3 Unit Revising Marathon HSC '10 (1 Viewer)

[jet]

OK, now that 2010 is here I am reviving the old Revision Marathon game.

Rules:
The OP posts a question, first person to get it correct asks the next question, and so on.

First Question


*Waits for person to shoot him*

 

[xV1P3R]

 

[cutemouse]

d/dx a^x = x.lna

Question: Find the volume generated when the area between the curve , the line x=16 and the x axis is rotated one revolution about the y axis.

Edit:

 

[addikaye03]

d/dx a^x = x.lna

Question: Find the volume generated when the area between the curve , the line x=16 and the x axis is rotated one revolution about the y axis.

d/dx (a^x)= lna . a^x

Spoiler

d/dx(a^x)=d/dx(e^(lna^x))=> d/dx (e^xlna)

therefore e^(xlna) d/dx (xlna) = a^xln(a)

y=3+rt(x) => rt(x)=y-3 therfore x^2=(y-3)^4

V=pi int (b->a) x^2 dy [remember to change limits]

=pi int (7->3) (y-3)^4 dy

=pi [1/5(y-3)^5] (7-->3)

=204.8 pi

I think that's right. Don't have calculator on hand lol

Question: If 4tan(a-b)=3tan(a), prove that tan(b)=sin2(a)/(7+cos2(a))

 

[cutemouse]

d/dx (a^x)= lna . a^x

Whoops... hahah... How could I forget? XD

d/dx (a^x)= lna . a^x
y=3+rt(x) => rt(x)=y-3 therfore x^2=(y-3)^4

V=pi int (b->a) x^2 dy [remember to change limits]

=pi int (7->3) (y-3)^4 dy

=pi [1/5(y-3)^5] (7-->3)

=204.8 pi

I think that's right. Don't have calculator on hand lol

What you have found is the volume generated when area enclosed between the curve, y=7 and the Y AXIS is rotated about the y axis... RTQ...

 

[addikaye03]

Question: If 4tan(a-b)=3tan(a), prove that tan(b)=sin2(a)/(7+cos2(a))

.

 

[shaon0]

.

4tan(a-b)=4((tan(a)-tan(b))/(1+tan(a)tan(b))
4tan(a)-4tan(b)=3(tan(a))^2.tan(b)+3tan(a)
tan(a)/(4+3(tan(a))^2)=tan(b)

sin(2a)/(7+cos(2a))=2sin(a)cos(a)/(2(cos(a))^2+6)
=sin(a)cos(a)/((cos(a))^2+3)
=tan(a)/((sec(a))^2+3)
=tan(a)/(4+3(tan(a))^2)
=tan(b)

 

[Drongoski]

 

[cutemouse]

 

[shaon0]

Let S(n) be the statement defined such that S(n): ln(n!)>n for {n E Z: n>=6}

Skipping base case where n=6

Let n=k:
Assume; ln(k!)>k

Let n=k+1:
LHS=ln((k+1)!)
=ln(k+1)+ln(k!)
>ln(k+1)+k
>k+1 (As ln(k+1)>1)

b) n!>e^n
1/n! <1/e^n

c) (1/1!+...+1/5!+(1/6!+...))=(1/1!+...+1/5!+(1/e^6+...))
1/1!+...=103/60+S{inf}
where s{inf}=(1/e^6)/(1-1/e)=1/e^5(e-1)
1/1!+...=103/60+1/e^5(e-1)

 

[Shadowdude]

I'm gonna wait here until I can see a question I can possibly attempt =P

That being said, anyone going to post the next question?

 

[shaon0]

I'm gonna wait here until I can see a question I can possibly attempt =P

That being said, anyone going to post the next question?

Same here, I'm waiting for another question.

 

[cutemouse]

 

[shaon0]

S (x-1)^n dx = S {sum} (nk) x^(n-k)(-1)^k dx
(x-1)^(n+1)/(n+1) = {sum} (nk) x^(n-k+1-1)^k/(n-k+1) +C
At x=0:
C=(-1)^(n+1)/(n+1)
(x-1)^(n+1)/(n+1) = {sum} (nk) x^(n-k+1-1)^k/(n-k+1)+(-1)^(n+1)/(n+1)

n(n-1)(x-1)^(n-2)={sum}(nk) (n-k)(n-k-1)(-1)^k.(x-1)^(n-k-2)
k=n-2,x=2:
n(n-1)=(n n-2).2.1
n(n-1)=(n n-2)*2

 

[cutemouse]

Given the product rule of differentiation, where u and v are functions of x, explain why where u, v and w are all functions of x.

 

[shaon0]

d/dx(dy/dx)=d/dx(uv'+u'v)
=u'v'+uv"+u"v+v'u'
y"=2u'v'+u"v+v"u

d/dx(uvw)=u'vw+uv'w+uvw'

Can't see how w variable somes into y'

 

[untouchablecuz]

d/dx(dy/dx)=d/dx(uv'+u'v)
=u'v'+uv"+u"v+v'u'
y"=2u'v'+u"v+v"u

d/dx(uvw)=u'vw+uv'w+uvw'

Can't see how w variable somes into y'

i think he means:

 

[xV1P3R]

 

[Dragonmaster262]

1

 

[Drongoski]

zero