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 a given constant and $f(0) = 1$. Then $f(2a - x)$ 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 $f(0) = 1$. Then $f(2a - x)$ is equal to
Updated On: Jul 5, 2022
- $f(-x)$
- $f(a)+f(a-x)$
- $f(x)$
- $-f(x)$
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The Correct Option is D
Solution and Explanation
$f(x - y) = f(x) f (y) - f (a - x) f (a + y)$
Put $x = 0 = y$.
$\therefore \:\: f(0) = f(0) f(0) -f(a) f(a) = (f(0))^2 - (f (a))^2 $
$\Rightarrow 1= 1-(f(a))^2 \Rightarrow f(a) = 0$
Now $f(2a-x) = f(a - (x - a)) $
$ = f(a)f(x - a)-f(a- a) f(a + x - a) $
$0\cdot f(x-a)-f(0) f(x) $
$= 0 - 1. f(x) = - f(x) $.
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Concepts Used:
Relations and functions
A relation R from a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.
Representation of Relation and Function
Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, represent this function in different forms.
- Set-builder form - {(x, y): f(x) = y2, x ∈ A, y ∈ B}
- Roster form - {(1, 1), (2, 4), (3, 9)}
- Arrow Representation
