Additional Math

[Jezla]

pLs explain the answer for this Q,
June 2007 ppr1 Q6

 

[Aathi]

I dont know ....i tried it with Jezla.

 

[Jezla]

Aachi really, i noe we tried it 2gthr and u ddnt know. no one seemz to know it here lets ash shenaz 2mrw!!!!!!!!

 

[usman]

(i) let t be the time taken b4 the impact (although it's already given as 1.8s but we've 2 show dis)
Now,
12j + t (15j) = (t - 0.5) 30j
12 + 15t = 30t - 15
15t = 27
t = 1.8s (shown)

(ii) 46i + 1.3 (ki + 30j) = 12j + 1.8 (40i + 15j)
46i + 1.3ki +39j = 12j + 72i + 27j
1.3ki + 46i = 72i
1.3ki = 26i
k = 20 (EUREKA)

may b my working is really difficult 2 understand. but try solving da question using my method, u'll get to the answer at last . dats 4 sure

the marking scheme is much easier. if u dun hav it dats here:
6. (i) In 1.8s , alien goes 27 cm up.
In 1.3 s missile goes 39 up.
But alien starts at 12 up.
→ 39 – 27 = 12
(ii) In 1.8s. alien goes 72 across
In 1.3 s, missile goes 1.3k
72 = 1.3k + 46 → k = 20.

 

[Jezla]

yeah itz hard to understand specially (ii) but hey thanx 4 your help!!

 

[iamjusme]

hmmm wow add math and all.................

 

[alisha]

HEY JEZBUG!!!!!!!!!!!!!!!!!!! wow. ad math!!! i hav a doubt in the oct/nov 2005 paper 1 question 8 and paper 2 question 5. well the list is much longer bt i want these 1st.

 

[PlanetMaster]

Oct/Nov 2005 Paper 01 Question 8:
y = (x+2) √(x-1)

u = (x+2)
u' = 1
v = √(x-1)
v' = 1 / (2√(x-1))

dy/dx = uv' + vu'
dy/dx = [(x+2) / 2√(x-1)] + [√(x-1)]
dy/dx = [x+2+2(x-1)] / 2√(x-1)
dy/dx = 3x / 2√(x-1)
Therefore k=3/2

For rest parts why don't you look up in marking scheme for the solution and lemme know where exactly you don't understand.

 

[usman]

dese are just da basics of calculus . in dis particular question u must be knowing atleast da product rule i.e.
d/dx (uv) = u . dv/dx + v . du/dx
& also dis simple rule
d/dx [sqrt(u)] = 1/[2 . sqrt(u)] . du/dx

 

[alisha]

ok...i gt it. i knw the product rule bt i gt confused after that. as for the integration we just need to get 2/3 of y rite? ya i gt it. thanx. and duz any1 knw an easy method for oct/nov 2005 paper 2 question 5?

 

[Zohaib Sherazi]

(1 + px)ⁿ = nC0(1)ⁿ(px)⁰ + nC1(1)ⁿ⁻¹(px)¹ + nC2(1)ⁿ⁻²(px)² + nC3(1)ⁿ⁻³(px)³ + . . .
(1 + px)ⁿ = nC0 + nC1 (px) + nC2 (p²x²) + nC3 (p³x³) + . . .

where nCr = "n choose r" = n!/(r! (n-r)!)

Matching each term, we get:
nC0 + nC1 (px) + nC2 (p²x²) + nC3 (p³x³) + . . . = 1 - 12x + 28p²x² + qx³ + . . .

----------

nC1 px = -12x
n px = -12x
np = -12 . . . . . . . [1]

----------

nC2 (p²x²) = 28p²x²
n(n-1)/2 = 28
(n²-n) = 56
n² - n - 56 = 0
(n+7)(n-8) = 0
n = -7 or 8 . . . . . but n > 0
n = 8

----------

Replacing n into equation [1] we get:
np = -12
8p = -12
p = -12/8
p = -3/2

----------

nC3 (p³x³) = qx³
8C3 p³x³ = qx³
q = 8C3 p³
q = 56 p³ = 56 (-3/2)³ = 56(-27/8)
q = -189

 

[alisha]

OMG! itz so easy! thanx alot!

 

[alisha]

ok................i 4gt everythin and nw im confused!

hw can we change nC2 into n(n-1)/2 ?

nC2= n!/(n-2)!2! rite????????

 

[M.Usman Khan]

n! = n(n-1)(n-2)!

v then cancel (n-2)! with the (n-2)! in the denominator

 

[alisha]

ok...............thanx.


nw these:

O/N/05 paper 02 Q10 part ii) and Q11 -b

M/J/06 paper 02 Q8 part ii)

O/N/o6 paper 01 Q6

 

[Jezla]

Nov 2006 ppr1 Q6 Ans:
(2-x/2)^6 Expand only for x and x^2 coz u'll be needin only those 2 values..
...6C1*2^5(-x/2)+6C2*2^4*(-x/2)^2......
....-96x+60x^2....
(k+x)(...-96x+60x^2...)
=-96kx+60kx^2-96x^2+60x^3
Then,
60k-96=48(as said in the Q)
60k =180
k =3

 

[alisha]

thanx jez....................bt i gez no 1 knwz d rest

hw disappointin!

 

[Jezla]

too lazy to type men!!
Come early 2mrw! I'll teach u!