Practice paper - Difficult, Answers needed. (1 Viewer)

[forstudents]

HEy guys this is quite a difficult practice paper, see if u can get the answers and post them up. Some questions are really challenging. Tell me wat u think.

Alrite guys, lets put our soloutions together and upload this to resources.

 

[Riviet]

I'll start us off:

Q3

(c) (i) Let p(x) = x³-15x+4
p(-4)=-64+60+4=0
.'. x=-4 is the other root.

(ii) Let the other two roots be a and 1/a.
Sum of roots = a + 1/a - 4 = 0
a² - 4a + 1 = 0
(a-2)² = 3
a=2+sqrt3
.'. two other roots are x=2+sqrt3 and x=2-sqrt3

1/(2+sqrt3)=(2-sqrt3)/(4-3) by rationalising denominator
=2-sqrt3
.'. 2+sqrt3 and 2-sqrt3 are reciprocals.

 

[mrzeidan1]

I'll do the easy ones

Question 1 b)

A (-4,2) B(2,-1) P(x,y) ratio 5:2
since external division, take ratio as -5:2

(x,y) = (mx₁+nx₂) / (m+n) , (my₁+ny₂) / (m+n)
(x,y) = [ (-5)(-4)+(2)(2) ] / -3 , [ (-5)(2)+(2)(-1) ] / -3
(x,y) = (24/-3 , -12/-3)
(x,y) = (-8,4)
P = (-8,4)

Question 1 c)

(x-1)^2 < 4(x-1)
x^2 - 2x + 1 < 4x - 4
x^2 - 6x + 5 < 0
(x-5)(x-1) < 0
1<x<5

Question 1 e)

let t = tan @/2

=1- [(1-t^2) / (1+t^2)] / 1+ [(1-t^2) / (1+t^2)]
=(1+t^2 -1 + t^2) / (1+t^2) / (1+t^2 -1 - t^2) / (1+t^2)
=(1+t^2 -1 + t^2) / (1+t^2 -1 - t^2)
=2t^2/2
=t^2

I dont know how to use all the fancy formatting .. hope im right

Edit: added question 1 e)

 

[SoulSearcher]

Didn't have to expand the last one.
(x-1)² < 4(x-1)
(x-1)² - 4(x-1) < 0
(x-1)[x-1-4] < 0
(x-1)(x-5) < 0
1 < x < 5

 

[mrzeidan1]

Didn't have to expand the last one.
(x-1)² < 4(x-1)
(x-1)² - 4(x-1) < 0
(x-1)[x-1-4] < 0
(x-1)(x-5) < 0
1 < x < 5

i see

 

[Rax]

Are there any certain questions you would like us to put up answers too, I may have a dig at that projectile one right now

 

[forstudents]

yeh guys, help me out with question 4b, 2c and all of 6, for now, thanx alot

 

[haque]

ist part of q7 is easy enough so i'll go to the next part
X=V^2sin(2@)/g
X'=2V^2cos(2@)/g times d@/dt (curtesy of the chain rule)
but w=d@/dt so the answer is proved.
at @=pi/6 X' is positive so X is increasing, at @=pi/3 X' is negative so X is decreasing(remember we are comparing this to time-i.e how it changes as @ varies with time)

X'^2=4w^2V^4cos^2(2@)/g^2 =4w(V^4/g^2 times(1-sin^2(2@)))=4w(V^4/g^2 -V^4sin^2(2@)/g^2) =4w(V^4/g^2 -X^2) the siginificance of this is the fact that when we take the square root,the absolute value of X must be less than V^2/g^2 so in otherr words this is the maximum range of values X can take. i know my responses are brief but u'll get the idea jst by thinking bout them.

 

[followme]

Q2
a) e^x=4x+8
let f(x)=e^x-4x-8
f'(x)=e^x-4
3-[(e³-12-8)/(e³-4)]
=2.9947 (4dp)
b)
i) asin(x-α)=asinxcosα-acosxsinα
acosα=1
asinα=√3
a=2
α=pi/3
sinx-√3cosx=2sin(x-pi/3)

ii) 2sin(x-pi/3)=2/√2
sin(x-pi/3)=1/√2
x-pi/3=pi(n)+(-1)ⁿ (pi/4)
x=pi(n)+(-1)ⁿ (pi/4)+pi/3 where n is an integer
c) 4 males 5 females group of 3
⁴C₁x⁵C₂+⁴C₂X⁵C₁=70

