Addmaths paper 1

[Ishrar Afrida]

radians man..

Many of my friends got 3.(something), I was sure that mine was wrong untill now

 

[Ishrar Afrida]

864 m.
324 m/s.

Damn I got distance right but speed I got 72.. -_-

 

[*Anonymous*]

The differential of y = 3 tan x - 2 was dy/dx = 3 sec^2 x, when you put x = 3pi/4, this becomes 6.
Gradient of normal = -1/6.
y + 5 = -1/6 (x-3pi/4)
6y + 30 = -x + 3pi/4 --> Eq. of normal.
Put x = 0
6y = 3pi/4 - 30
y = (3pi/4 - 30)/6 = -4.61

lol, i got the gradient of curve wrong so eventually..... all wrong...
and btw, can you explain the binomial question and trhe first trignometric one which was of 3 marks?

 

[Moiz Awan]

What was the answer to first question of kinematics part?

 

[Saad Mughal]

What was the answer to first question of kinematics part?

6 s.

 

[Bilal Khan]

864 m.
324 m/s.

speed was 72 m/s ..

 

[Moiz Awan]

Range of values for K?

 

[Moiz Awan]

Speed was 324m/s. You had to put distance equal to ZERO because it was at O.

 

[Snowysangel]

well , there was only one REAL root..
binomial one , i myslef got wrong..
hope i get method marks..
Coordinates of P were like (0 , 3.35)

Yeah I was getting 11/16 when I used that method..so I left it like that and plugged in calues from 0 till 10 in the x2 equation till I got the answer..th

 

[Saad Mughal]

lol, i got the gradient of curve wrong so eventually..... all wrong...
and btw, can you explain the binomial question and trhe first trignometric one which was of 3 marks?

25/4 = nC2 (1/4) - nC1(1/2)
25/4 = n(n-1)(n-2)!/2!(n-2)! (1/4) - n(n-1)!/1!(n-1)! (1/2)
25/4 = n^2 - n/8 - n/2
50 = n^2 - n - 4n
n^2 - 5n - 50 = 0
Simplifying gives n=10 and n=-5 (rej.).

Please give the question for the trigonometry one, I'll solve.

 

[Bilal Khan]

Range of values for K?

4<k<12

 

[Saad Mughal]

Range of values for K?

4<k<12

 

[Bilal Khan]

Speed was 324m/s. You had to put distance equal to ZERO because it was at O.

no , he said it was again at O..

 

[Saad Mughal]

no , he said it was again at O..

Yeah, again at O.
It was at O at t=0.
Then, s = 18t^2 - t^3.
When s = 0 (i.e. it is AGAIN at O).
t^3 = 18t^2
t = 18 s
v = -324 m/s.

 

[Snowysangel]

no , he said it was again at O..

So displacement was 0

 

[Bilal Khan]

Yeah, again at O.
It was at O at t=0.
Then, s = 18t^2 - t^3.
When s = 0 (i.e. it is AGAIN at O).
t^3 = 18t^2
t = 18 s
v = -324 m/s.

it means u got time = 2 sec..
i did speed = distance/time..

 

[Bilal Khan]

now i am in deep shit man..
u guz , got my answers wrong...

 

[Ishrar Afrida]

I think there will be no permutations in paper 2 and it will be full of vectors and circular measure.

 

[bilawalss]

What was the answer to permutations one ??

 

[Saad Mughal]

it means u got time = 2 sec..
i did speed = distance/time..

The DISTANCE was 0.

 

[Saad Mughal]

What was the answer to permutations one ??

360
60
36