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Differential Calculus
Comments (9)
put x= -x,
f(sin(-x)) = f(cos(-x)).
so, f(-Sinx)) = f(cosx)
we get, f(-Sinx)) = f(Sinx)
It implies, it can be true only when f(x) is constant, beacuse it's giving f(-1/2) =f(1/2) and f(-1) =f(1)
and so on...
Possible for constant function only. => f(x) =c.
@ Himanshu Jain
u got f(-x)= f(x)
f(x) is an even fn, not always a constant fn
Here i think f(x) is not a unique function
f(sin x)= f(sin (pi/2 - x))
implies f(x) is symmetric about sin pi/4 and also - sin pi/4 ( as f is even)
i.e. 1/root 2 and -1/root2
so f(x) is any even fn symm abt above pts. try finding examples
@Filokak: it's not f(-x) = f(x)
it's actually, f(g(-x)) = f(g(x)). So, you can't say, it's an even function.
Both are different things. Here, Sin and Cos can be treated as a different funtion.
Ok, try this way:
put x=x+pi
f(-Sinx) = f(-Cosx).
Also, we have, f(-Sinx) = f(Sinx) = f(Cosx)
=> f(Cosx) = f(-Cosx).
So, it can be said as constant function b'cos to satisfy the same values for Cos and Sin, there are some particular values are possible only.
@Himanshu Jain: Here both are satisfied i.e. f(g(-x))=f(g(x)) and f(-x)=f(x) .....
[u already have f(sinx)=f(-sinx)..............also note sin(-x)=sinx ...
Try to verify the fn below.....it clearly satisfies the given condition
f(x) = a*sin (pi x^2) + b
---- f(cosx) =a*[sin pi * (cosx)^2]+ b
=a*[sin pi( 1- (sinx)^2)] +b
=a*[sin{ pi - pi (sinx)^2}]+b
=a*[sin pi ( sin x)^2] +b
= f(sinx)
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A constant function? i.e. f(x) =c .Pls verify.