..
@nick_baba centre will trace a hexagon. small hexagon side is 1. altitude will be rt(3)/2. so for bigger altitude will be rt(3)/2 + 1/2. found out side frm that and then perimeter. actually, the ans in this case comes out to be 9.46, nt 9.14 xactly. so mayb wrng. whats the OA?
@nick_baba said:A coin with diameter 1 cm rolls around the outside of a regular hexagon with edges of length 1 cm until it returns to its original position. In centimeters, what is the length of the path traced out by the centre of the coin?(1) 6 +pi/2(2) 12 + pi (3) 6 + pi (4) 12 + 2pi (5) 6 + 2pi
6+2pi
@gautham87 said:@krumThanks so much sir. But could you please explain the concept behind this ?I'm sorry if its really basic.. but I couldn't find solutions to this or concept explanation anywhere Also, pls explain what are the conditions in which such functions exist / don't exist
@krum bhai, concepts plz... link etc diya to bhi chalega...
@gautham87 said:@krumThanks so much sir. But could you please explain the concept behind this ?I'm sorry if its really basic.. but I couldn't find solutions to this or concept explanation anywhere Also, pls explain what are the conditions in which such functions exist / don't exist
@audiq7 said:@krum bhai, concepts plz... link etc diya to bhi chalega...
Set A has m elements Set B has n elements.
(1)
If m>n One to One function =0
If n>m One to One function = P(n,m)
(2)
Number of functions = n^m
Number of relations = 2^(m*n)
(3)
No of Many to One functions = Number of func- Number of One to One=n^m -nPm
(4)
If n>m Number of Onto functions =0
If m>n Number of Onto functions= sigma ( r = 1 to n) (-1)^(n-r) *C(n,r)*r^m
(1)
If m>n One to One function =0
If n>m One to One function = P(n,m)
(2)
Number of functions = n^m
Number of relations = 2^(m*n)
(3)
No of Many to One functions = Number of func- Number of One to One=n^m -nPm
(4)
If n>m Number of Onto functions =0
If m>n Number of Onto functions= sigma ( r = 1 to n) (-1)^(n-r) *C(n,r)*r^m
@nick_baba said:A coin with diameter 1 cm rolls around the outside of a regular hexagon with edges of length 1 cm until it returns to its original position. In centimeters, what is the length of the path traced out by the centre of the coin?(1) 6 +pi/2(2) 12 + pi (3) 6 + pi (4) 12 + 2pi (5) 6 + 2pi
6+pi ?
@nick_baba said:(1) 5/6 (2) 6/7 (3) 1 (4) 7/8 (5) none of thesebut i don think it'll be of any help anyways.....@ScareCrow28@adwaitjw@grkkrg try with them...
7/8?
@maddy2807 said:6+pi ?
@deedeedudu bhai ne diya hai...6+2pi a rah hai...edges pe kitne degree ka arc liya hai??
@nick_baba said:@deedeedudu bhai ne diya hai...6+2pi a rah hai...edges pe kitne degree ka arc liya hai??
120
@krum said:7/8?
@ScareCrow28 said:None of these..7/2??
yar jo OA diya hai..its 5/6...but ho skta hai its not right...u post ur approaches..
@maddy2807 said:oh yes. edges par 60 ka arc ayega. 6+2pi will be the final ans.
plz explain yar.. bouncer 
@audiq7 said:plz explain yar.. bouncer
total movement of the center will be.
perimeter of the hexagon. + movement of the coin at the edges.{ this will be an arc with 120 degree. i have given wrng.}
6+ 6*120/360*2*pi*1/2 [6 is the no edges of hexagon]
@audiq7 said:
Let f(f(x)) = 2f(x) €“ x for all real x. If f(f(f(f(f(f(7)))))) = 0 then f(7) = ??
@nick_baba said:(1) 5/6 (2) 6/7 (3) 1 (4) 7/8 (5) none of thesebut i don think it'll be of any help anyways.....@ScareCrow28@adwaitjw@grkkrg try with them...
5) none of these?
f(x) = (x - a)
f(f(x)) = x - 2a
2f(x) - x = x - 2a
f(f(f(f(f(f(7)))))) = 0
7 - 6a = 0
a = 7/6
f(7) = 7 - 7/6 = 35/6 ?
f(x) = (x - a)
f(f(x)) = x - 2a
2f(x) - x = x - 2a
f(f(f(f(f(f(7)))))) = 0
7 - 6a = 0
a = 7/6
f(7) = 7 - 7/6 = 35/6 ?
@krum 
Thanks so much sir.Awesome list.
Some doubts,
What is P(n,m) ?
What about the Bijective case ( do we just have to sum up one-one and onto) ?
@krum said:Set A has m elements Set B has n elements.(1)If m>n One to One function =0If n>m One to One function = P(n,m)(2)Number of functions = n^mNumber of relations = 2^(m*n)(3)No of Many to One functions = Number of func- Number of One to One=n^m -nPm(4)If n>m Number of Onto functions =0If m>n Number of Onto functions= sigma ( r = 1 to n) (-1)^(n-r) *C(n,r)*r^m
Thanks so much sir.Awesome list.
Some doubts,
What is P(n,m) ?
What about the Bijective case ( do we just have to sum up one-one and onto) ?
@gautham87 said:@krumThanks so much sir.Awesome list. Some doubts,What is P(n,m) ?What about the Bijective case ( do we just have to sum up one-one and onto) ?
P(n,m)=nPm, permutation
for bijective, n=m , and no. of bijective functions = n!
for bijective, n=m , and no. of bijective functions = n!