JEE Mains Previous Years Questions with Solutions Android App

Download our App

JEE Mains Previous Years Questions with Solutions

4.5 
Star 1 Star 2 Star 3 Star 4
Star 5
  (100k+ )
1

AIEEE 2003

MCQ (Single Correct Answer)
Let $${Z_1}$$ and $${Z_2}$$ be two roots of the equation $${Z^2} + aZ + b = 0$$, Z being complex. Further , assume that the origin, $${Z_1}$$ and $${Z_2}$$ form an equilateral triangle. Then
A
$${a^2} = 4b$$
B
$${a^2} = b$$
C
$${a^2} = 2b$$
D
$${a^2} = 3b$$

Explanation

Given quadratic equation,
$${Z^2} + aZ + b = 0$$
and two roots are $${Z_1}$$ and $${Z_2}$$.

$$\therefore$$ $${Z_1}$$ + $${Z_2}$$ = $$-a$$ and $${Z_1}$$$${Z_2}$$ = $$b$$

Question says,
There are three complex numbers:
1. Origin (0)
2. $${Z_1}$$
3. $${Z_2}$$
and they form an equilateral triangle. So They are the vertices of the triangle.

[ Important Point : If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]

In this question,
$${Z_1}$$ = 0, $${Z_2}$$ = $${Z_1}$$ and $${Z_3}$$ = $${Z_2}$$

By putting those values in the equation we get,

$${0^2}$$ + $$Z_1^2$$ + $$Z_2^2$$ = $$0$$ + $${Z_1}{Z_2}$$ + 0

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $${Z_1}{Z_2}$$

$$ \Rightarrow $$ $$Z_1^2$$ + $$Z_2^2$$ = $$b$$ [ as $${Z_1}$$$${Z_2}$$ = $$b$$ ]

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2{Z_1}{Z_2}$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ - $$2b$$ = $$b$$

$$ \Rightarrow $$ $${\left( {{Z_1} + {Z_2}} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${\left( { - a} \right)^2}$$ = $$3b$$

$$ \Rightarrow $$ $${a^2}$$ = $$3b$$

So Option (D) is correct.

[ Note : This question is asked to check if you know the following formula -

"If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -
$$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$" ]
2

AIEEE 2003

MCQ (Single Correct Answer)
If $$z$$ and $$\omega $$ are two non-zero complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$Arg(z) - Arg(\omega ) = {\pi \over 2},$$ then $$\,\overline {z\,} \omega $$ is equal to
A
$$- i$$
B
1
C
- 1
D
$$i$$

Explanation

Given that,
$$\left| {z\omega } \right| = 1$$
$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1
$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$
$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,
$${{z \over \omega }}$$ = $$ki$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$       [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$       [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

As $${{z \over \omega }}$$ is imaginary so we can write,

$${{z \over \omega }}$$ = $$ - {{\overline z } \over {\overline \omega }}$$
[ When $$z$$ is imaginary then $$z$$ = $$-\overline z $$ ]

$$ \Rightarrow $$ $$\overline z \omega $$ = $$ - z\overline \omega $$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$$\overline \omega $$.$$\omega $$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-{{z \over \omega }}$$.$${\left| \omega \right|^2}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${\left( {{1 \over {\sqrt k }}} \right)^2}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-ki$$.$${1 \over k}$$

$$ \Rightarrow $$ $$\overline z \omega $$ = $$-i$$

Method 2 :

Given that,
$$\left| {z\omega } \right| = 1$$
$$ \Rightarrow $$ $$\left| z \right|\left| \omega \right|$$ = 1
$$ \Rightarrow $$ $$\left| z \right|$$ = $${1 \over {\left| \omega \right|}}$$

and $$Arg(z) - Arg(\omega ) = {\pi \over 2}$$
$$ \Rightarrow $$ $$Arg\left( {{z \over \omega }} \right)$$ $$= {\pi \over 2}$$

When argument of a complex number is $${\pi \over 2}$$, it means it is making an angle of $${\pi \over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.

So, $${{z \over \omega }}$$ is a purely imaginary number that means there is no real part in this complex number.

