MCQ
$f (x) =$ $\frac{x}{{\ln x}}\,$ and $g (x) =$ $\frac{{\ln x}}{x}\,$. Then identify the $CORRECT $ statement
  • $\frac{1}{{g(x)}}\,$ and $f (x)$ are identical functions
  • B
    $\frac{1}{{f(x)}}\,$ and $g (x)$ are identical functions
  • C
    $f (x) . g (x) = 1$  $\forall \,\,x\, > \,0$
  • D
    $\frac{1}{{f(x)\,.\,g(x)}}\,\,\, = \,\,1\,$  $\forall \,\,x\, > \,0$ 

Answer

Correct option: A.
$\frac{1}{{g(x)}}\,$ and $f (x)$ are identical functions
a
$(A)$ $\frac{1}{{g(x)}}\, = \,\frac{1}{{\ln x/x}}\,\,\,;\,\,f(x)\, = \,\frac{x}{{\ln x}}$          $x > 0 , x \ne 1$ for both
$(B)$ $\frac{1}{{f(x)}}\, = \,\frac{1}{{x/\ln x}}\,\,\,;\,\,g(x)\, = \,\frac{{\ln x}}{x}\,$           

$\frac{1}{{f(x)}}$ is not defined at $x=1$ but $g(1)=0$
$(C)$ $f(x) . g(x)$ = $\frac{x}{{\ln x}}\,.\,\,\frac{{\ln x}}{x}\,\,$ $=1$             

if $x > 0 , x\,\, \ne \,\,1$ $\Rightarrow\,\, N. I. $
$(D)$ $\frac{1}{{f(x)\,.\,g(x)}}\,\,\, = \,\,\,\frac{1}{{\frac{x}{{\ln x}}\,.\,\,\frac{{\ln x}}{x}}}\,\,\, = \,1$             

 only for $x > 0$ and  $x\,\, \ne \,\,1 $

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