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 . Then - Mathematics MCQ AIEEE 2026 | ExamTest.live

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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 . Then f (2 a - x) is equal to:

Solution & Step-by-step Explanation

Put x = 0 and y = 0 : f (0) = f (0) f (0) - f (a) f (a) ⟹ 1 = 1 - [f (a)]^{2} ⟹ f (a) = 0 .Now put y = a - x : f (x - (a - x)) = f (x) f (a - x) - f (a - x) f (a + (a - x)) f (2 x - a) = f (x) f (a - x) - f (a - x) f (2 a - x) This doesn't seem to lead directly. Let's try putting x = a in the original equation: f (a - y) = f (a) f (y) - f (0) f (a + y) Since f (a) = 0 and f (0) = 1 : f (a - y) = - f (a + y) .Let a + y = T ⟹ y = T - a : f (a - (T - a)) = - f (T) f (2 a - T) = - f (T) .Replacing T with x : f (2 a - x) = - f (x) .

Try it yourself before checking the explanation above.

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 . Then f (2 a - x) is equal to:

- f (x)

f (x)

f (a) + f (a - x)

f (- x)

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