Q3
a) y=0
tan^-1(ax+b)=0
ax+b=0
x=-b/a
dy/dx = a/[1+(ax+b)²]
at x=-b/a, m_tangent=a/(1+0)=a
so y-0=a(x+b/a)
y=ax+b

 

[Riviet]

Q4 (b) (i)

1. 3⁵=243

2.
(α) p=(1/3)³(2/3)=2/81

(β) p=1.(1/3)²(2/3)=2/27

(γ) p=(1/3)³(2/3)².⁵C₃=40/243

(ii) p= p(3 lights the same colour)=(⁵C₃.3)/243=30/243

.'. p(3 lights the same colour on two OR three occasions)=p(2 occasions) + p(3 occasions)
=(30/243)².(1 - 30/243) + (30/243)³
=0.015 (3dp)

 

[little master]

for the binomial one:
i am guessin this one

5Ck 1 ^5-k (x+x^2)^k
5C0+5C1 (x+x^2) + 5C2(x+x^2)^2 + 5C3 (x+x^2)^3 +5C4 (x+x^2)^4 +5C5(x+x^2)^5

1+5C1 (x+x^2) + 5C2(x+x^2)^2 + 5C3 (x+x^2)^3 +5C4 (x+x^2)^4 +(x+x^2)^5

is this the right way??

 

[SoulSearcher]

[1+(x+x²)]⁵
= 1 + ⁵C₁(x+x²) + ⁵C₂(x+x²)² + ⁵C₃(x+x²)³ ...
= 1 + 5(x+x²) + 10(x²+2x³+x⁴) + 20(x³+...)
= 1 + 5x + 5x² + 10x² + 20x³ + 20x³ + ...
= 1 + 5x + 15x² + 40x³ + ...

 

[Riviet]

little master - The important thing to note from SoulSearcher's solution is that he only expanded terms up to the ones in x³, which is what the question asked.

Q1 (a)

Integral = [ln(x+sqrt(x²+9)]⁴₀
= ln(4+5) - ln3
= ln3

 

[forstudents]

guys for 1d i got:
-1/60, tell me if thats right. mt working out is scrambled a bit but its below.

 

[little master]

little master - The important thing to note from SoulSearcher's solution is that he only expanded terms up to the ones in x³, which is what the question asked.

oh alrite ...i guess i ignored tht part; was on the right track though. thx for pointing tht out Rivet.

 

[Riviet]

forstudents - your solution to 1 (d) looks right to me.

 

[acmilan]

I'm tired and my eyes are half shut, but ill give 6 a go:

6(b)(iii)
Area between f(x) and f^-1(x) = 2*Area between f(x) and the line y = x

Since f(x) and y = x are odd, the area on the positive and negative sides of the x-axis are equal, so:

Area between f(x) and f^-1(x)
= 2*Area between f(x) and the line y = x
= 2*2 int (0 to 1) 2x/(x²+1) - x dx
= 4 [ln(x² + 1) - x²/2]¹₀
= 4ln 2 - 2

(iv)
int (0 to 1) f^-1(x)
= int (0 to 1) f(x) - Area between f(x) and f^-1(x)
= int (0 to 1) 2x/(x²+1) dx - (2ln2 - 1)
= [ln(x²+1)]¹₀ - (2ln2 - 1)
= 1 - ln2

(v) If f^-1(x) = 1/2 then f(f^-1(x)) = f(1/2)

So x = f(1/2) = 2*(1/2)/((1/2)²+1) = 4/5

 

[muttiah]

Didn't have to expand the last one.
(x-1)² < 4(x-1)
(x-1)² - 4(x-1) < 0
(x-1)[x-1-4] < 0
(x-1)(x-5) < 0
1 < x < 5

nice.. thats good.. cant watse time expanding in 3 unit..

 

[followme]

guys for 1d i got:
-1/60, tell me if thats right. mt working out is scrambled a bit but its below.

0.5 --> 0
0 --> 1
⁰∫₁(1-x)(x)⁴ (-1/2) dx
= (1/2) ¹∫₀(1-x)(x)⁴dx
answer is +1/60

 

[lil_a2n_boi]

0.5 --> 0
0 --> 1
⁰∫₁(1-x)(x)⁴ (-1/2) dx
= (1/2) ¹∫₀(1-x)(x)⁴dx
answer is +1/60

Why did your limits change position?