IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

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!!!!