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please solve it !!!!!!!!!!!!

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  • YES
    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
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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 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.
       
      Thanks in advance.
       
      regards,
       
      Avdhesh.
       


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    • mahathi chitta
      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
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        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.
         
        Thanks in advance.
         
        regards,
         
        Avdhesh.
         

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      • Kunal_Naik
        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
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          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.

           

          Thanks in advance.

           

          regards,

           

          Avdhesh.

           


          Did you know? You can CHAT without downloading messenger. Click here

           

           


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