Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 44350 by Necxx last updated on 27/Sep/18

If f(x+y)=f(x).f(y) for real x,y  and f(0)≠0.Let F(x)=((f(x))/(1+(f(x))^2 ))  then F(x) is  a)even b)odd c)neither even nor odd

Iff(x+y)=f(x).f(y)forrealx,yandf(0)0.LetF(x)=f(x)1+(f(x))2thenF(x)isa)evenb)oddc)neitherevennorodd

Commented by maxmathsup by imad last updated on 27/Sep/18

⇒f(x).f(−x)=f(0) ⇒f(0)^2 =f(0) ⇒f(0) =1  ⇒F(0)=((f(0))/(1+(f(0))^2 )) =(1/2)  due to F(0)≠0 F can t be odd let see if F is even  F(−x) =((f(−x))/(1+(f(−x))^2 )) =((1/(f(x)))/(1+((1/(f(x))))^2 )) = (1/(f(x){((f^2 (x)+1)/(f^2 (x)))})) =((f(x))/(1+f^2 (x)))=F(x)  so  F is even.

f(x).f(x)=f(0)f(0)2=f(0)f(0)=1F(0)=f(0)1+(f(0))2=12duetoF(0)0FcantbeoddletseeifFisevenF(x)=f(x)1+(f(x))2=1f(x)1+(1f(x))2=1f(x){f2(x)+1f2(x)}=f(x)1+f2(x)=F(x)soFiseven.

Commented by Necxx last updated on 27/Sep/18

thank you sir

thankyousir

Commented by Necxx last updated on 27/Sep/18

that′s it

thatsit

Terms of Service

Privacy Policy

Contact: info@tinkutara.com