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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 ?

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 ?

Grade:12

5 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
10 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
askIITians Faculty
ritika
21 Points
5 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
askIITians Faculty 256 Points
4 years ago
Hello Student

Please see the solution in the attachment.
643-2117_ck_588afa2282884.png
I hope this solution will help you.
Vikas TU
14149 Points
4 years ago
643-2117_ck_588afa2282884.png
Dear student 
Please refer the clear image of this solution 
Feel free to ask querry 
We will happy to help you.
Good Luck 
Cheers 
ankit singh
askIITians Faculty 614 Points
4 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)

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