0

Guest

Courses New

A real valued functional equation f(x-y)=f(x)f(y)--(a-x)f(a+y), where is given constant and f(0)=1 then f(2a-x)=

A real valued functional equation f(x-y)=f(x)f(y)--(a-x)f(a+y), where is given constant and f(0)=1 then f(2a-x)=

Grade:11

1 Answers

Arun
25750 Points
5 years ago
Dear student
 
I have solved it in the following image, please check and in case of any difficulty please feel free to ask.
 
f(x-y) = f(x)f(y)-f(a-x)f(a+y)
y = 0
f(x-0) = f(x)f(0)-f(a-x)f(a+0)
f(0) = 1
f(x) = f(x)-f(a-x)f(a)
f(a-x)f(a) = 0
\Rightarrow f(a) = 0
f(2a-x) = f(a-(x-a)) = f(a)f(x-a)-f(a-a)f(a+x-a)f(2a-x) = f(a)f(x-a)-f(0)f(x)
f(a) = 0, f(0) = 1
f(2a-x) = -f(x)

Think You Can Provide A Better Answer ?

Other Related Questions on IIT JEE Entrance Exam

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More