Let a curve y = f (x) pass through the point (2, ( _ e 2) ^ 2 ) and have slope 2y x _ e x for all positive real values of x . Then the value of f(e) is equal to _____.

Correct answer: 0 Let a curve y = f(x) pass through the point (2, ( _ e 2) ^ 2 ) and have slope 2y x _ e x for all positive real values of x . Therefore, y ^ ' = (2y)/(xlnx) ⇒ (dy)/(y) = (2dx)/(xℓnx) ⇒ ℓn|y| = 2ℓn|ℓnx| + C Put x =2 , y = (ℓn2) ^ 2 ⇒ c = 0 ⇒ y = (ℓnx) ^ 2 ⇒ f(e) = 1 .

JEE Main 202125 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2021 · 25 July · Evening Shift · Mathematics · Application of Derivatives

Let a curve y = f (x) pass through the point (2, ( _ e 2) ^ 2 ) and...

Let a curve

y = f

\left(x\right)

pass through the point \left(2, {\left(\log_{e}2\right)}

^{2}\right) and have slope

\dfrac{2y}{x{\log}_{e}x}

for all positive real values of

x .

Then the value of

f\left(e\right)

is equal to _____.

Correct answer

Step-by-step solution

Let a curve

y = f\left(x\right)

pass through the point \left(2, {

\left(\log_{e}2\right)}^{2}\right)

and have slope

\dfrac{2y}{x{\log}_{e}x}

for all positive real values of

x

Therefore, ${y}^{'} = \dfrac{2y}{xlnx}$

$⇒ \dfrac{dy}{y} = \dfrac{2dx}{xℓnx}$

$⇒ ℓn\left|y\right| = 2ℓn\left|ℓnx\right| + C$

Put $x =2 , y = {\left(ℓn2\right)}^{2}$

$⇒ c = 0$

$⇒ y = {\left(ℓnx\right)}^{2}$

$⇒ f\left(e\right) = 1 .$

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