Maths, Addmaths and Statistics: Post your doubts here!

[asd]

i tr

i tried the same method as ashiqbal did

Im getting 180 finally, would you please tell the year of the paper ?

 

[HarisLatif]

2*6*5*4 = 240
is this the answer?

can you tell me what natural log 0 equals to?

 

[asd]

can you tell me what natural log 0 equals to?

natural log as in ln(0) right? it will be 1 :O

 

[HarisLatif]

Im getting 180 finally, would you please tell the year of the paper ?

Im getting 180 finally, would you please tell the year of the paper ?

yes it is from nov 2002 p2question 5 i can't see this question in the pastpapers. it is in the add maths classified

 

[asd]

natural log as in ln(0) right? it will be 1 :O

No, its not possible..

 

[HarisLatif]

natural log as in ln(0) right? it will be 1 :O

right ok so see this question 12 either first part http://www.xtremepapers.com/papers/...matics - Additional (4037)/4037_w11_qp_13.pdf and tell me what the range would be

 

[HarisLatif]

Plz man, help in the question above

is my answer right?

 

[ashiqbal]

HarisLatif or syed1995 or multixamza01, I need help here in this question (only the first part):
Given that y=kx^2 -4x +3k, express y in the form a(x-p)^2+q, where a, p and q are in terms of k. Hence find the value of k if the maximum value of y is 4.

are u sure it is value of k and not values of k?
anyway, see below and tell me:

 

[Haris Bin Zahid]

is my answer right?

Well......the answer in the book is a little different. Rest is right, but instead of 3k^2, it is only 3k.

 

[ashiqbal]

its 200

which year? which question? are u sure it 200?

We will do permutations oye

doesnt matter if we do Permutation or combination there.

 

[syed1995]

please solve this one.
how many different odd4 digit numbers less than 4000 can be formed from the digits 1,2,3,4,5,6,7 if no digit may be repeated. please hurry

Answer 200 ??

3 Choices for the first number .. Choose 2 out of remaining 5.... 4 choices for the last number.

First Number: 1,2,3
Last Number: 1,3,5,7
Middle 2 Numbers: 2 from the rest.

in case of 1 as first number

In case of 1> 1P1*5P2*3P1 <-- Since 1 was already used. 3 remaining to choose the last from.
In case of 2> 1P1*5P2*4P1 <-- Since all the odd numbers were available. 4 remaining to choose the last from.
In case of 3> 1P1*5P2*3P1 <-- Since 3 was already used. 3 remaining to choose the last from.

Multiply them all, then add all those possibilities

(1*5P2*3P1)+(1*5P2*3P1)+(1*5P2*4P1) = 200 possible digits can be formed.

 

[HarisLatif]

No, its not possible..

and the range in OR part is making me confuse too. so clear me if you understand that one

 

[Haris Bin Zahid]

are u sure it is value of k and not values of k?
anyway, see below and tell me:
View attachment 11591

Thnx a lot. But why do we get 2 values of k?

 

[asd]

Answer 200 ??

3 Choices for the first number .. Choose 2 out of remaining 5 4 choices for the last number.

in case of 1 as first number

In case of 1> 1P1*5P2*3P1 <-- Since 1 was already used. 3 remaining to choose the last from.
In case of 2> 1P1*5P2*4P1 <-- Since all the odd numbers were available. 4 remaining to choose the last from.
In case of 3> 1P1*5P2*3P1 <-- Since 3 was already used. 3 remaining to choose the last from.

Multiply them all, then add all those possibilities

(1*5P2*3P1)+(1*5P2*3P1)+(1*5P2*4P1) = 200 possible digits can be formed.

you are good.

 

[Haris Bin Zahid]

are u sure it is value of k and not values of k?
anyway, see below and tell me:
View attachment 11591

I'm dead sure, it's Exercis 4.1 Question 17 In "New Additional Mathematic" book!

 

[ashiqbal]

Answer 200 ??

3 Choices for the first number .. Choose 2 out of remaining 5 4 choices for the last number.

in case of 1 as first number

In case of 1> 1P1*5P2*3P1 <-- Since 1 was already used. 3 remaining to choose the last from.
In case of 2> 1P1*5P2*4P1 <-- Since all the odd numbers were available. 4 remaining to choose the last from.
In case of 3> 1P1*5P2*3P1 <-- Since 3 was already used. 3 remaining to choose the last from.

Multiply them all, then add all those possibilities

(1*5P2*3P1)+(1*5P2*3P1)+(1*5P2*4P1) = 200 possible digits can be formed.

kk! almost forgot that 5P2. instead was using 2!

 

[syed1995]

you are good.

the only 2 topics i am good at are permutation and circular measure, because i have practiced them like hell!

 

[syed1995]

kk! almost forgot that 5P2. instead was using 2!

dude 5 digits mein say 2 lenay hain .. 2! kahan say aagaya ?

 

[ashiqbal]

Thnx a lot. But why do we get 2 values of k?

I'm dead sure, it's Exercis 4.1 Question 17 In "New Additional Mathematic" book!

ok! found out the reason!
y has a maximum value, right? so the coeffecient of (x-h)^2 must be negative, remember? so we will reject k = 2 as k is the coeffecient of (x-h)^2.

 

[HarisLatif]

Answer 200 ??

3 Choices for the first number .. Choose 2 out of remaining 5.... 4 choices for the last number.

First Number: 1,2,3
Last Number: 1,3,5,7
Middle 2 Numbers: 2 from the rest.

in case of 1 as first number

In case of 1> 1P1*5P2*3P1 <-- Since 1 was already used. 3 remaining to choose the last from.
In case of 2> 1P1*5P2*4P1 <-- Since all the odd numbers were available. 4 remaining to choose the last from.
In case of 3> 1P1*5P2*3P1 <-- Since 3 was already used. 3 remaining to choose the last from.

Multiply them all, then add all those possibilities

(1*5P2*3P1)+(1*5P2*3P1)+(1*5P2*4P1) = 200 possible digits can be formed.

but for selecting the first number we'll right like this 3C1 as we can chose 1 digit from the three as first digit?