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

Hint:
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.