Joint Entrance Examination

Graduate Aptitude Test in Engineering

NEW

New Website Launch

Experience the best way to solve previous year questions with **mock
tests** (very detailed analysis), **bookmark your favourite questions**, **practice** etc...

1

MCQ (Single Correct Answer)

Let $$f(x)$$ be a polynomial of degree four having extreme values

at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :

at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :

A

$$0$$

B

$$4$$

C

$$-8$$

D

$$-4$$

$$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3 \Rightarrow \mathop {Lim}\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2$$

So, $$f(x)$$ contains terms in $$x{}^2,{x^3}$$ and $${x^4}$$

Let $$f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

Since $$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2$$

Hence, $$f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

$$f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3}$$

As given: $$f'\left( 1 \right) = 0$$ and $$f'\left( 2 \right) = 0$$

Hence, $$4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $$8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 \right)$$

By $$4x\left( {eq1} \right) - eq\left( 2 \right),$$ we get

$$16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0$$

$$ \Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2$$

and by eqn. $$\left( 1 \right),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2$$

$$ \Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4}$$

$$f\left( 2 \right) = 2 \times 4 - 2 \times 8 + {1 \over 2} \times 16 = 0$$

So, $$f(x)$$ contains terms in $$x{}^2,{x^3}$$ and $${x^4}$$

Let $$f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

Since $$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2$$

Hence, $$f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

$$f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3}$$

As given: $$f'\left( 1 \right) = 0$$ and $$f'\left( 2 \right) = 0$$

Hence, $$4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $$8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 \right)$$

By $$4x\left( {eq1} \right) - eq\left( 2 \right),$$ we get

$$16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0$$

$$ \Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2$$

and by eqn. $$\left( 1 \right),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2$$

$$ \Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4}$$

$$f\left( 2 \right) = 2 \times 4 - 2 \times 8 + {1 \over 2} \times 16 = 0$$

2

MCQ (Single Correct Answer)

If $$f$$ and $$g$$ are differentiable functions in $$\left[ {0,1} \right]$$ satisfying

$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$

$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$

A

$$f'\left( c \right) = g'\left( c \right)$$

B

$$f'\left( c \right) = 2g'\left( c \right)$$

C

$$2f'\left( c \right) = g'\left( c \right)$$

D

$$2f'\left( c \right) = 3g'\left( c \right)$$

Since, $$f$$ and $$g$$ both are continuous function on $$\left[ {0,1} \right]$$

and differentiable on $$\left( {0,1} \right)$$ then $$\exists c \in \left( {0,1} \right)$$ such that

$$f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4$$

and $$g'\left( c \right) = {{g\left( 1 \right) - g\left( 0 \right)} \over 1} = {{2 - 0} \over 1} = 2$$

Thus, we get $$f'\left( c \right) = 2g'\left( c \right)$$

and differentiable on $$\left( {0,1} \right)$$ then $$\exists c \in \left( {0,1} \right)$$ such that

$$f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4$$

and $$g'\left( c \right) = {{g\left( 1 \right) - g\left( 0 \right)} \over 1} = {{2 - 0} \over 1} = 2$$

Thus, we get $$f'\left( c \right) = 2g'\left( c \right)$$

3

MCQ (Single Correct Answer)

The intercepts on $$x$$-axis made by tangents to the curve,

$$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $$ which are parallel to the line $$y=2x$$, are equal to :

$$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $$ which are parallel to the line $$y=2x$$, are equal to :

