Mathematics: Post your doubts here!

[Rutzaba]

i did this in a rather long way
ending up in a simultaneous eq bringing the right answers

 

[Dug]

Thanks a lot.. I have one more question.
What if the functions were:
f(x) = 3 - 2sinx for 0 ≤ x < pi
f(x) = 3 - 2sinx for 0 ≤ x < 2pi
Would there be any difference in the ranges? (Notice the domain has changed)..

Yes, it will be different!
For the first, 1 ≤ f(x) ≤ 3
For second, 1≤ f(x) ≤ 5

 

[syed1995]

i did this in a rather long way
ending up in a simultaneous eq bringing the right answers

That's how I solved it as well.. i think that's the correct way.

 

[salvatore]

I see you haven't understood my answer through PM's.
Anyways, the range won't change.

It would change.

there is no sin 270 here.. so no sinx=-1..

so the limits would be 3-2(0) and 3-2(1)
so
3 to 1 would be the limits.. instead of the 5 to 1 in the case of 0<x<2pi.

PANDA- I didn't wanna bother you with all those questions

Omg.. I'm even more confused now. Which is which?

 

[Rutzaba]

guys arithematic progression dont have 'r' they have d...
solve this q sumone

 

[MustafaMotani]

guys arithematic progression dont have 'r' they have d...
solve this q sumone

which.?

 

[Rutzaba]

which.?

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_12.pdf
7...
the ms shows opposite

 

[Rutzaba]

PANDA- I didn't wanna bother you with all those questions

Omg.. I'm even more confused now. Which is which?

relax... take a break and cum bak later. study sumthing other than math fr a whiile

 

[salvatore]

Yes, it will be different!
For the first, 3 ≤ f(x) < 1
For second, 1≤ f(x) ≤ 5

Alright.. thanks
So in summary, the domain of the function affects the min & max values while the period eg. sin(ax), sin(bx) etc will not affect it. Is that right?

 

[salvatore]

relax... take a break and cum bak later. study sumthing other than math fr a whiile

I'm freaking out for the exam!!

 

[MustafaMotani]

In general
Sn = n/2 (2a + (n-1)d)
therefore
n/2 (2a + (n-1)d) = n^2 + 8n

n/2 (2a + dn -d) = n^2 + 8n
na + n^2(d/2) - dn/2 = n^2 + 8n
Comparing...

d/2 = 1
d=2


a-d/2 = 8
a=9

are answers correct..? Rutzaba

 

[Rutzaba]

I'm freaking out for the exam!!

the more u freak the less u cn study... try doing nothing for complete ten minutes
and try to ignore the studies... then try studying with out even taking breATH for half AN hour ignoring the tension ^_^
i kno its not easy... but have a go... a lil practice will help

 

[Rutzaba]

In general
Sn = n/2 (2a + (n-1)d)
therefore
n/2 (2a + (n-1)d) = n^2 + 8n

n/2 (2a + dn -d) = n^2 + 8n
na + n^2(d/2) - dn/2 = n^2 + 8n
Comparing...

d/2 = 1
d=2


a-d/2 = 8
a=9

are answers correct..?

nhi a=32 and r where did r cum from? r= 0.75
although i remember bringing these values in ppr sum how

 

[MustafaMotani]

nhi a=32 and r where did r cum from? r= 0.75
although i remember bringing these values in ppr sum how

i have done part a
are us asking for part b..?

 

[Dug]

Alright.. thanks
So in summary, the domain of the function affects the min & max values while the period eg. sin(ax), sin(bx) etc will not affect it. Is that right?

I put wrong inequalities there!! I am so sorry :/
For the first, it's 1 ≤ f(x) ≤ 3


Period does affect the range!!
E.g
sin(x) will have a range -1 ≤ f(x) ≤ 1 in the domain 0 to 2π
sin(½x) will have a range 0 ≤ f(x) ≤ 1 in the same domain!!

 

[Rutzaba]

i have done part a
are us asking for part b..?

no dude part a
the ms is definitely wrong or may be the way theyy have done it is wrong

 

[MustafaMotani]

no dude part a
the ms is definitely wrong or may be the way theyy have done it is wrong

there is no r in part a... u must have seen wrong ..?? check it again...

 

[Rutzaba]

there is no r in part a... u must have seen wrong ..?? check it again...

that is wat i am saying.... lemme c

 

[MustafaMotani]

7 (a) S
n = n² + 8n.
S
1 = 9 → a = 9
S2 = 20 → a + d = 11 → d = 2
(or equating n² + 8n with S
n and comparing
coefficients)
B1
M1 A1
[3]
co
Realises that S
2 is a + (a + d). co
(b) ara =− 9
30 2
arar =+
Eliminates a → 010133
2
rr =−+
or → 081572
2
aa =+−
→ r = ⅔
→ a = 27

Marking scheme Rutzaba

 

[Rutzaba]

7 (a) S
n = n² + 8n.
S
1 = 9 → a = 9
S2 = 20 → a + d = 11 → d = 2
(or equating n² + 8n with S
n and comparing
coefficients)
B1
M1 A1
[3]
co
Realises that S
2 is a + (a + d). co
(b) ara =− 9
30 2
arar =+
Eliminates a → 010133
2
rr =−+
or → 081572
2
aa =+−
→ r = ⅔
→ a = 27

Marking scheme Rutzaba

yes you are right i was seeing the november answer sheet sorry