If x _ e ( _ e x)-x^ 2 +y^ 2 =4(y>0), then (d y)/(d x) at x=e is equal to :

Correct answer: B. (2 e-1) 2 4+e^ 2 Consider the equation, x _ e ( _ e x)-x^ 2 +y^ 2 =4 Differentiate both sides w.r.t. x _ e ( _ e x)+x · 1 x · _ e x -2 x+2 y (d y)/(d x)=0 _ e ( _ e x)+ 1 _ e x -2 x+2 y (d y)/(d x)=0 When x=e, y= 4+e^ 2 . Put these values in (1), 0+1-2 e+2 4+e^ 2 (d y)/(d x)=0 (d y)/(d x)= 2 e-1 2 4+e^ 2

JEE Main 201911 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differentiation

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Differentiation

If x _ e ( _ e x)-x^ 2 +y^ 2 =4(y>0), then (d y)/(d x) at x=e is...

x \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4(y>0),

then

\frac{d y}{d x}

x=e

is equal to :

Options

A $\frac{(1+2 e)}{2 \sqrt{4+e^{2}}}$
B $\frac{(2 e-1)}{2 \sqrt{4+e^{2}}}$
C $\frac{(1+2 e)}{\sqrt{4+e^{2}}}$
D $\frac{e}{\sqrt{4+e^{2}}}$

Correct answer

B. $\frac{(2 e-1)}{2 \sqrt{4+e^{2}}}$

Step-by-step solution

Consider the equation,

x \log _{e}

\left(\log _{e} x\right)-x^{2}+y^{2}=4

Differentiate both sides w.r.t.

x \log _{e}\left(\log _{e} x\right)+x \cdot \frac{1}{x \cdot \log _{e} x}-2 x+2 y \frac{d y}{d x}=0 \log _{e}\left(\log _{e} x\right)+\frac{1}{\log _{e} x}-2 x+2 y \frac{d y}{d x}=0

When

x=e, y=\sqrt{4+e^{2}}

. Put these values in (1),

0+1-2 e+2 \sqrt{4+e^{2}} \frac{d y}{d x}=0 \frac{d y}{d x}=\frac{2 e-1}{2 \sqrt{4+e^{2}}}

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