Question

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 the constant function f(0)=1,f(2a-x) is

Answer 1

hope the answer helps

Answer 2

The value of f(2a-x) is -f(x)

Therefore f(2a-x)=-f(x)

Step-by-step explanation:

• Given that 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 the constant function and f(0)=1

• To find the value of f(2a-x) :
• f(x-y)=f(x)f(y)-f(a-x)f(a+y)
• put y=0 in above function we get
• f(x-0)=f(x)f(0)-f(a-x)f(a+0)
• f(x)=f(x)(1)-f(a-x)f(a) ( since f(0)=1 )

• Subtract f(x) on both sides we get
• f(x)-f(x)=f(x)-f(a-x)f(a)-f(x)
• 0=-f(a-x)f(a)+0
• f(a)f(a-x)=0
• $f(a)=\frac{0}{f(a-x)}$
• Therefore f(a)=0
• Now we can write f(2a-x)=f(a-(x-a))

• =f(a)f(x-a)-f(a-a)f(a+(x-a)) ( by given f(x-y)=f(x)f(y)-f(a-x)f(a+y) )
• =f(a)f(x-a)-f(0)f(x)
• =(0)f(x-a)-(1)f(x) ( by f(a)=0 and f(0)=1 )
• =0-f(x)
• =-f(x)

• Therefore f(2a-x)=-f(x)
• The value of f(2a-x) is -f(x)