1

### JEE Main 2019 (Online) 9th January Morning Slot

If y = y(x) is the solution of the differential equation,

x$dy \over dx$ + 2y = x2, satisfying y(1) = 1, then y($1\over2$) is equal to :
A
${{7} \over {64}}$
B
${{49} \over {16}}$
C
${{1} \over {4}}$
D
${{13} \over {16}}$

## Explanation

Given,

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

$\Rightarrow$  ${{dy} \over {dx}} + \left( {{2 \over x}} \right)y = x$

This is a linear differential equation.

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

$= {e^{2\ln x}}$

$= {x^2}$

$\therefore$  Solution is,

$y \cdot {x^2} = \int {x \cdot {x^2}dx}$

$\Rightarrow$  $y{x^2} = {{{x^4}} \over 4} + C$

given  $y\left( 1 \right) = 1$

$\therefore$  $1.1 = {4 \over 4} + C$

$\Rightarrow$  $C = {3 \over 4}$

$\therefore$  Equation is

$y{x^2} = {{{x^4}} \over 4} + {3 \over 4}$

$\therefore$  $y\left( {{1 \over 2}} \right)$   means   $x = {1 \over 2}$

$\therefore$  $y \cdot {\left( {{1 \over 2}} \right)^2} = {1 \over 4} \times {\left( {{1 \over 2}} \right)^4} + {3 \over 4}$

$\Rightarrow$  ${y \over 4} = {1 \over {64}} + {3 \over 4}$

$\Rightarrow$   ${y \over 4} = {{1 + 48} \over {64}}$

$\Rightarrow$  y = ${{49} \over {16}}$
2

### JEE Main 2019 (Online) 9th January Morning Slot

The maximum volume (in cu.m) of the right circular cone having slant height 3 m is :
A
2$\sqrt3$$\pi B 3\sqrt3$$\pi$
C
6$\pi$
D
${4 \over 3}\pi$

## Explanation

$\therefore$   h = 3 cos$\theta$

r = 3 sin$\theta$

We know volume of right circular cone,

V = ${1 \over 3}\pi {r^2}h$

= ${1 \over 3}\pi$(3 sin$\theta$)2 3 cos$\theta$

= 9 $\pi$ sin2$\theta$ cos$\theta$

For maximum or minimum value of volume,

${{dv} \over {d\theta }}$ = 0

$\therefore$   (2sin$\theta$ cos$\theta$) cos$\theta$ + 3sin2$\theta$ .($-$ sin$\theta$) = 0

$\Rightarrow$  2 sin$\theta$ cos2$\theta$ $-$ sin3$\theta$ = 0

$\Rightarrow$  2 sin$\theta$(1 $-$ sin2$\theta$) $-$ sin3 $\theta$ = 0

$\Rightarrow$  2 sin$\theta$ $-$ 2 sin3$\theta$ $-$ sin3$\theta$ = 0

$\Rightarrow$  3 sin3$\theta$ = 2 sin$\theta$

$\Rightarrow$   sin2$\theta$ = ${2 \over 3}$

$\Rightarrow$  sin$\theta$ = $\sqrt {{2 \over 3}}$

${{{d^2}v} \over {d{\theta ^2}}}$ = 2cos$\theta$ $-$ 3(3sin$\theta$ cos$\theta$)

= 2 cos$\theta$ $-$ 9 sin$\theta$ cos$\theta$

= 2 $\times$ ${1 \over {\sqrt 3 }}$ $-$ 9 $\times$ ${{\sqrt 2 } \over {\sqrt 3 }}$ $\times$ ${1 \over {\sqrt 3 }}$

= ${2 \over {\sqrt 3 }} - 3\sqrt 2 \, < 0$

$\therefore$  Volume is maximum

when sin$\theta$ = $\sqrt {{2 \over 3}}$

$\therefore$  Maximum volume is

= 9 $\pi$ ${\left( {\sqrt {{2 \over 3}} } \right)^2} \times {1 \over {\sqrt 3 }}$

= 9 $\pi$ $\times$ ${2 \over 3} \times {1 \over {\sqrt 3 }}$

= $2\sqrt 3 \,\pi$
3

### JEE Main 2019 (Online) 9th January Evening Slot

Let f : [0,1] $\to$ R be such that f(xy) = f(x).f(y), for all x, y $\in$ [0, 1], and f(0) $\ne$ 0. If y = y(x) satiesfies the differential equation, ${{dy} \over {dx}}$ = f(x) with y(0) = 1, then y$\left( {{1 \over 4}} \right)$ + y$\left( {{3 \over 4}} \right)$ is equal to :
A
3
B
4
C
2
D
5

## Explanation

If f(xy) = f(x) f(y) $\forall$ x, y $\in$ R and f(0) $\ne$ 0

put x = y = 0

$\Rightarrow$  f(0) = [f(0)]2

$\Rightarrow$  f(0) = 1

put y = 0 $\Rightarrow$ f(0) = f(x) f(0)

$\Rightarrow$ f(x) = 1

given that ${{dy} \over {dx}}$ = f(x)

$\therefore$  ${{dy} \over {dx}}$ = 1 $\Rightarrow$ y = x + k

given that y(0) = 1

$\therefore$  k = 1

hence y = x + 1

y$\left( {{1 \over 4}} \right)$ + y$\left( {{3 \over 4}} \right)$ = $\left( {{1 \over 4} + 1} \right)$ + $\left( {{3 \over 4} + 1} \right)$ = 3
4

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

The curve amongst the family of curves represented by the differential equation, (x2 – y2)dx + 2xy dy = 0 which passes through (1, 1) is
A
a circle with centre on the y-axis
B
an ellipse with major axis along the y-axis
C
a circle with centre on the x-axis
D
a hyperbola with transverse axis along the x-axis

## Explanation

(x2 $-$ y2) dx + 2xy dy = 0

${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$

Put   $y = vx \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$

Solving we get,

$\int {{{2v} \over {{v^2} + 1}}dv = \int { - {{dx} \over x}} }$

ln(v2 + 1) = $-$ ln x + C

(y2 + x2) = Cx

1 + 1 = C $\Rightarrow$ C = 2

y2 + x2 = 2x