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• Dear all, Could someone help me out in solving this problem- Let f be a function defined on the set of integers. Assume that f satisfies the following
Message 1 of 3 , Jul 18, 2007
Dear all,

Could someone help me out in solving this problem-

Let "f" be a function defined on the set of integers. Assume that "f" satisfies the following properties-

f(0) ¹ 0
f(1) = 3
f(x) f(y) = f(x+y) + f(x-y)

for all integers x & y.

Q. What is the value of  f(3)
Options are 7, 18, 123, 322 & none of these.

Q. What is the value of f(7)
Options are 123, 322, 843, 1126 & none of these.

Please tell me the best way to solve such kind of questions.

regards,

Avdhesh.

• hi i hope this may help u...but m not sure whether this is d easiest way out..nyways when u put x=y=0..this gives eqn whose solution is f(0)= 0 or 2 as its
Message 2 of 3 , Aug 16, 2007
hi
i hope this may help u...but 'm not sure whether this is d easiest way out..nyways
when u put x=y=0..this gives eqn whose solution is f(0)= 0 or 2
as its already stated that its not 0..it is 2;

x=y
[ {f(x)}^2-2] =f(2x) ... * (I)

n
y=1
f(x+1)+f(x-1) =3f(x).... *(II)

f(1) = 3
f(2) = f(1)^2 - 2 =7                     .......  by I
f(3) = 3f(2) - f(1) = 18                 .......  by II
f(4) =  7^2  -  2= 47                   ......   by I
f(6) =  18^2 - 2 = 322                 ......   by I
f(8) =  47^2 - 2 = 2207               .......  by I
f(7) = 1/3 * (322 +2207) = 843   ......... by II

hence these are my solutions...correct me if i'm wrong..
MKC

YES <avdheshtomar@...> wrote:
Dear all,

Could someone help me out in solving this problem-

Let "f" be a function defined on the set of integers. Assume that "f" satisfies the following properties-

f(0) ¹ 0
f(1) = 3
f(x) f(y) = f(x+y) + f(x-y)

for all integers x & y.

Q. What is the value of  f(3)
Options are 7, 18, 123, 322 & none of these.

Q. What is the value of f(7)
Options are 123, 322, 843, 1126 & none of these.

Please tell me the best way to solve such kind of questions.

regards,

Avdhesh.

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• My approach is as follows: F(1,0) == f(1)*f(0) = f(1+0) + f(1-0) 3*f(0) = 2*3 == f(0) = 2 F(1,1) == f(1)*f(1) = f(1+1) + f(1-1) 3*3 = f(2) + 2 == f(2) = 7
Message 3 of 3 , Sep 9, 2007

My approach is as follows:

F(1,0) è f(1)*f(0) = f(1+0) + f(1-0)

3*f(0) = 2*3 è f(0) = 2

F(1,1) è f(1)*f(1) = f(1+1) + f(1-1)

3*3 = f(2) + 2 è f(2) = 7

F(2,1) è f(2)*f(1) = f(2+1) + f(2-1)

7*3 = f(3) + 3 è f(3) = 18

F(2,2) è f(2) * f(2) = f(2+2) + f(2-2)

7*7 = f(4) + 2     è f(4) = 47

F(4,3) è f(4) * f(3) = f(4+3) + f(4-3)

47*18 = f(7) + 3 è f(7) = 843

Regards,

Kunal Naik.

From: ascent4cat@yahoogroups.com [mailto:ascent4cat@yahoogroups.com] On Behalf Of mahathi chitta
Sent: Thursday, August 16, 2007 11:20 PM
To: ascent4cat@yahoogroups.com
Subject: Re: [2IIM CAT Prep] please solve it !!!!!!!!!!!!

hi

i hope this may help u...but 'm not sure whether this is d easiest way out..nyways

when u put x=y=0..this gives eqn whose solution is f(0)= 0 or 2

as its already stated that its not 0..it is 2;

x=y

[ {f(x)}^2-2] =f(2x) ... * (I)

n

y=1

f(x+1)+f(x-1) =3f(x).... *(II)

f(1) = 3

f(2) = f(1)^2 - 2 =7                     .......  by I

f(3) = 3f(2) - f(1) = 18                 .......  by II

f(4) =  7^2  -  2= 47                   ......   by I

f(6) =  18^2 - 2 = 322                 ......   by I

f(8) =  47^2 - 2 = 2207               .......  by I

f(7) = 1/3 * (322 +2207) = 843   ......... by II

hence these are my solutions... correct me if i'm wrong..

MKC

YES <avdheshtomar@ yahoo.co. in> wrote:

Dear all,

Could someone help me out in solving this problem-

Let "f" be a function defined on the set of integers. Assume that "f" satisfies the following properties-

f(0) ¹ 0

f(1) = 3

f(x) f(y) = f(x+y) + f(x-y)

for all integers x & y.

Q. What is the value of  f(3)

Options are 7, 18, 123, 322 & none of these.

Q. What is the value of f(7)

Options are 123, 322, 843, 1126 & none of these.

Please tell me the best way to solve such kind of questions.

regards,

Avdhesh.