IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

kislay

let f(x+y) +f(x-y)=2f(x).f(y),for all x,y belonging to real no.s
and f(0)=K; then
a
f is even if K =1
b
f is odd if K=0
c
f is always odd
d
f is neither odd nor even for any value of K

ans given is c

ramyadiamond

Given,

f(x+y) +f(x-y) =2f(x).f(y)

Differenciating both sides, w.r.t x, we get

f^'(x+y) (1+y^') +f^'(x-y)(1-y^') = 2 f '(x).f(y) +2f(x).f '(y) .y'

Now since y is independent of x at any value, hence dy/dx will be zero.

Hence the above expression becomes,

f '(x+y) +f'(x-y) = 2f '(x).f(y)

now put x=0, and y=0 in the eqn

f '(0) + f '(0) = 2f '(0).f(0)

2f '(0) =2f '(0) .K

therefore, K=0

Now in

f(x+y) +f(x-y) =2f(x).f(y)

put, x=0, and u get

f(y) +f(-y) = 2f(0).f(y)

f(y) +f(-y) = 2 K f(y)

f(y) +f(-y) =0

Hence the given function is always odd for all real no.s

Cheers!!

abhujabal

Ramya was going the right till she made a small mistake.....and went the wrong way....

f(x+y) +f(x-y) =2f(x).f(y)

Differenciating both sides, w.r.t x, we get

f '(x+y) (1+y') +f '(x-y)(1-y') = 2 f '(x).f(y) +2f(x).f '(y) .y'

y is independent of x at any value, hence dy/dx will be zero.

Hence the above expression becomes,

f '(x+y) +f'(x-y) = 2f '(x).f(y)

now put x=0, and y=0 in the eqn

f '(0) + f '(0) = 2f '(0).f(0)

2f '(0) =2f '(0) .K

f(0)=K

hence , f'(0)=0

therefore,2f'(0)[K-1] =0

K=1 (This was where she made mistake)

Now in

f(x+y) +f(x-y) =2f(x).f(y)

put, x=0, and u get

f(y) +f(-y) = 2f(0).f(y)

f(y) +f(-y) = 2 K f(y)

f(y) +f(-y) = 2f(y)

Now putting -y instead we have

f(-y) +f(y) = 2f(-y)

Hence f(y)=f(-y)

Hence the given function is always even for all real nos.

I hope now I am correct!!!!

ramyadiamond

oopsie, yes, i went wrong there. Thanks for correcting!!

kislay

but yaar ans. is c ........

bvsatyaram

Hi All,
Even Abhujaba is wrong

------------------
The Mistake
------------------
If ab=ac then b=c. this is wrong.
If ab=ac, then either b=c or a=0.
So, when u ended up with 2f '(0) =2f '(0) .K
we have, either f '(0)=0 or K=0
So, 2 cases arise
------------------
End of mistake
------------------

A better solution is as follows:
f(x+y) +f(x-y) =2f(x).f(y)
put x=0 and y=0
so we get, K+K=2*K^2
There fore, either K=0 or K=1 ..................................( 1 )

Now put x=0 in the given equation. U get,
f(y) +f(-y) = 2 K f(y)

If K=0, then the function is odd.
If K=0, then the funstion is even.
So, the answer is A and B

bvsatyaram

A mistake in "The Mistake"

------------------
The Mistake
------------------
If ab=ac then b=c. this is wrong.
If ab=ac, then either b=c or a=0.
So, when u ended up with 2f '(0) =2f '(0) .K
we have, either f '(0)=0 or K=1
So, 2 cases arise
------------------
End of mistake
------------------

ayshwarya

hey y ve u done such a big process yaaron its so simple wen v find d value of f[1] by putting x=1,y=0 v get f(0)=1 only hence v must take f(0)=1 only hence by solving v wud get f(n)=2f(n-1)^2-1 which is an odd hence f is an odd funcn

ayshwarya

im srry its not odd fn but it must b always odd
printing mistake rate me if im writ plz

Ask & Discuss Questions with Community & Experts