1

### JEE Main 2019 (Online) 11th January Morning Slot

If  xloge(logex) $-$ x2 + y2 = 4(y > 0), then ${{dy} \over {dx}}$ at x = e is equal to :
A
${{\left( {1 + 2e} \right)} \over {2\sqrt {4 + {e^2}} }}$
B
${{\left( {1 + 2e} \right)} \over {\sqrt {4 + {e^2}} }}$
C
${{\left( {2e - 1} \right)} \over {2\sqrt {4 + {e^2}} }}$
D
${e \over {\sqrt {4 + {e^2}} }}$

## Explanation

Differentiating with respect to x,

$x.{1 \over {\ell nx}}.{1 \over x} + \ell n(\ell nx) - 2x + 2y.{{dy} \over {dx}} = 0$

at   $x = e$  we get

$1 - 2e + 2y{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2y}}$

$\Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2\sqrt {4 + {e^2}} }}\,\,$

as   $y(e) = \sqrt {4 + {e^2}}$
2

### JEE Main 2019 (Online) 11th January Evening Slot

The solution of the differential equation,

${{dy} \over {dx}}$ = (x – y)2, when y(1) = 1, is :
A
$-$ loge $\left| {{{1 + x - y} \over {1 - x + y}}} \right|$ = x + y $-$ 2
B
loge $\left| {{{2 - x} \over {2 - y}}} \right|$ = x $-$ y
C
loge $\left| {{{2 - y} \over {2 - x}}} \right|$ = 2(y $-$ 1)
D
$-$ loge $\left| {{{1 - x + y} \over {1 + x - y}}} \right|$ = 2(x $-$ 1)

## Explanation

x $-$ y = t

$\Rightarrow$ ${{dy} \over {dx}} = 1 - {{dt} \over {dx}}$

$\Rightarrow$  1 $-$ ${{dt} \over {dx}}$ = t2 $\Rightarrow$  $\int {{{dt} \over {1 - {t^2}}}}$ = $\int {1dx}$

$\Rightarrow$  ${1 \over 2}\ell n\left( {{{1 + t} \over {1 - t}}} \right) = x + \lambda$

$\Rightarrow$  ${1 \over 2}\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right) = x + \lambda$ given y(1) = 1

$\Rightarrow$  ${1 \over 2}\ell n(1) = 1 + \lambda \Rightarrow \lambda = - 1$

$\Rightarrow$  $\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right)$ = 2(x $-$ 1)

$\Rightarrow$  $- \ell n\left( {{{1 - x + y} \over {1 + x - y}}} \right)$ = 2(x $-$ 1)
3

### JEE Main 2019 (Online) 12th January Morning Slot

Let y = y(x) be the solution of the differential equation, x${{dy} \over {dx}}$ + y = x loge x, (x > 1). If 2y(2) = loge 4 $-$ 1, then y(e) is equal to :
A
$- {e \over 2}$
B
$- {{{e^2}} \over 2}$
C
${{{e^2}} \over 4}$
D
${e \over 4}$

## Explanation

${{dy} \over {dx}} = {y \over x} = \ell nx$

${e^{\int {{1 \over x}dx} }} = x$

$xy = \int {x\ell nx + C}$

$\ell nx{{{x^2}} \over 2} - \int {{1 \over x}.{{{x^2}} \over 2}}$

$xy = {x \over 2}\ell nx - {{{x^2}} \over 4} + C,$

for   $2y\left( 2 \right) = 2\ell n2 - 1$

$\Rightarrow$  $C = 0$

$y = {x \over 2}\ell nx - {x \over 4}$

$y\left( e \right) = {e \over 4}$
4

### JEE Main 2019 (Online) 12th January Evening Slot

If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, y) on it as ${{{x^2} - 2y} \over x}$, then the curve also passes through the point :
A
(–1, 2)
B
$\left( { - \sqrt 2 ,1} \right)$
C
$\left( { \sqrt 3 ,0} \right)$
D
(3, 0)

## Explanation

${{dy} \over {dx}} = {{{x^2} - 2y} \over x}$                  (Given)

${{dy} \over {dx}} + 2{y \over x} = x$

I.F = ${e^{\int {{2 \over x}dx} }} = {x^2}$

$\therefore$  y.x2 = $\int {x.{x^2}} dx + C$

= ${{{x^4}} \over y} + C$

hence bpasses through (1, $-$ 2) $\Rightarrow$ C= $-$ ${9 \over 4}$

$\therefore$  yx2 = ${{{x^4}} \over 4} - {9 \over 4}$

Now check option(s), Which is satisly by option (ii)