Mathematics: Post your doubts here!

[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

 

[MustafaMotani]

Rutzaba i think u were lookin in wrong ms.. xD

 

[salvatore]

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!!

Oh yeah.. I think I got it.
Sin(1/2x) is half a cycle so it does not reach the negative axis! Don't tell me I'm wrong now

 

[Rutzaba]

Rutzaba i think u were lookin in wrong ms.. xD

lol sorry for wasting ur tym ^_^

 

[MustafaMotani]

lol sorry for wasting ur tym ^_^

anytime..

 

[daredevil]

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

wth!! :O
and how did they come up with S1=9 ???? Gosh i'm gonna kill those freakking examiners!!! :O

 

[Dug]

Oh yeah.. I think I got it.
Sin(1/2x) is half a cycle so it does not reach the negative axis! Don't tell me I'm wrong now

Exactly!! For trig functions, you have to think like that.. It doesn't go lower towards the -1. If you keep the shape, domain and period in mind, you will never be wrong. A simple sketch can also prove very helpful!

 

[syed1995]

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

You're freaking out

Yea that's the answers i got.. I just shifted the 2 of n/2 to the other side to make it easier to work with

i got

d=2
16=2a-d
16=2a-2
2a=18
a=9

Answer

 

[syed1995]

Damn.. so many replies within minutes! No fair.. taking screenshots takes time

 

[syed1995]

wth!! :O
and how did they come up with S1=9 ???? Gosh i'm gonna kill those freakking examiners!!! :O

They plugged in n=1 in the n^2 + 8n equation.