even........odd
3 posters
Science for Fun :: Level 1 , 2 and 3 :: JEE :: Maths
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even........odd
by sakura kinomoto Sat Nov 22, 2008 12:48 am
F(x) is a real valued function, satisfying f(x+y) + f(x-y) = 2f(x).f(y) for all y Є R. Then:
A) f(x) is an even function
B) f(x) is an odd function
C) F(x) is odd if f(0)=0
D) F(x) is even if f(0)=0
A) f(x) is an even function
B) f(x) is an odd function
C) F(x) is odd if f(0)=0
D) F(x) is even if f(0)=0
sakura kinomoto- Number of posts : 77
Age : 33
Location : Science City
points :
Registration date : 2008-11-09
sakura kinomoto- Number of posts : 77
Age : 33
Location : Science City
points :
Registration date : 2008-11-09
Infinity squared- Admin
- Number of posts : 270
Age : 33
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Registration date : 2008-10-03 -
Re: even........odd
by sakura kinomoto Thu Nov 27, 2008 10:34 pm
that's correct 
this question is concept+application.
all the answers are correct.
a,b,c,d
this question is concept+application.
all the answers are correct.
a,b,c,d
sakura kinomoto- Number of posts : 77
Age : 33
Location : Science City
points :
Registration date : 2008-11-09
Re: even........odd
by kabi Tue Dec 09, 2008 9:40 am
Hello
Here is my solution::
well by putting x=y=0.it will give f(0)=0,1 this condition is independent of the fact that func is even or odd. c and d are right.
put x=0 and y=-a
Taking f(0)=1
f(a)=f(-a)
Taking f(0)=0
f(a)=-f(-a)
Thus all the options are correct.
Thank u.
Here is my solution::
well by putting x=y=0.it will give f(0)=0,1 this condition is independent of the fact that func is even or odd. c and d are right.
put x=0 and y=-a
Taking f(0)=1
f(a)=f(-a)
Taking f(0)=0
f(a)=-f(-a)
Thus all the options are correct.
Thank u.
kabi- Number of posts : 63
points :
Registration date : 2008-12-07 -
Science for Fun :: Level 1 , 2 and 3 :: JEE :: Maths
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