A

$$ \pm 1$$

B

$$ \pm 2$$

C

$$ \pm 3$$

D

$$ \pm 4$$

Since, $$y = \int\limits_0^x {\left| t \right|} dt,x \in R$$

therefore $${{dy} \over {dx}} = \left| x \right|$$

But from $$y = 2x,{{dy} \over {dx}} = 2$$

$$ \Rightarrow \left| x \right| = 2 \Rightarrow x = \pm 2$$

Points $$y = \int\limits_0^{ \pm 2} {\left| t \right|dt} = \pm 2$$

$$\therefore$$ equation of tangent is

$$y - 2 = 2\left( {x - 2} \right)$$ or $$y + 2 = 2\left( {x + 2} \right)$$

$$ \Rightarrow $$ $$x$$-intercept $$ = \pm 1.$$

therefore $${{dy} \over {dx}} = \left| x \right|$$

But from $$y = 2x,{{dy} \over {dx}} = 2$$

$$ \Rightarrow \left| x \right| = 2 \Rightarrow x = \pm 2$$

Points $$y = \int\limits_0^{ \pm 2} {\left| t \right|dt} = \pm 2$$

$$\therefore$$ equation of tangent is

$$y - 2 = 2\left( {x - 2} \right)$$ or $$y + 2 = 2\left( {x + 2} \right)$$

$$ \Rightarrow $$ $$x$$-intercept $$ = \pm 1.$$

4

MCQ (Single Correct Answer)

A line is drawn through the point $$(1, 2)$$ to meet the coordinate axes at $$P$$ and $$Q$$ such that it forms a triangle $$OPQ,$$ where $$O$$ is the origin. If the area of the triangle $$OPQ$$ is least, then the slope of the line $$PQ$$ is :

A

$$-{1 \over 4}$$

B

$$-4$$

C

$$-2$$

D

$$-{1 \over 2}$$

Equation of a line passing through $$\left( {{x_1},{y_1}} \right)$$ having

slope $$m$$ is given by $$y - {y_1} = m\left( {x - {x_1}} \right)$$

Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its

equation is

$$\left( {y - 2} \right) = m\left( {x - 1} \right)$$

where $$m$$ is the slope of the line $$PQ$$.

Now, point $$P\left( {x,0} \right)$$ will also satisfy the equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow 0 - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow x - 1 = {{ - 2} \over m}$$

$$ \Rightarrow x = {{ - 2} \over m} + 1$$

Also, $$OP = \sqrt {\left( {x - 0} \right){}^2 + {{\left( {0 - 0} \right)}^2}} = x = {{ - 2} \over m} + 1$$

Similarly, point $$Q\left( {0,y} \right)$$ will satisfy equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow y - 2 = m\left( { - 1} \right) \Rightarrow y = 2 - m$$b

and $$OQ = y = 2 - m$$

Area of $$\Delta POQ = {1 \over 2}\left( {OP} \right)\left( {OQ} \right) = {1 \over 2}\left( {1 - {2 \over m}} \right)\left( {2 - m} \right)$$

( As Area of $$\Delta = {1 \over 2} \times $$ base $$\,\, \times \,\,$$ height )

$$ = {1 \over 2}\left[ {2 - m - {4 \over m} + 2} \right] = {1 \over 2}\left[ {4 - \left( {m + {4 \over m}} \right)} \right]$$

$$ = 2 - {m \over 2} - {2 \over m}$$

Let Area $$ = f\left( m \right) = 2 - {m \over 2} - {2 \over m}$$

Now, $$f'\left( m \right) = {{ - 1} \over 2} + {2 \over {{m^2}}}$$

Put $$f'\left( m \right) = 0$$

$$ \Rightarrow {m^2} = 4 \Rightarrow m = \pm 2$$

Now, $$f''\left( m \right) = {{ - 4} \over {{m^3}}}$$

$${\left. {f''\left( m \right)} \right|_{m = 2}} = - {1 \over 2} < 0$$

$${\left. {f''\left( m \right)} \right|_{m = - 2}} = {1 \over 2} > 0$$

Area will be least at $$m=-2$$

Hence, slope of $$PQ$$ is $$-2.$$

slope $$m$$ is given by $$y - {y_1} = m\left( {x - {x_1}} \right)$$

Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its

equation is

$$\left( {y - 2} \right) = m\left( {x - 1} \right)$$

where $$m$$ is the slope of the line $$PQ$$.