So we can assume,
$${{z \over \omega }}$$ = $$ki$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| {ki} \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$\left| k \right|\left| i \right|$$

$$ \Rightarrow $$ $${\left| {{z \over \omega }} \right|}$$ = $$k$$       [ as $$\left| i \right|$$ = 1 ]

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$$${1 \over {\left| \omega \right|}}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$$$ \times $$ $$\left| z \right|$$ = $$k$$       [ as $${1 \over {\left| \omega \right|}}$$ = $$\left| z \right|$$ ]

$$ \Rightarrow $$ $${\left| z \right|^2}$$ = $$k$$

$$ \Rightarrow $$ $$\left| z \right|$$ = $$\sqrt k $$

$$\therefore$$ $$\left| \omega \right|$$ = $${1 \over {\sqrt k }}$$

(1) Magnitude of $$\overline z \omega $$

= $$\left| {\overline z } \right|\left| \omega \right|$$

= $$\left| z \right|\left| \omega \right|$$ [ as $$\left| z \right|$$ = $$\left| {\overline z } \right|$$ ]

= $$\sqrt k $$.$${{1 \over {\sqrt k }}}$$

= 1

$$\therefore$$ The distance from the origin of $${\overline z \omega }$$ is 1.

(2) Argument of $${\overline z \omega }$$ = $$Arg\left( {\overline z \omega } \right)$$

= $$Arg\left( {\overline z } \right) + Arg\left( \omega \right)$$

= $$-Arg\left( z \right) + Arg\left( \omega \right)$$

= $$ - \left( {Arg\left( z \right) - Arg\left( \omega \right)} \right)$$

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

$$\therefore$$ $${\overline z \omega }$$ is at (0, -1) on the negative side of imaginary axis and making an angle of $${\pi \over 2}$$ clockwise.

$$\therefore$$ $${\overline z \omega }$$ = 0 + (-1)$$ \times $$$$i$$ = $$-i$$
3

AIEEE 2002

MCQ (Single Correct Answer)
The locus of the centre of a circle which touches the circle $$\left| {z - {z_1}} \right| = a$$ and$$\left| {z - {z_2}} \right| = b\,$$ externally

($$z,\,{z_1}\,\& \,{z_2}\,$$ are complex numbers) will be
A
an ellipse
B
a hyperbola
C
a circle
D
none of these

Explanation

Let the circle be $$\left| {z - {z_3}} \right| = r.$$

Then according to given conditions

$$\left| {{z_3} - {z_1}} \right| = r + a$$ (Shown in the image)

and $$\left| {{z_3} - {z_2}} \right| = r + b.$$ (Shown in the image)

Eliminating $$r,$$ we get

$$\left| {{z_3} - {z_1}} \right| - \left| {{z_3} - {z_2}} \right| = a - b.$$

$$\therefore$$ Locus of center $${z_3}$$ is

$$\left| {z - {z_1}} \right| - \left| {z - {z_2}} \right| = a - b$$ = constant.

Definition of hyperbola says, when difference of distance between two points is constant from a particular point then that particular point will lie on a hyperbola.

Here distance of z1 from z3 is = $$r + a$$ and distance of z2 from z3 is = $$r + b$$

Now their difference = ($$r + a$$) - ($$r + b$$) = $$a - b$$ = a constant

$$\therefore$$ Locus of z3 is a hyperbola.
4

AIEEE 2002

MCQ (Single Correct Answer)
If $$\left| {z - 4} \right| < \left| {z - 2} \right|$$, its solution is given by
A
$${\mathop{\rm Re}\nolimits} (z) > 0$$
B
$${\mathop{\rm Re}\nolimits} (z) < 0$$
C
$${\mathop{\rm Re}\nolimits} (z) > 3$$
D
$${\mathop{\rm Re}\nolimits} (z) > 2$$

