p2 june 2004,please help

I hae the paper but I don't have the answers please I need them if you know from where I can download them or if you can send the answers to me.. :rolleyes:

Reply 1

habosh

I still need this...pleaaaase anyone :frown:

Reply 2

Gaz031

habosh

I hae the paper but I don't have the answers please I need them if you know from where I can download them or if you can send the answers to me.. :rolleyes:

I don't think there are answers about. Hrm.
Here's my answers for the p2 June 2004 paper, there's nothing saying they are all correct though:

Q1:
Factorises to:
(x-3)(x-5)/(x-3)(x+3) x 2x(x+3)/(x-5)
=2x.

Q2i)
Need to find 1/cos2x, thus use:
cos2x = 1-2(sinx)^2. = 1 - 18/25 = 7/25
Hence sec2x = 1/7/25 = 25/7.
ii) cot2x + cosec2x = [1-(tanx)^2]/2tanx + 1/(2sinxcosx)
= 1/2tanx - tanx/2 + 1/2sinxcosx
= cosx/2sinx - sinx/2cosx + 1/2sinxcosx
= (cosx)(cosx)/(2sinx)(cosx) - (sinx)(sinx)/(2sinxcosx) + 1/2sinxcosx
= [(cosx)^2 - (sinx)^2 + 1]/2sinxcosx
= [ (cosx)^2 - 1 + (cosx)^2 + 1 ]/2sinxcosx
= 2cosx/2sinxcosx
= cosx/sinx
= cotx.

Q3a) [x^3 - 1/2x]^12
= [x^3(1- 1/2x^4)]^12
= x^36[1 - 1/2x^4]^12
= x^36 [ 1 - 12/2x^4 + 66/4x^8 - (220/8x^12) ]
= x^36 - 6x^32 + 16.5x^28 + 27.5x^24.
b) The term independent of x will be the term when x is cancelled - ie, the (1/2x^4) will cancel the x^36 - 36/4 = 9, hence it will be the ninth term.
x^36(-220/512x^36)
=-55/128.

Q4a) Square y and integrate, with limits of 4, 1 and multiply by pi:
Gives an answer of pi[19.5 + ln256]
b) At x=1, y=3, at x=4, y also =3
Hence both radiuses are the same, 3.
c) You need to multiply by a constant to go from the actual radius to the real radius:
3k=6
k=2,
but we are working in 3 dimensions, so you need to multiply by k^3=8
So the actual volume = 8[original volume]
=Approx 630cm^3.

Reply 3

Gaz031

Q5a)
x^3 + x^2 - 4x - 1 =0
x^3 + x^2 = 4x+1
x^2(x+1) = 4x+1
x^2 = (4x+1)/(x+1)
x = SQR[(4x+1)/(x+1)]

b)
x1=1
x2=1.58
x3=1.68
x4=1.70.

c)
f(1.695) = -0.0372
f(1.705) = 0.0435
Root lies between 1.695 and 1.705 = 1.70(2dp)

d)If x=-1 the iteration formulae produces a divide by zero - which goes to infinity and is not processable.

Q6)a) log(5)[x^2/y] = log(5)[x^2] - log(5)[y]
=2a - b
b) log(5)[25xrooty] = log(5)25 + log(5)x + 0.5log(5)y
= 2 + a + 0.5b

c) You hence have:
2a - b = 1
2 + a + 0.5b = 1
d)
You can solve the quadratics to get a = 0.25, b=-1.5.

e) log(5)X = -.25
x = 5^-.25 = 0.669

log(5)Y = -1.5
y= 5^-1.5 = 0.089.

Q7a) f(x) = x + (1/5)e^x
f'(x) = 1 + (1/5)e^x

b) Crosses y axis at A, at A x=0, y=1/5.
f'(x) = 2/5
y - (1/5) = (2/5)x
5y-1 = 2x
5y=2x+1

c) Under 1 you have 1.2424, under 1.5 you have 1.548, under 2 you have 1.865.
d) h=0.5
Estimation = 9.7176/4
=2.43 units^2.

Q8
a)

y=ln(3x-6)
x=ln(3y-6)
e^x = 3y-6
3y = e^x + 6
y = f^-1(x) = (1/3)(e^x) + 2

b) For the domain i put xER
For the range i put f^-1(x) > 2, as (1/3)e^x is always positive.

c) ln(3x-6) = 3
e^3 = 3x-6
3x = e^3 + 6
x = 8.70 (3sf)

d) Can't do on screen, you can plug some values to check it.
e) (5/3, 0), (7/3, 0) i think, without checking via a graph.

Hope that helped. Feel free to contest any answers.