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General Aptitude

1

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

2

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

3

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

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

4

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

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

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Trigonometric Functions & Equations *keyboard_arrow_right*

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Inverse Trigonometric Functions *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*