1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

The curve satifying the differeial equation, (x2 $$-$$ y2) dx + 2xydy = 0 and passing through the point (1, 1) is :
A
a circle of radius one.
B
a hyperbola.
C
an ellipse.
D
a circle of radius two.

Explanation

(x2 $$-$$ y2) dx + 2xydy = 0

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

Let y = vx

$${{dy} \over {dx}}$$ = v + x $${{dv} \over {dx}}$$

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

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

$$ \Rightarrow $$  $${{2vdv} \over {{v^2} + 1}}$$ = $$-$$ $${{dx} \over x}$$

After intergrating, we get

$$\ln \left| {{v^2} + 1} \right|$$ = $$-$$ ln$$\left| x \right|$$ + lnc

$${{y{}^2} \over {{x^2}}}$$ + 1 = $${c \over x}$$

As curve passes through the point (1, 1), so 1 + 1 = c

$$ \Rightarrow $$  c = 2

x2 + y2 $$-$$ 2x = 0, which is a circle of radius one.
2
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

The differential equation representing the family of ellipse having foci eith on the x-axis or on the $$y$$-axis, center at the origin and passing through the point (0, 3) is :
A
xy y'' + x (y')2 $$-$$ y y' = 0
B
x + y y'' = 0
C
xy y'+ y2 $$-$$ 9 = 0
D
xy y' $$-$$ y2 + 9 = 0

Explanation

Equation of ellipse,

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

As ellipse passes through (0, 3)

$$\therefore\,\,\,$$ $${{{0^2}} \over {{a^2}}} + {{{3^2}} \over {{b^2}}} = 1$$

$$ \Rightarrow $$    b2 = 9

$$\therefore\,\,\,$$ Equation of ellipse becomes,

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$$

Differentiating w.r.t    x, we get,

$${{2x} \over {a{}^2}}$$ + $${{2y} \over 9}$$ . $${{dy} \over {dx}} = 0$$

$$ \Rightarrow $$    $${x \over {{a^2}}}$$ = $$-$$ $${y \over 9}.{{dy} \over {da}}$$

$$ \Rightarrow $$    $${x \over {{a^2}}} = - {y \over 9}.y'......$$ (1)

We got earlier,

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9}$$ = 1

$$ \Rightarrow $$   $${x \over {{a^2}}}.x + {{{y^2}} \over 9} = 1$$

putting value of equation (1) here,

$$ {-{y\,y'} \over 9}.x + {{{y^2}} \over 9} = 1$$

$$ \Rightarrow $$    $$-$$ xyy' + y2 = 9

$$ \Rightarrow $$   xyy' $$-$$ y2 + 9 = 0
3
MCQ (Single Correct Answer)

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 = $${{49} \over {16}}$$
4
MCQ (Single Correct Answer)

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

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

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