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

If f(x) is a differentiable function in the interval (0, $$\infty $$) such that f (1) = 1 and

$$\mathop {\lim }\limits_{t \to x} $$ $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$ equal to :

$$\mathop {\lim }\limits_{t \to x} $$ $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$ equal to :

A

$${{13} \over 6}$$

B

$${{23} \over 18}$$

C

$${{25} \over 9}$$

D

$${{31} \over 18}$$

$$\mathop {\lim }\limits_{t \to x} {{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1$$

It is in $${0 \over 0}$$ form

So, applying L' Hospital rule,

$$\mathop {\lim }\limits_{t \to x} {{2tf\left( x \right) - {x^2}f'\left( x \right)} \over 1} = 1$$

$$ \Rightarrow $$ 2xf(x) $$-$$ x^{2}f '(x) = 1

$$ \Rightarrow $$ f '(x) $$-$$ $${2 \over x}$$f(x) $$=$$ $${1 \over {{x^2}}}$$

$$ \therefore $$ I.F = $${e^{\int {{{ - 2} \over x}dx} }} = {e^{ - 2\log x}} = {1 \over {{x^2}}}$$

$$ \therefore $$ Solution of equation,

f(x)$${1 \over {{x^2}}}$$ = $$\int {{1 \over x}\left( { - {1 \over {{x^2}}}} \right)} \,dx$$

$$ \Rightarrow $$ $${{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + C$$

Given that,

f(1) = 1

$$ \therefore $$ $${1 \over 1}$$ = $${1 \over 3}$$ + C

$$ \Rightarrow $$ C $$=$$ $${2 \over 3}$$

$$ \therefore $$ f(x) $$=$$ $${2 \over 3}$$ x^{2} + $${1 \over {3x}}$$

$$ \therefore $$ f$$\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}$$

It is in $${0 \over 0}$$ form

So, applying L' Hospital rule,

$$\mathop {\lim }\limits_{t \to x} {{2tf\left( x \right) - {x^2}f'\left( x \right)} \over 1} = 1$$

$$ \Rightarrow $$ 2xf(x) $$-$$ x

$$ \Rightarrow $$ f '(x) $$-$$ $${2 \over x}$$f(x) $$=$$ $${1 \over {{x^2}}}$$

$$ \therefore $$ I.F = $${e^{\int {{{ - 2} \over x}dx} }} = {e^{ - 2\log x}} = {1 \over {{x^2}}}$$

$$ \therefore $$ Solution of equation,

f(x)$${1 \over {{x^2}}}$$ = $$\int {{1 \over x}\left( { - {1 \over {{x^2}}}} \right)} \,dx$$

$$ \Rightarrow $$ $${{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + C$$

Given that,

f(1) = 1

$$ \therefore $$ $${1 \over 1}$$ = $${1 \over 3}$$ + C

$$ \Rightarrow $$ C $$=$$ $${2 \over 3}$$

$$ \therefore $$ f(x) $$=$$ $${2 \over 3}$$ x

$$ \therefore $$ f$$\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}$$

2

The solution of the differential equation

$${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$$

where 0 $$ \le $$ x < $${\pi \over 2}$$, and y (0) = 1, is given by :

$${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$$

where 0 $$ \le $$ x < $${\pi \over 2}$$, and y (0) = 1, is given by :

A

y = 1 $$-$$ $${x \over {\sec x + \tan x}}$$

B

y^{2} = 1 + $${x \over {\sec x + \tan x}}$$

C

y^{2} = 1 $$-$$ $${x \over {\sec x + \tan x}}$$

D

y = 1 + $${x \over {\sec x + \tan x}}$$

Given,

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

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

Now, let

y^{2} $$=$$ t

$$ \Rightarrow $$ 2y$${{dy} \over {dx}} = {{dt} \over {dx}}$$

$$ \therefore $$ New equation,

$${{dt} \over {dx}} + t\sec x = \tan x$$

$$ \therefore $$ I.F $$=$$ $${e^{\int {\sec xdx} }}$$

$$=$$ $${e^{\ln \left( {\sec x + \tan x} \right)}}$$

$$=$$ sec x + tan x

$$ \therefore $$ Solution is,

t(sec x + tan x) $$=$$ $$\int {\tan x} $$ (sec x + tan x) dx

$$ \Rightarrow $$ t(sec x + tan x) $$=$$ sec x + tan x $$-$$ x + c

$$ \Rightarrow $$ t $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}} + c$$

$$ \Rightarrow $$ y^{2} $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}} + c$$

