1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

If |z $$-$$ 3 + 2i| $$ \le $$ 4 then the difference between the greatest value and the least value of |z| is :
A
$$2\sqrt {13} $$
B
8
C
4 + $$\sqrt {13} $$
D
$$\sqrt {13} $$

Explanation

$$\left| {z - \left( {3 - 2i} \right)} \right| \le 4$$ represents a circle whose center is (3, $$-$$2) and radius = 4.

$$\left| z \right|$$ = $$\left| z -0\right|$$  represents the distance of point 'z' from origin (0, 0)



Suppose RS is the normal of the circle passing through origin 'O' and G is its center (3, $$-$$2).

Here, OR is the least distance

and OS is in the greatest distance

OR = RG $$-$$ OG and OS = OG + GS . . . . .(1)

As, RG = GS = 4

OG = $$\sqrt {{3^2} + \left( { - 2{)^2}} \right)} $$ = $$\sqrt {9 + 4} $$ = $$\sqrt {13} $$

From (1), OR = 4 $$-$$ $$\sqrt {13} $$ and OS = 4 + $$\sqrt {13} $$

So, required difference = $$\left( {4 + \sqrt {13} } \right)$$ $$-$$ $$\left( {4 - \sqrt {13} } \right)$$

= $$\sqrt {13} + \sqrt {13} $$ = $$2\sqrt {13} $$
2
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

The least positive integer n for which $${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1,$$ is :
A
2
B
3
C
5
D
6

Explanation

$${\left( {{{1 + i\sqrt 3 } \over {1 - i\sqrt 3 }}} \right)^n} = 1$$

$$ \Rightarrow \,\,\,\,\,{\left( {{{{{1 + i\sqrt 3 } \over 2}} \over {{{1 - i\sqrt 3 } \over 2}}}} \right)^n} = 1$$

We know, $$\omega $$ = $$-$$ $${{1 - i\sqrt 3 } \over 2}$$

and $$\omega $$2 = $$-$$ $${{\left( {1 + i\sqrt 3 } \right)} \over 2}$$

$$ \Rightarrow $$ $$\,\,\,\,$$ $${\left( {{{ - {\omega ^2}} \over { - \omega }}} \right)^n} = 1$$

$$ \Rightarrow $$ ($$\omega $$)n = 1 = $$\omega $$3

$$ \Rightarrow $$ $$\,\,\,\,$$ n = 3
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Let $$\alpha $$ and $$\beta $$ be two roots of the equation x2 + 2x + 2 = 0 , then $$\alpha ^{15}$$ + $$\beta ^{15}$$ is equal to :
A
-256
B
512
C
-512
D
256

Explanation

Given equation,

x2 + 2x + 2 = 0

$$ \therefore $$  x = $${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$$

x = $$-$$ 1 $$ \pm $$ i

$$ \therefore $$  $$\alpha $$ = $$-$$ 1 + i

and $$\beta $$ = $$-$$ 1 $$-$$ i

Note :

x + iy = r (cos$$\theta $$ + isin$$\theta $$)

$$ \therefore $$  (x + iy)n = rn (cosn$$\theta $$ + isinn$$\theta $$)

$$ \therefore $$  $$-$$ 1 + i = $$\sqrt 2 $$ [cos$${{3\pi } \over 4}$$ + isin$${{3\pi } \over 4}$$ ]

$$ \Rightarrow $$  ($$-$$ 1 + i)15 = $${\left( {\sqrt 2 } \right)^{15}}$$ [cos$$\left( {{{15.3\pi } \over 4}} \right) + i\sin \left( {{{15.3\pi } \over 4}} \right)$$]

And $$-$$1 $$-$$ i = $$\sqrt 2 $$ $$\left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right]$$

= $$\sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} \right]$$

$$ \therefore $$  ($$-$$1 $$-$$ i)15 = $${\left( {\sqrt 2 } \right)^{15}}\left[ {\cos \left( {{{15.3\pi } \over 4}} \right) - i\sin \left( {{{15.3\pi } \over 4}} \right)} \right]$$

Now

$$\alpha $$15 + $$\beta $$15

= ($$-$$1 + i)15 + ($$-$$ 1 $$-$$ i)15

=  $${\left( {\sqrt 2 } \right)^{15}}$$ $$\left[ {2\cos \left( {{{15.3\pi } \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} \right)} \right]$$

= $${\left( {\sqrt 2 } \right)^{15}} \times 2 \times - {1 \over {\sqrt 2 }}$$

= $$ - {\left( {\sqrt 2 } \right)^{14}}.2$$

= $$-$$ 27 $$ \times $$ 2

= $$-$$ 28

= $$-$$ 256
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Let
A = $$\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}$$
. Then the sum of the elements in A is :
A
$${5\pi \over 6}$$
B
$$\pi $$
C
$${3\pi \over 4}$$
D
$${{2\pi } \over 3}$$

Explanation

Given complex number,

$${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$$

$$ = {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$

$$ = {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$

$$ = {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$

As complex number is purely imaginary, So real part of this complex number is zero.

$$ \therefore $$   $${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$$ = 0

$$ \Rightarrow $$   $$3 - 4{\sin ^2}\theta = 0$$

$$ \Rightarrow $$   $$\sin \theta = \pm {{\sqrt 3 } \over 2}$$

as   $$\theta $$ $$ \in $$ $$\left( { - {\pi \over 2},\pi } \right)$$

$$ \therefore $$   $$\theta $$ $$=$$ $$-$$ $${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$$

$$ \therefore $$   Sum of those values of A is

$$ = - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$$

$$ = {{2\pi } \over 3}$$

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