Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

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

2

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

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

3

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

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

put x = y = 0

$$ \Rightarrow $$ f(0) = [f(0)]

$$ \Rightarrow $$ 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

The curve amongst the family of curves represented by the differential equation, (x^{2} – y^{2})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

(x^{2} $$-$$ y^{2}) 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(v^{2} + 1) = $$-$$ ln x + C

(y^{2} + x^{2}) = Cx

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

y^{2} + x^{2} = 2x

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

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(v

(y

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

y

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