Now, point $$P\left( {x,0} \right)$$ will also satisfy the equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow 0 - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow x - 1 = {{ - 2} \over m}$$

$$ \Rightarrow x = {{ - 2} \over m} + 1$$

Also, $$OP = \sqrt {\left( {x - 0} \right){}^2 + {{\left( {0 - 0} \right)}^2}} = x = {{ - 2} \over m} + 1$$

Similarly, point $$Q\left( {0,y} \right)$$ will satisfy equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow y - 2 = m\left( { - 1} \right) \Rightarrow y = 2 - m$$b

and $$OQ = y = 2 - m$$

Area of $$\Delta POQ = {1 \over 2}\left( {OP} \right)\left( {OQ} \right) = {1 \over 2}\left( {1 - {2 \over m}} \right)\left( {2 - m} \right)$$

( As Area of $$\Delta = {1 \over 2} \times $$ base $$\,\, \times \,\,$$ height )

$$ = {1 \over 2}\left[ {2 - m - {4 \over m} + 2} \right] = {1 \over 2}\left[ {4 - \left( {m + {4 \over m}} \right)} \right]$$

$$ = 2 - {m \over 2} - {2 \over m}$$

Let Area $$ = f\left( m \right) = 2 - {m \over 2} - {2 \over m}$$

Now, $$f'\left( m \right) = {{ - 1} \over 2} + {2 \over {{m^2}}}$$

Put $$f'\left( m \right) = 0$$

$$ \Rightarrow {m^2} = 4 \Rightarrow m = \pm 2$$

Now, $$f''\left( m \right) = {{ - 4} \over {{m^3}}}$$

$${\left. {f''\left( m \right)} \right|_{m = 2}} = - {1 \over 2} < 0$$

$${\left. {f''\left( m \right)} \right|_{m = - 2}} = {1 \over 2} > 0$$

Area will be least at $$m=-2$$

Hence, slope of $$PQ$$ is $$-2.$$

On those following papers in MCQ (Single Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

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

JEE Main 2021 (Online) 27th August Evening Shift (1)

JEE Main 2021 (Online) 27th August Morning Shift (1)

JEE Main 2021 (Online) 20th July Morning Shift (1)

JEE Main 2021 (Online) 16th March Evening Shift (1)

JEE Main 2021 (Online) 26th February Evening Shift (2)

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

JEE Main 2021 (Online) 25th February Evening Shift (1)

JEE Main 2021 (Online) 25th February Morning Shift (1)

JEE Main 2021 (Online) 24th February Evening Shift (3)

JEE Main 2021 (Online) 24th February Morning Shift (2)

JEE Main 2020 (Online) 6th September Evening Slot (2)

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

JEE Main 2020 (Online) 5th September Evening Slot (2)

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

JEE Main 2020 (Online) 4th September Evening Slot (2)

JEE Main 2020 (Online) 3rd September Evening Slot (1)

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

JEE Main 2020 (Online) 2nd September Evening Slot (2)

JEE Main 2020 (Online) 2nd September Morning Slot (3)

JEE Main 2020 (Online) 9th January Morning Slot (1)

JEE Main 2019 (Online) 12th April Morning Slot (2)

JEE Main 2019 (Online) 10th April Evening Slot (2)

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

JEE Main 2019 (Online) 9th April Morning Slot (3)

JEE Main 2019 (Online) 8th April Evening Slot (2)

JEE Main 2019 (Online) 8th April Morning Slot (2)

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

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

JEE Main 2019 (Online) 11th January Morning Slot (2)

JEE Main 2019 (Online) 10th January Evening Slot (1)

JEE Main 2019 (Online) 10th January Morning Slot (2)

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

JEE Main 2018 (Online) 15th April Morning Slot (2)

JEE Main 2017 (Online) 9th April Morning Slot (2)

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

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

JEE Main 2016 (Online) 9th April Morning Slot (2)

JEE Main 2016 (Offline) (2)

JEE Main 2015 (Offline) (1)

JEE Main 2014 (Offline) (1)

JEE Main 2013 (Offline) (1)

AIEEE 2012 (3)

AIEEE 2011 (2)

AIEEE 2010 (3)

AIEEE 2009 (2)

AIEEE 2008 (2)

AIEEE 2007 (3)

AIEEE 2006 (2)

AIEEE 2005 (4)

AIEEE 2004 (4)

AIEEE 2003 (1)

AIEEE 2002 (2)

Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations