First slide

Functions (XII)

Question

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

Moderate

A

$f (x) f^{- 1} (x) = x^{2} - 4$
B

$f (x) f^{- 1} (x) = x^{2} - 6$
C

$f (x) f^{- 1} (x) = x^{2} - 1$
D

None of these

Solution

$\begin{array}{l} Sol. (c) Taking x = y = 1, we get \\ f (1) f (1) - f (1) = 2 \\ \Rightarrow f^{2} (1) - f (1) - 2 = 0 \\ \Rightarrow (f (1) - 2) (f (1) + 1) = 0 \\ \Rightarrow f (1) = 2 [As f (1) > 0] \end{array}$
$\begin{array}{l} Taking y = 1, we get \\ \begin{array}{ll} f (x) \cdot f (1) - f (x) = x + 1 \\ \Rightarrow & f (x) = x + 1 \\ \Rightarrow & f^{- 1} (x) = x - 1 \\ ∴ & f (x) \cdot f^{- 1} (x) = x^{2} - 1 \end{array} \end{array}$

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