Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

Structural Analysis

Construction Material and Management

Reinforced Cement Concrete

Steel Structures

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

Hydrology

Irrigation

Geomatics Engineering Or Surveying

Environmental Engineering

Transportation Engineering

Engineering Mathematics

General Aptitude

1

The curve satifying the differeial equation, (x^{2} $$-$$ y^{2}) 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.

(x^{2} $$-$$ y^{2}) 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

x^{2} + y^{2} $$-$$ 2x = 0, which is a circle of radius one.

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

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

x

2

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'+ y^{2} $$-$$ 9 = 0

D

xy y' $$-$$ y^{2} + 9 = 0

Equation of ellipse,

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

As ellipse passes through (0, 3)

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

$$ \Rightarrow $$ b^{2} = 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' + y^{2} = 9

$$ \Rightarrow $$ xyy' $$-$$ y^{2} + 9 = 0

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

As ellipse passes through (0, 3)

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

$$ \Rightarrow $$ b

$$\therefore\,\,\,$$ 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' + y

$$ \Rightarrow $$ xyy' $$-$$ y

3

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

x$$dy \over dx$$ + 2y = x^{2}, satisfying y(1) = 1, then y($$1\over2$$) is equal
to :

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

A

$$ {{7} \over {64}}$$

B

$$ {{49} \over {16}}$$

C

$$ {{1} \over {4}}$$

D

$$ {{13} \over {16}}$$

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 = $${{49} \over {16}}$$

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

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

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

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

$${4 \over 3}\pi $$

$$ \therefore $$ 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$$\theta $$)

= 9 $$\pi $$ sin

For maximum or minimum value of volume,

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

$$ \therefore $$ (2sin$$\theta $$ cos$$\theta $$) cos$$\theta $$ + 3sin

$$ \Rightarrow $$ 2 sin$$\theta $$ cos

$$ \Rightarrow $$ 2 sin$$\theta $$(1 $$-$$ sin

$$ \Rightarrow $$ 2 sin$$\theta $$ $$-$$ 2 sin

$$ \Rightarrow $$ 3 sin

$$ \Rightarrow $$ sin

$$ \Rightarrow $$ 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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Trigonometric Functions & Equations *keyboard_arrow_right*

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Circle *keyboard_arrow_right*

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