MCQ
Let $f :(0, \infty) \rightarrow(0, \infty)$ be a differentiable function such that $f(1)= e$ and $\lim \limits_{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}=0$ If $f ( x )=1,$ then $x$ is equal to
  • A
    $2e$
  • B
    $\frac{1}{2 e }$
  • C
    $e$
  • $\frac{1}{ e }$

Answer

Correct option: D.
$\frac{1}{ e }$
d
$L=\lim _{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}$

using L.H. rule

$L =\lim _{t \rightarrow x} \frac{2 tf ^{2}( x )- x ^{2} \cdot 2 f ^{\prime}( t ) \cdot f ( t )}{1}$

$\Rightarrow L =2 xf ( x )\left( f ( x )- x f ^{\prime}( x )\right)=0$ (given)

$\Rightarrow f(x)=x f^{\prime}(x) \Rightarrow \int \frac{f^{\prime}(x) d x}{f(x)}=\int \frac{d x}{x}$

$\Rightarrow \ell n | f ( x )|=\ell n | x |+ C$

$\because f (1)= e , x >0, f ( x )>0$

$\Rightarrow f ( x )= ex , \quad$ if $f ( x )=1 \Rightarrow x =\frac{1}{ e }$

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