Question

$Iff:R\to R$ be a function satisfying the functional Rule $f(x+f(y\left)\right)=f\left(x\right)+x+f(x-y);\forall x,y\in R$ then
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \left(P\right)& f\left(0\right)& \left(A\right)& 1\\ \left(Q\right)& |f\left(1\right)+f\left(2\right)|& \left(B\right)& 3\\ \left(R\right)& |f\left(2\right)+f(-3)|& \left(C\right)& 0\\ \left(S\right)& |f\left(1\right)+f(-3)|& \left(D\right)& 2\end{array}$

• A. $P\to B,Q\to A,R\to C,S\to D$
• B. $P\to A,Q\to B,R\to D,S\to C$
• C. $P\to C,Q\to B,R\to A,S\to D$
• D. $P\to C,Q\to B,R\to D,S\to A$

Answer 1

The correct option is C $P\to C,Q\to B,R\to A,S\to D$

Put y = 0 in the given equation and let f(0) = k
f(x+k) = 2f(x)+x
Put x =-k, k =2 f(-k) -k
f(-k) = k
Put y = -k in functional definition, then
f(x+k) =f(x) + x+ f(x+k)
f(x) =-x