Let $f : R^{+} \to R^{+}$ be a function satisfying the relation $f (x \cdot f (y)) = f (xy) + x$ for all $x, y \in R^{+}$ . Then $\lim_{x \to 0} (\frac{(f (x))^{1 / 3} - 1}{(f (x))^{1 / 2} - 1}) =$

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$\frac{1}{2}$

$\frac{2}{3}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

Given relation is $f (x \cdot f (y)) = f (xy) + x$

Interchanging x and y in Eq. (1.56), we have

$f (y \cdot f (x)) = f (yx) + y$

Again replacing x with f (x) in Eq. (1.56) we get

$f (f (x) \cdot f (y)) = f (y \cdot f (x)) + f (x)$

Therefore, Eqs. (1.56) - (1.58) imply

$f (f (x) \cdot f (y)) = f (xy) + y + f (x)$

Again interchanging x and y in Eq. (1.59), we have

$f (f (y) \cdot f (x)) = f (yx) + x + f (y)$

Equations (1.59) and (1.60) imply

$f (xy) + y + f (x) = f (yx) + x + f (y)$

Suppose $f (x) - x = f (y) - y = λ$

Substituting $f (x) = λ + x$ we have

$\begin{array}{r} x \cdot f (y) + λ = (xy + λ) + x \Rightarrow x \cdot f (y) = xy + x Therefore x (y + λ) = xy + x [∵ f (y) = λ + y] \\ \Rightarrow λx = x \Rightarrow λ = 1 (∵ x > 0) So f (x) = x + λ = x + 1 Hence \lim_{x \to 0} \frac{(f (x))^{1 / 3} - 1}{(f (x))^{1 / 2} - 1} = \lim_{x \to 0} \frac{(1 + x)^{1 / 3} - 1}{(1 + x)^{1 / 2} - 1} \\ = \lim_{x \to 0} (\frac{(1 + x)^{1 / 3} - 1}{1 + x - 1}) \cdot (\frac{1 + x - 1}{(1 + x)^{1 / 2} - 1}) = \frac{1 / 3}{1 / 2} = \frac{2}{3} \end{array}$