while doing it...

Yes Ross pays for the bday but your calculations are a bit off

Spoiler

i) $40 508 is the amount Ross wil receive

ii) Difference: $40 508 - $38 997 = $1511 .'. Ross will pay for the bday

 

[followme]

Yes Ross pays for the bday but your calculations are a bit off

Mmm.. just wondering where did i go wrong?