Q97 / 155 · WBJEE 2017 · Mathematics

Let for all $x>0, f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right),$ then

  • A
    $f(x)+f\left(\frac{1}{x}\right)=1$
  • Correct answer
    B
    $f(x y)=f(x)+f(y)$
  • C
    $f(x y)=x f(y)+y f(x)$
  • D
    $f(x y)=x f(x)+y f(y)$

Solution

We have, $f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right)$ $=\lim _{n \rightarrow \infty} \frac{x^{1 / n}-1}{1 / n}$ Let $\quad \frac{1}{n}=y$ $\begin{aligned} \therefore \quad f(x)=& \lim _{y \rightarrow 0} \frac{x^{y}-1}{y} \\ &=\log x \end{aligned}$ $\therefore \quad f(x y)=\log (x, y)$ $=\log x+\log y$ $=f(x)+f(y)$

Hence, the correct answer is option: 'B'.

Relevant Questions