A function f:R→R satisfies the equation $$f(x)f(y)$$−$$f(xy)=x+y$$∀x,y∈R and $$f(1)$$>$$0$$, then

Correct option is 3f(x)f−1(x)=x2−1

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$A function f : R \to R satisfies the equation f (x) f (y) - f (x y) = x + y \forall x, y \in R and f (1) > 0, then$

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f(x)f−1(x)=x2−4

f(x)f−1(x)=x2−6

f(x)f−1(x)=x2−1

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detailed solution

Correct option is C

Sol. (c) Taking x=y=1 , we get f(1)f(1)−f(1)=2⇒ f2(1)−f(1)−2=0⇒ (f(1)−2)(f(1)+1)=0⇒ f(1)=2 [ As f(1)>0] Taking y=1, we get f(x)⋅f(1)−f(x)=x+1⇒f(x)=x+1⇒f−1(x)=x−1∴f(x)⋅f−1(x)=x2−1

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A function f:R→R satisfies the equation f(x)f(y)−f(xy)=x+y∀x,y∈R and f(1)>0, then

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Ready to Test Your Skills?

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$A function f : R \to R satisfies the equation f (x) f (y) - f (x y) = x + y \forall x, y \in R and f (1) > 0, then$