If fx-y,fxfyand fx+y are in AP for all x,y and f0≠0, then
f3=f-3
f3+f-3=0
f'2+f'-2=0
f'2=f'-2
Explanation for the correct option:
Finding the relation:
Since, fx-y,fxfyand fx+y are in AP
fx-y+fx+y=2fxfy...1
Substitute x=0,y=0 in 1
2f0=2f0f0⇒f0f0-1=0⇒f0=1or0⇒f0=1[f(0)≠0]
Now, substitute x=0,y=x in 1
f-x+fx=2f0fx⇒f-x=fx...2[∵f(0)=1]
By differentiating 2 with respect to x.
f'-x-1=f'x⇒f'x+f'-x=0
Substitute x=2, we get
Hence, the correct option is C.
If x,y and z are in AP and tan-1x,tan-1yand tan-1z are also in AP, then