# A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to ?

Jitender Singh IIT Delhi
8 years ago
Ans:
$f(x-y) = f(x)f(y)-f(a-x)f(a+y)$
$y = 0$
$f(x-0) = f(x)f(0)-f(a-x)f(a+0)$
$f(0) = 1$
$f(x) = f(x)-f(a-x)f(a)$
$f(a-x)f(a) = 0$
$\Rightarrow f(a) = 0$
$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)$
$f(a) = 0, f(0) = 1$
$f(2a-x) = -f(x)$
Thanks & Regards
Jitender Singh
IIT Delhi
ritika
21 Points
4 years ago
f(x-y) = f(x)f(y) – f(a-x) f(a+y)   …...(1)
Replace y = -x
(1) becomes:   f(2x) = f(x) f(-x) – f(a-x) f(a-x)
f(2x) = f(x) f(-x) – f^2(a-x)   …..(2)
Now, put x = 0
(2) becomes:   f(0) = f(0) f(0) – f^2(a)
f(0) = f^2(0) –  f^2(a)
1 = 1 – f^2(a)   …...(Given that f(0) = 1)
f^2(a) = 0
f(a) = 0   …..(3)
f(2a-x) = f(2a) – f(x)
= f(a+a) – f(x)
= f(a) + f(a) – f(x)
= 0 – f(x)   …......(from (3))
= – f(x)
Yash Chourasiya
3 years ago
Hello Student

Please see the solution in the attachment.

Vikas TU
14149 Points
2 years ago
Dear student
Please refer the clear image of this solution
Good Luck
Cheers
ankit singh
2 years ago
f(2x) = f(x) f(-x) – f(a-x) f(a-x)
f(2x) = f(x) f(-x) – f^2(a-x)   …..(2)
Now, put x = 0
(2) becomes:   f(0) = f(0) f(0) – f^2(a)
f(0) = f^2(0) –  f^2(a)
1 = 1 – f^2(a)   …...(Given that f(0) = 1)
f^2(a) = 0
f(a) = 0   …..(3)
f(2a-x) = f(2a) – f(x)
= f(a+a) – f(x)
= f(a) + f(a) – f(x)
= 0 – f(x)   …......(from (3))
= – f(x)