If fx2−6x+6+fx2−4x+4=2x∀x∈R then f(−3)+f(9)−5f(1)=
7
8
9
10
Let x2−6x+6=x2−4x+4
Then x = 1
Putting x=1 in the given relation, we get f(1)=1 .
For x2−6x+6=1,x=1 or 5 For x2−4x+4=1,x=1 or 3 Put x=3 and 5 in the given relation.
f(−3)+f(1)=6⇒f(−3)=5 and f(1)+f(9)=10⇒f(9)=9∴ f(−3)+f(9)−5f(1)=9