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Functional equation f(x)f(y) + f(3/x)f(3/y) = 2f(xy) | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Consider f:R+→Rf:\mathbb{R}^{+}\to\mathbb{R} such that f(1)=1f\left(1\right) = 1 and
f(x)f(y)+f(3x)f(3y)=2f(xy)∀ x,y∈R+.f\left(x\right)f\left(y\right)+f\left(\frac3x\right)f\left(\frac3y\right) = 2f\left(xy\right) \qquad\forall\,x,y \in \mathbb{R}^{+}.
 Find f(99)f\left(99\right).
Solution
Answer: 1
Step 1: Putting x=y=1x = y = 1,
f(1)2+f(3)2=2f(1)⟹1+f(3)2=2⟹f(3)=±1f\left(1\right)^2+f\left(3\right)^2 = 2f\left(1\right) \quad\Longrightarrow\quad 1+f\left(3\right)^2 = 2 \quad\Longrightarrow\quad f\left(3\right) = \pm1
 Write ε=f(3)\varepsilon = f\left(3\right), so ε2=1\varepsilon^2 = 1. Step 2: Putting y=3xy = \dfrac3x, the two products on the left become the same:
f(x)f(3x)+f(3x)f(x)=2f(3)⟹f(x)f(3x)=ε.(1)f\left(x\right)f\left(\tfrac3x\right)+f\left(\tfrac3x\right)f\left(x\right) = 2f\left(3\right) \quad\Longrightarrow\quad f\left(x\right)f\left(\tfrac3x\right) = \varepsilon . \tag{1}
 Step 3: Putting y=1y = 1,
f(x)f(1)+f(3x)f(3)=2f(x)⟹f(x)+εf(3x)=2f(x)f\left(x\right)f\left(1\right)+f\left(\tfrac3x\right)f\left(3\right) = 2f\left(x\right) \quad\Longrightarrow\quad f\left(x\right)+\varepsilon f\left(\tfrac3x\right) = 2f\left(x\right)
εf(3x)=f(x)⟹f(3x)=εf(x)(2)\varepsilon f\left(\tfrac3x\right) = f\left(x\right) \quad\Longrightarrow\quad f\left(\tfrac3x\right) = \varepsilon f\left(x\right) \tag{2}
 (using ε−1=ε\varepsilon^{-1} = \varepsilon). From (1) and (2), on dividing by ε≠0\varepsilon \ne 0,
f(x)⋅εf(x)=ε⟹f(x)2=1⟹f(x)=±1 for every x>0f\left(x\right)\cdot\varepsilon f\left(x\right) = \varepsilon \quad\Longrightarrow\quad f\left(x\right)^2 = 1 \quad\Longrightarrow\quad f\left(x\right) = \pm1 \ \text{for every } x>0
 Step 4: Putting x=y=tx = y = \sqrt t with t>0t>0,
f(t)2+f(3t)2=2f(t)f\left(\sqrt t\right)^2+f\left(\tfrac3{\sqrt t}\right)^2 = 2f\left(t\right)
 and by Step 3 each square on the left is 11, so
1+1=2f(t)⟹f(t)=1∀ t>01+1 = 2f\left(t\right) \quad\Longrightarrow\quad f\left(t\right) = 1 \qquad\forall\,t>0
 Step 5: f≡1f \equiv 1 gives 1⋅1+1⋅1=2=2⋅11\cdot1+1\cdot1 = 2 = 2\cdot1 and f(1)=1f\left(1\right) = 1 ; also f(3)=1f\left(3\right) = 1, so ε=+1\varepsilon = +1 and the branch ε=−1\varepsilon = -1 does not occur.
f(99)=1f\left(99\right) = 1
 Answer: 11
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