Question

A real valued function $f\left( x \right)$satisfies the functional equation $f\left( {x - y} \right) = f(x)f(y) - f(a - x)(a + y)$, where a is given constant and $f\left( 0 \right) = 1$, then
A $f\left( {2a - x} \right) = f(x),f(x)$ is symmetric about $x = a$
B $f\left( {2a - x} \right) = 0 = f(x),f(x)$ is symmetric about $x = a$
C $f\left( {2a - x} \right) + f(x) = 0,f(x)$ is not symmetric about $x = a$
D $f(x) = 0,f(x)$is symmetric about $x = a$

Answer 1

Here we have $f\left( x \right)$as a real valued function and a real valued function is a function that assigns a real number to each member in its domain and also we are given $f\left( {x - y} \right) = f(x)f(y) - f(a - x)(a + y)$and a is a given constant and another condition we are provided in the question is $f\left( 0 \right) = 1$so in this question we can apply a approach to the solution by putting $y = 0$and solving it using the given condition $f\left( 0 \right) = 1$and finding the relation between $f\left( {2a - x} \right),f(x)$ and checking that the resultant answer satisfies which condition among the given option and if it is symmetric about $x = a$as a function is symmetric about any axis if the graph of that function can be reflected about that axis that can be done by replacing y with $\left( { - y} \right)$ and checking if we get the same function back.

Complete step by step solution:
Here we are given with a real valued function $f\left( x \right)$ which satisfies the functional equation $f\left( {x - y} \right) = f(x)f(y) - f(a - x)(a + y)$, where a is given constant
So as a real valued function is a function that assigns a real number to each member in its domain.
Also we are provided with the given condition $f\left( 0 \right) = 1$
We can start solving the question by finding the relation between $f\left( {2a - x} \right),f(x)$
Which can be done by putting $y = 0$in the given condition that is $f\left( {x - y} \right) = f(x)f(y) - f(a - x)(a + y)$
Putting $y = 0$we get –:
$f\left( x \right) = f(x)f\left( 0 \right) - f\left( {a - x} \right)f(a)$
Now we are given $f\left( 0 \right) = 1$
So the resultant equation becomes –
$f\left( x \right) = f(x) - f\left( {a - x} \right)f(a)$
$0 = f\left( {a - x} \right)f(a)$ --------a
Now from above condition we get the fact that $0 = f(a)$and $0 = f(a - x)$
But $0 \ne f(a - x)$as it will become 0 for every value of x which is not possible as we are given $f\left( x \right)$as a real valued function and a real valued function is a function that assigns a real number to each member in it’s domain
So we are left with $0 = f(a - x)$
Now for finding the relation between $f\left( {2a - x} \right),f(x)$ we can write $f\left( {2a - x} \right)$as $f\left( {a - \left( {x - a} \right)} \right)$
Now we are given $f\left( {x - y} \right) = f(x)f(y) - f(a - x)(a + y)$, applying this on $f\left( {a - \left( {x - a} \right)} \right)$
We get $f(a)f(x - a) - f(a - a)f(x)$
$f(a)f(x - a) - f(0)f(x)$
Now we are given $f\left( 0 \right) = 1$also we have found earlier in a that $0 = f\left( {a - x} \right)f(a)$
So the resulting equation becomes -:
$f\left( {2a - x} \right) = - f(x)$
$f\left( {2a - x} \right) + f(x) = 0$
So the above equation gives us the relation between $f\left( {2a - x} \right),f(x)$.
Now checking for symmetry of $f\left( x \right)$ about $x = a$that can be done by substituting $x = a - x$in $f(2a - x)$as a symmetric function is that function whose graph of can be reflected about that axis that can be done by replacing y with $\left( { - y} \right)$ and checking if we get the same function back
So now putting the value we get
$f(2a - (a - x)) = - f(a - x)$
$f\left( {a + x} \right) \ne - f\left( {a - x} \right)$
So $f\left( x \right)$is not symmetric about $x = a$
Hence the relation of $f\left( {2a - x} \right),f(x)$is $f\left( {2a - x} \right) + f(x) = 0$and $f\left( x \right)$is not symmetric about $x = a$

So from the above options option C is correct answer that is $f\left( {2a - x} \right) + f(x) = 0,f(x)$ is not symmetric about $x = a$

Note:
Another approach to this question can be used is that taking $x = a$ and $y = (x - a)$
And then finding the relationship between $f\left( {2a - x} \right),f(x)$ and then checking the further symmetry either of the methods one discussed in solution and other discussed in this note would take similar amounts of the time.