A real valued function f(x) satisfies the functional equation f(x-y)=f(x) f(y)-f(a-x)…
Mathematics · JEE Main · NTA Exams — Sets, Relations and Functions
A real valued function \(f(x)\) satisfies the functional equation \(f(x-y)=f(x) f(y)-f(a-x) f(a+y)\) where ' \(\mathbf{a}\)' is a given constant and \(f(0)=1\), then \(f(2 a-x)\)is equal to
- \(f(x)\)
- \(-f(x)\)
- \(f(-x)\)
- \(f(a]+f(a-x)\)
Answer
(B) -f(x)
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- The number of real solutions of the equation 1 + |e x - 1| = e x (e x - 2) is :
- Range of the function f( x )= x 1+ x ^ 2 is
- [ ( x _ 1 - x _ 2 ]^ 2 + [12+ 1- x _ 1 ^ 2 - 4 x _ 2 ]^ 2 ] x _ 1 , x _ 2 R is
- Find the no. of solutions of |[x]-2 x|=4 , where [ x ] is the greatest integer x
- If ( ^ -1 x )^ 2 - ( ^ -1 x )^ 2 =a ; 0 x a ≠ 0, then the value of 2 x 2 – 1 is
- Let f be a function from R to R given by f (x) = x^ 2 -4 x^ 2 +1 . Then f (x) is.
- The function f : R → R defined by f (x) = (x – 1) (x – 2) (x – 3) is
- The number of values of x , where the function f(x)= x+ ( 2 x) attains its maximum, is
More Sets, Relations and Functions questions · Browse all practice questions