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 $$\theta $$ denotes the acute angle between the curves, y = 10 – x^{2} and y = 2 + x^{2} at a point of their intersection, the |tan $$\theta $$| is equal to :

A

$$8 \over 15$$

B

$$4 \over 9$$

C

$$7 \over 17$$

D

$$8 \over 17$$

Angle between the curves is the acute angle between the tangents at the point of intersection.

y = 10 $$-$$ x

and y = 2 + x

$$ \therefore $$ 10 $$-$$ x

$$ \Rightarrow $$ 2x

$$ \Rightarrow $$ x

$$ \Rightarrow $$ x = 2, $$-$$ 2

$$ \therefore $$ points of intersection (2, 6) and ($$-$$ 2, 6)

$${{dy} \over {dx}}$$ for curve 1 = $$-$$ 2x

$$ \therefore $$ Slope(m

$${{dy} \over {dx}}$$ for curve 2 = 2x

$$ \therefore $$ slope (m

$$ \therefore $$ tan$$\theta $$ = $$\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$

= $$\left| {{{ - 4 - 4} \over {1 + \left( { - 16} \right)}}} \right|$$

= $${8 \over {15}}$$

2

Equation of a common tangent to the circle, x^{2} + y^{2} – 6x = 0 and the parabola, y^{2} = 4x is :

A

$$2\sqrt 3 $$y = 12x + 1

B

$$\sqrt 3 $$y = x + 3

C

$$2\sqrt 3 $$y = -x - 12

D

$$\sqrt 3 $$y = 3x + 1

We know,

Equation of tangent to the parabola y^{2} = 4ax is,

y = mx + $${a \over m}$$

$$ \therefore $$ Equation of tangent to the parabola y^{2} = 4x is,

y = mx + $${1 \over m}$$

$$ \Rightarrow $$ m^{2}x $$-$$ ym + 1 = 0

This tangent is also the tangent to the circle x^{2} + y^{2} $$-$$ 6x = 0

So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3

$$ \therefore $$ $$\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3$$

$$ \Rightarrow $$ (3m^{2} + 1)^{2} = 9(m^{4} + m^{2})

$$ \Rightarrow $$ 9m^{4} + 6m^{2} + 1 = 9m^{4} + 9m^{2}

$$ \Rightarrow $$ 3m^{2} = 1

$$ \Rightarrow $$ m = $$ \pm $$ $${1 \over {\sqrt 3 }}$$

So, possible tangents are

y = $${1 \over {\sqrt 3 }}$$x + $$\sqrt 3 $$

$$ \Rightarrow $$ $$\sqrt 3 $$y = x + 3

or y = $$-$$ $${x \over {\sqrt 3 }}$$ $$-$$ $$\sqrt 3 $$

$$ \Rightarrow $$ $$\sqrt 3 y$$ = $$-$$ x $$-$$ 3

Equation of tangent to the parabola y

y = mx + $${a \over m}$$

$$ \therefore $$ Equation of tangent to the parabola y

y = mx + $${1 \over m}$$

$$ \Rightarrow $$ m

This tangent is also the tangent to the circle x

So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3

$$ \therefore $$ $$\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3$$

$$ \Rightarrow $$ (3m

$$ \Rightarrow $$ 9m

$$ \Rightarrow $$ 3m

$$ \Rightarrow $$ m = $$ \pm $$ $${1 \over {\sqrt 3 }}$$

So, possible tangents are

y = $${1 \over {\sqrt 3 }}$$x + $$\sqrt 3 $$

$$ \Rightarrow $$ $$\sqrt 3 $$y = x + 3

or y = $$-$$ $${x \over {\sqrt 3 }}$$ $$-$$ $$\sqrt 3 $$

$$ \Rightarrow $$ $$\sqrt 3 y$$ = $$-$$ x $$-$$ 3

3

A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :

A

$${3 \over 2}$$

B

$$\sqrt 3 $$

C

2

D

$${2 \over {\sqrt 3 }}$$

Let the equation of hyperbola

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

Given 2a = 4

$$ \Rightarrow $$ $$a$$ = 2

It passes through (4, 2)

$$ \therefore $$ $${{16} \over 4} - {4 \over {{b^2}}}$$ = 1

$$ \Rightarrow $$ b^{2} = $${4 \over 3}$$

e = $$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$ = $$\sqrt {1 + {{4/3} \over 4}} $$

= $$\sqrt {1 + {1 \over 3}} $$ = $${2 \over {\sqrt 3 }}$$

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

Given 2a = 4

$$ \Rightarrow $$ $$a$$ = 2

It passes through (4, 2)

$$ \therefore $$ $${{16} \over 4} - {4 \over {{b^2}}}$$ = 1

$$ \Rightarrow $$ b

e = $$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$ = $$\sqrt {1 + {{4/3} \over 4}} $$

= $$\sqrt {1 + {1 \over 3}} $$ = $${2 \over {\sqrt 3 }}$$

4

Let A(4, $$-$$ 4) and B(9, 6) be points on the parabola, y^{2} = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of $$\Delta $$ACB is maximum. Then, the area (in sq. units) of $$\Delta $$ACB, is :

A

$$31{1 \over 4}$$

B

$$30{1 \over 2}$$

C

32

D

$$31{3 \over 4}$$

$$\Delta ABC = {1 \over 2}\left| {\matrix{ 4 & { - 4} & 1 \cr 9 & 6 & 1 \cr {{t^2}} & {2t} & 1 \cr } } \right|$$

D = 60 + 10t $$-$$ 10t

$${{d\Delta } \over {dt}} = 0 \Rightarrow t = {1 \over 2}$$

$${{{d^2}\Delta } \over {d{t^2}}} = - 20 < 0$$

$$ \therefore $$ max at $$t = {1 \over 2}$$

max area $$\Delta = 65 - {5 \over 2}$$

$$ = {{125} \over 2} = 31{1 \over 4}$$

Number in Brackets after Paper Name Indicates No of Questions

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

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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*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*