1

### JEE Main 2019 (Online) 10th January Evening Slot

A helicopter is flying along the curve given by y – x3/2 = 7, (x $\ge$ 0). A soldier positioned at the point $\left( {{1 \over 2},7} \right)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is -
A
${1 \over 6}\sqrt {{7 \over 3}}$
B
${{\sqrt 5 } \over 6}$
C
${1 \over 2}$
D
${1 \over 3}$$\sqrt {{7 \over 3}}$

## Explanation $y - {x^{3/2}} = 7\left( {x \ge 0} \right)$

${{dy} \over {dx}} = {3 \over 2}{x^{1/2}}$

$\left( {{3 \over 2}\sqrt x } \right)\left( {{{7 - y} \over {{1 \over 2} - x}}} \right) = - 1$

$\left( {{3 \over 2}\sqrt x } \right)\left( {{{ - {x^{3/2}}} \over {{1 \over 2} - x}}} \right) = - 1$

${3 \over 2}.{x^2} = {1 \over 2} - x$

$3{x^2} = 1 - 2x$

$3{x^2} + 2x - 1 = 0$

$3{x^2} + 3x - x - 1 = 0$

$\left( {x + 1} \right)\left( {3x - 1} \right) = 0$

$\therefore$  $x = - 1$  (rejected)

$x = {1 \over 3}$

$y = 7 + {x^{3/2}} = 7 + {\left( {{1 \over 3}} \right)^{3/2}}$

${\ell _{AB}} = \sqrt {{{\left( {{1 \over 2} - {1 \over 3}} \right)}^2} + {{\left( {{1 \over 3}} \right)}^3}} = \sqrt {{1 \over {36}} + {1 \over {27}}}$

$= \sqrt {{{3 + 4} \over {9 \times 12}}}$

$= \sqrt {{7 \over {108}}} = {1 \over 6}\sqrt {{7 \over 3}}$
2

### 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   $y(e) = \sqrt {4 + {e^2}}$
3

### 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)
4

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

$\Rightarrow$  $C = 0$

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

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