functions problem

[tkt]

IF f:R->R, such that f(x+y)+f(x-y)=2f(x)f(y) and f(0) is not equal to 0, then f(x) is :

(A) even function
(B) odd function
(C) periodic function
(D) none

pls help me...

[MI]

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

putting x=y=0

2f(0) = 2 square(f(0))

Given f(0) != 0

f(0)=1

putting x=0 in topmost given equation

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

=> f is an even function

[tkt]

hmmm... got it .. thanks :) and another quest. :

f: R to Q such that f(3)=10 then f(x) is
(A) odd
(B) even
(C) increasing function
 (D) nothing can be said about it being even or odd

[MI]

f(x) is a constant function as its range is Q i.e. rational numbers

So its even function also.

[tkt]

ummm.. couldnt understand that... :/

[tkt]

ok.. this is a very silly question :p are all rational numbers even??

[MI]

Even function is f(x) = f(-x). It has nothing to do with even numbers.

[tkt]

ok.. got it.. i hope its right..
since f(3) = 10 or 3(square) =1,
which  means f(x)= x(square) + 1
 so f(x)= f(-x) in any case,
so f is even function.

and yeah i was confusing even nos. with even functions.. thanks a lot :)

[Sumit]

@MI: I don't think f(x) should be constant function only it can be f(x)=x^2+1 like taani mention...this function has domain=real no. and range=rational no..however answer will be same in both the case i.e even function...plz let me know If i'm wrong...

[tkt]

sumit, even i think f(x) wont be constant. but yaa, ans. remains the same.

[Sinistral]

What is the answer mentioned??

"one" of the ways in which we can construct such a function is:
f(x)= 3x+1, if x belongs to Q
     = 0     , elsewhere.

now, it wont be even or odd.
it definitely wont be continuous because we can always find an irrational no between any two rational numbers.
now we can construct specific examples wherein the function CAN become even (I cant right now think of a way to make such a function odd). Nevertheless, in general we cant comment whether it will ALWAYS be even or odd.

in this case it is increasing but then we can always make it decreasing (eg f(x)= -3x +19, x belongs to Q; = 0 e.w.).

so the only certain thing about a function whose co-domain is Q is, it is never continous. but its not an option.
the only feasible correct option then (according to me) should be (D) nothing can be said about it being even or odd

Please tell me if I missed something.

[Amit Goyal]

Problem:
f: R to Q such that f(3)=10 then f(x) is
(A) odd
(B) even
(C) increasing function
(D) nothing can be said about it being even or odd

Definitions: 
Let f(x) be a real-valued function of a real variable. Then f is even if the following equation holds for all x and -x in the domain of f:
f(x) = f(-x)

Let f(x) be a real-valued function of a real variable. Then f is odd if the following equation holds for all x and -x in the domain of f:
-f(x) = f(-x)

Let f(x) be a real-valued function of a real variable. Then f is increasing if the following holds for all x and x' in the domain of f:
x ≥ x'  implies f(x) ≥ f(x')

Solution:
Consider the following f: R to Q
f(x) =
-10 for x ≤ -3
20 for -3 < x < 3
10 for x ≥ 3

Note that f satisfies f(3) = 10,
f is not increasing because f(2) > f(3),
f is not odd because f(-2) ≠ -f(2),
f is not even because f(-3) ≠ f(3).
Thus, we rule out options (A), (B) and (C).

[MI]

Thanks.... I was making a mistake there

[tkt]

thanks sir and @sinistral, the answer mentioned there was option b.

[Amit Goyal]

Do the following problem:
f: R --> R with f(R)=Q (i.e. f takes on rational values) and f is continuous such that f(3)=10 then f(x) is
(A) odd
(B) even
(C) increasing function
(D) nothing can be said about it being even or odd

[wolverine]

Sir nothing can be said about it being even or odd.

[tkt]

sir,
since the function can be constructed in many ways like
f(x)=x(square) + 1, or,
f(x)=3x  + 1, or,
like as you said
f(x) =
 -10 for x ≤ -3
  20 for -3 < x < 3
 10 for x ≥ 3

so, it is not clear if any of the following condition holds true for all f(R)=Q
 f(x) = f(-x) [for even function] or
 f(-x) = -f(x) [for odd function] or
 f(x) ≥ f(x'), given x ≥ x' [for increasing function]

Therefore nothing can be said about it being even or odd.

[Sinistral]

the way you defined f got me confused. what I understand is f: R --> Q
In that case, f can be continuous only if it is a constant function ie if f(x) = k , x ∈ R & k ∈ Q.

So automatically f becomes an even function.
did I miss something?

this ques raised a (silly) doubt in my mind: are constant functions increasing/decreasing? (though they are definitely not strictly increasing/decreasing).?

[tkt]

hey, @ sinistral i think constant functions are neither decreasing nor increasing, because for every x , u get the same f(x). pls see sir's defination of increasing function. and i too have a very stupid question- are all rational numbers even ? coz thats what i can recapitulate from ur ans.

[duck]

Hi.. :)

Its given that a function is continuous and can take only rational values and f(3) =10
So, it must be the case that f(x) = 10 which is an even function.

@Sinistral:
Constant function is both increasing and decreasing.

@taani:
A rational number is any number that can be expressed as a fraction p/q of two integers, with the denominator q not equal to zero. Since q may be equal to 1, every integer is a rational number.
So, we have odd rational numbers also.
Eg: 3, 5, 7, ....