Let f : (0 , ) (0 , ) be a differentiable function such that f (1 )=eand limt x t^2f^2 (x )-x^2f^2 (t )t-x=0. If f (x )=1, then x is equal to:

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Correct answer: D. (d) JEE Main 2020 limt x t^2f^2 (x )-x^2f^2 (t )t-x=0 ⇒ limt x 2tf^2 (x )-2x^2f (t ) ∙f^ ' (t )1=0 (Using L’ Hospital’s Rule) ⇒f (x )=xf^ ' (x ) ⇒ f^ ' (x )f (x )= 1x Integrating w.r.t. x, we get ⇒ lnf (x )= lnx+ lnC ⇒f (x )= Cx ∵f (1 )=e ⇒C=e ;so f (x )= ex When f (x )=1= ex ⇒x= 1e

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

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