Given,

y(0) $$=$$ 1

$$ \therefore $$ 1 $$=$$ 1 $$-$$ 0 + c

$$ \Rightarrow $$ c $$=$$ 0

$$ \therefore $$ y^{2} $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}}$$

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

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

Now, let

y

$$ \Rightarrow $$ 2y$${{dy} \over {dx}} = {{dt} \over {dx}}$$

$$ \therefore $$ New equation,

$${{dt} \over {dx}} + t\sec x = \tan x$$

$$ \therefore $$ I.F $$=$$ $${e^{\int {\sec xdx} }}$$

$$=$$ $${e^{\ln \left( {\sec x + \tan x} \right)}}$$

$$=$$ sec x + tan x

$$ \therefore $$ Solution is,

t(sec x + tan x) $$=$$ $$\int {\tan x} $$ (sec x + tan x) dx

$$ \Rightarrow $$ t(sec x + tan x) $$=$$ sec x + tan x $$-$$ x + c

$$ \Rightarrow $$ t $$=$$ 1 $$-$$ $${x \over {\sec x + \tan x}} + c$$

$$ \Rightarrow $$ y

Given,

y(0) $$=$$ 1

$$ \therefore $$ 1 $$=$$ 1 $$-$$ 0 + c

$$ \Rightarrow $$ c $$=$$ 0

$$ \therefore $$ y

3

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the
maximum area (in sq. m) of the flower-bed, is:

A

10

B

25

C

30

D

12.5

We have

Total length = r + r + r$$\theta $$ = 20

$$ \Rightarrow $$ 2r + r$$\theta $$ = 20

$$ \Rightarrow $$ $$\theta = {{20 - 2r} \over r}$$ .......(1)

A = Area = $${\theta \over {2\pi }} \times \pi {r^2}$$ = $${1 \over 2}{r^2}\theta $$ = $${1 \over 2}{r^2}\left( {{{20 - 2r} \over r}} \right)$$

$$ \Rightarrow $$ A = 10r – r^{2}

For A to be maximum

$${{dA} \over {dr}} = 0$$

$$ \Rightarrow $$ 10 – 2r = 0

$$ \Rightarrow $$ r = 5

$${{{d^2}A} \over {d{r^2}}} = - 2 < 0$$

$$ \therefore $$ For r = 5, A is maximum From (1)

$${\theta = {{20 - 2\left( 5 \right)} \over r}}$$ = 2

A = $${2 \over {2\pi }} \times \pi {\left( 5 \right)^2}$$ = 25 sq. m

Total length = r + r + r$$\theta $$ = 20

$$ \Rightarrow $$ 2r + r$$\theta $$ = 20

$$ \Rightarrow $$ $$\theta = {{20 - 2r} \over r}$$ .......(1)

A = Area = $${\theta \over {2\pi }} \times \pi {r^2}$$ = $${1 \over 2}{r^2}\theta $$ = $${1 \over 2}{r^2}\left( {{{20 - 2r} \over r}} \right)$$

$$ \Rightarrow $$ A = 10r – r

For A to be maximum

$${{dA} \over {dr}} = 0$$

$$ \Rightarrow $$ 10 – 2r = 0

$$ \Rightarrow $$ r = 5

$${{{d^2}A} \over {d{r^2}}} = - 2 < 0$$

$$ \therefore $$ For r = 5, A is maximum From (1)

$${\theta = {{20 - 2\left( 5 \right)} \over r}}$$ = 2

A = $${2 \over {2\pi }} \times \pi {\left( 5 \right)^2}$$ = 25 sq. m

4

The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes
through the point:

A

$$\left( {{1 \over 2},{1 \over 2}} \right)$$

B

$$\left( {{1 \over 2}, - {1 \over 3}} \right)$$

C

$$\left( {{1 \over 2},{1 \over 3}} \right)$$

D

$$\left( { - {1 \over 2}, - {1 \over 3}} \right)$$

Given $$y = {{x + 6} \over {\left( {x - 2} \right)\left( {x - 2} \right)}}$$

At y-axis, x = 0 $$ \Rightarrow $$ y = 1

On differentiating, we get

$${{dy} \over {dx}} = {{\left( {{x^2} - 5x + 6} \right)\left( 1 \right) - \left( {x + 6} \right)\left( {2x - 5} \right)} \over {{{\left( {{x^2} - 5x + 6} \right)}^2}}}$$

$${{dy} \over {dx}} = 1$$ at point (0, 1)

$$ \therefore $$ Slope of normal = – 1

Now equation of normal is y – 1 = –1 (x – 0)

$$ \Rightarrow $$ y – 1 = – x

x + y = 1 ......(1)

By checking each option you can see point $$\left( {{1 \over 2},{1 \over 2}} \right)$$ satisfy equation (1).

