Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

Transportation Engineering

Irrigation

Engineering Mathematics

Construction Material and Management

Fluid Mechanics and Hydraulic Machines

Hydrology

Environmental Engineering

Engineering Mechanics

Structural Analysis

Reinforced Cement Concrete

Steel Structures

Geomatics Engineering Or Surveying

General Aptitude

1

Let [x] denote the greatest integer less than or equal to x. Then $$\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$

A

equals $$\pi $$ + 1

B

equals 0

C

does not exist

D

equals $$\pi $$

R.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$

(as x $$ \to $$ 0^{+} $$ \Rightarrow $$ [x] $$=$$ 0)

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}$$

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$$

L.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}$$

(as x $$ \to $$ 0^{$$-$$} $$ \Rightarrow $$ [x] $$=$$ $$-$$1)

$$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi $$

R.H.L. $$ \ne $$ L.H.L.

(as x $$ \to $$ 0

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}$$

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$$

L.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}$$

(as x $$ \to $$ 0

$$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi $$

R.H.L. $$ \ne $$ L.H.L.

2

Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x – $$\pi $$) cos |x| is not differentiable. Then the set K is equal to :

A

{0, $$\pi $$}

B

$$\phi $$ (an empty set)

C

{ r }

D

{0}

f(x) = sin$$\left| x \right| - \left| x \right|$$ + 2(x $$-$$ $$\pi $$) cosx

$$ \because $$ sin$$\left| x \right|$$ $$-$$ $$\left| x \right|$$ is differentiable function at c = 0

$$ \therefore $$ k = $$\phi $$

$$ \because $$ sin$$\left| x \right|$$ $$-$$ $$\left| x \right|$$ is differentiable function at c = 0

$$ \therefore $$ k = $$\phi $$

3

$$\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}$$ is equal to :

A

0

B

4

C

1

D

2

$$\mathop {\lim }\limits_{x \to 0} {{x{{\tan }^2}2x} \over {\tan 4x{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{x\left( {{{{{\tan }^2}2x} \over {4{x^2}}}} \right)4{x^2}} \over {\left( {{{\tan 4x} \over {4x}}} \right)4x\left( {{{{{\sin }^2}x} \over {{x^2}}}} \right){x^2}}} = 1$$

4

Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = 4, then $$\int\limits_0^a \, $$f(x) g(x) dx is equal to :

A

4$$\int\limits_0^a \, $$f(x)dx

B

$$-$$ 3$$\int\limits_0^a \, $$f(x)dx

C

$$\int\limits_0^a \, $$f(x)dx

D

2$$\int\limits_0^a \, $$f(x)dx

$${\rm I} = \int_0^a {f\left( x \right)g\left( x \right)dx} $$

$${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} $$

$${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx} $$

$${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}} $$

$$ \Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx} $$

$${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} $$

$${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx} $$

$${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}} $$

$$ \Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx} $$

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (9) *keyboard_arrow_right*

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *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*

Mathematical Reasoning *keyboard_arrow_right*

Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

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*