Explanation

Given $$\left| {z - 4} \right| < \left| {z - 2} \right|$$

Let $$\,\,\,z = x + iy$$

$$ \Rightarrow \left| {\left. {\left( {x - 4} \right) + iy} \right)} \right| < \left| {\left( {x - 2} \right) + iy} \right|$$

$$ \Rightarrow {\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2}$$

$$ \Rightarrow {x^2} - 8x + 16 < {x^2} - 4x + 4$$

$$ \Rightarrow 12 < 4x$$

$$ \Rightarrow x > 3$$

$$ \Rightarrow {\mathop{\rm Re}\nolimits} \left( z \right) > 3$$

Questions Asked from Complex Numbers

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
JEE Main 2021 (Online) 31st August Evening Shift (1)
JEE Main 2021 (Online) 27th August Morning Shift (1)
JEE Main 2021 (Online) 26th August Evening Shift (1)
JEE Main 2021 (Online) 26th August Morning Shift (1)
JEE Main 2021 (Online) 27th July Evening Shift (1)
JEE Main 2021 (Online) 27th July Morning Shift (1)
JEE Main 2021 (Online) 22th July Evening Shift (1)
JEE Main 2021 (Online) 20th July Morning Shift (1)
JEE Main 2021 (Online) 18th March Evening Shift (1)
JEE Main 2021 (Online) 18th March Morning Shift (1)
JEE Main 2021 (Online) 17th March Evening Shift (1)
JEE Main 2021 (Online) 17th March Morning Shift (1)
JEE Main 2021 (Online) 16th March Evening Shift (1)
JEE Main 2021 (Online) 16th March Morning Shift (1)
JEE Main 2021 (Online) 25th February Evening Shift (1)
JEE Main 2021 (Online) 25th February Morning Shift (1)
JEE Main 2020 (Online) 6th September Evening Slot (1)
JEE Main 2020 (Online) 6th September Morning Slot (1)
JEE Main 2020 (Online) 5th September Evening Slot (1)
JEE Main 2020 (Online) 5th September Morning Slot (1)
JEE Main 2020 (Online) 4th September Evening Slot (1)
JEE Main 2020 (Online) 4th September Morning Slot (1)
JEE Main 2020 (Online) 3rd September Evening Slot (1)
JEE Main 2020 (Online) 2nd September Evening Slot (1)
JEE Main 2020 (Online) 2nd September Morning Slot (1)
JEE Main 2020 (Online) 9th January Evening Slot (1)
JEE Main 2020 (Online) 9th January Morning Slot (1)
JEE Main 2020 (Online) 8th January Morning Slot (1)
JEE Main 2020 (Online) 7th January Evening Slot (1)
JEE Main 2020 (Online) 7th January Morning Slot (1)
JEE Main 2019 (Online) 12th April Evening Slot (1)
JEE Main 2019 (Online) 12th April Morning Slot (1)
JEE Main 2019 (Online) 10th April Evening Slot (1)
JEE Main 2019 (Online) 10th April Morning Slot (1)
JEE Main 2019 (Online) 9th April Evening Slot (1)
JEE Main 2019 (Online) 9th April Morning Slot (1)
JEE Main 2019 (Online) 8th April Evening Slot (1)
JEE Main 2019 (Online) 8th April Morning Slot (1)
JEE Main 2019 (Online) 12th January Evening Slot (1)
JEE Main 2019 (Online) 12th January Morning Slot (1)
JEE Main 2019 (Online) 11th January Evening Slot (1)
JEE Main 2019 (Online) 11th January Morning Slot (1)
JEE Main 2019 (Online) 10th January Evening Slot (1)
JEE Main 2019 (Online) 10th January Morning Slot (1)
JEE Main 2019 (Online) 9th January Evening Slot (1)
JEE Main 2019 (Online) 9th January Morning Slot (2)
JEE Main 2018 (Online) 16th April Morning Slot (1)
JEE Main 2018 (Offline) (1)
JEE Main 2018 (Online) 15th April Evening Slot (1)
JEE Main 2018 (Online) 15th April Morning Slot (1)
JEE Main 2017 (Offline) (1)
JEE Main 2016 (Online) 9th April Morning Slot (1)
JEE Main 2016 (Offline) (1)
JEE Main 2015 (Offline) (1)
JEE Main 2014 (Offline) (1)
JEE Main 2013 (Offline) (1)
AIEEE 2012 (1)
AIEEE 2011 (2)
AIEEE 2010 (1)
AIEEE 2008 (2)
AIEEE 2007 (1)
AIEEE 2006 (2)
AIEEE 2005 (3)
AIEEE 2004 (3)
AIEEE 2003 (3)
AIEEE 2002 (3)

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12