At y-axis, x = 0 $$ \Rightarrow $$ y = 1

On differentiating, we get

$${{dy} \over {dx}} = {{\left( {{x^2} - 5x + 6} \right)\left( 1 \right) - \left( {x + 6} \right)\left( {2x - 5} \right)} \over {{{\left( {{x^2} - 5x + 6} \right)}^2}}}$$

$${{dy} \over {dx}} = 1$$ at point (0, 1)

$$ \therefore $$ Slope of normal = – 1

Now equation of normal is y – 1 = –1 (x – 0)

$$ \Rightarrow $$ y – 1 = – x

x + y = 1 ......(1)

By checking each option you can see point $$\left( {{1 \over 2},{1 \over 2}} \right)$$ satisfy equation (1).

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (2) *keyboard_arrow_right*

AIEEE 2003 (2) *keyboard_arrow_right*

AIEEE 2004 (2) *keyboard_arrow_right*

AIEEE 2005 (2) *keyboard_arrow_right*

AIEEE 2006 (1) *keyboard_arrow_right*

AIEEE 2007 (1) *keyboard_arrow_right*

AIEEE 2008 (1) *keyboard_arrow_right*

AIEEE 2009 (1) *keyboard_arrow_right*

AIEEE 2010 (1) *keyboard_arrow_right*

AIEEE 2011 (2) *keyboard_arrow_right*

AIEEE 2012 (1) *keyboard_arrow_right*

JEE Main 2013 (Offline) (1) *keyboard_arrow_right*

JEE Main 2014 (Offline) (1) *keyboard_arrow_right*

JEE Main 2015 (Offline) (1) *keyboard_arrow_right*

JEE Main 2016 (Offline) (1) *keyboard_arrow_right*

JEE Main 2016 (Online) 9th April Morning Slot (1) *keyboard_arrow_right*

JEE Main 2016 (Online) 10th April Morning Slot (1) *keyboard_arrow_right*

JEE Main 2017 (Offline) (3) *keyboard_arrow_right*

JEE Main 2017 (Online) 8th April Morning Slot (1) *keyboard_arrow_right*

JEE Main 2018 (Offline) (4) *keyboard_arrow_right*

JEE Main 2018 (Online) 15th April Morning Slot (1) *keyboard_arrow_right*

JEE Main 2018 (Online) 15th April Evening Slot (1) *keyboard_arrow_right*

JEE Main 2018 (Online) 16th April Morning Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 9th January Morning Slot (2) *keyboard_arrow_right*

JEE Main 2019 (Online) 9th January Evening Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 10th January Evening Slot (3) *keyboard_arrow_right*

JEE Main 2019 (Online) 11th January Morning Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 11th January Evening Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 12th January Morning Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 12th January Evening Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 8th April Morning Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 9th April Morning Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 9th April Evening Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 10th April Morning Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 10th April Evening Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 12th April Morning Slot (1) *keyboard_arrow_right*

JEE Main 2019 (Online) 12th April Evening Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 7th January Morning Slot (2) *keyboard_arrow_right*

JEE Main 2020 (Online) 7th January Evening Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 8th January Morning Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 8th January Evening Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 9th January Evening Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 2nd September Evening Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 3rd September Morning Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 4th September Morning Slot (2) *keyboard_arrow_right*

JEE Main 2020 (Online) 4th September Evening Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 5th September Morning Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 5th September Evening Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 6th September Morning Slot (1) *keyboard_arrow_right*

JEE Main 2020 (Online) 6th September Evening Slot (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 24th February Morning Slot (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 25th February Morning Slot (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 26th February Morning Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 26th February Evening Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 16th March Morning Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 16th March Evening Shift (2) *keyboard_arrow_right*

JEE Main 2021 (Online) 17th March Morning Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 17th March Evening Shift (2) *keyboard_arrow_right*

JEE Main 2021 (Online) 18th March Morning Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 18th March Evening Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 20th July Morning Shift (2) *keyboard_arrow_right*

JEE Main 2021 (Online) 22th July Evening Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 25th July Morning Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 27th July Evening Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 25th July Evening Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 27th July Morning Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 26th August Morning Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 26th August Evening Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 27th August Morning Shift (2) *keyboard_arrow_right*

JEE Main 2021 (Online) 27th August Evening Shift (2) *keyboard_arrow_right*

JEE Main 2021 (Online) 31st August Morning Shift (1) *keyboard_arrow_right*

JEE Main 2021 (Online) 31st August Evening Shift (2) *keyboard_arrow_right*

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*

Mathematical Reasoning *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*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*