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JEE Main

Explore popular questions from Complex Numbers and Quadratic Equations for JEE Main. This collection covers Complex Numbers and Quadratic Equations previous year JEE Main questions hand picked by experienced teachers.

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Complex Numbers and Quadratic Equations

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Q 1. If {tex} \left( \frac { 1 + i } { 1 - i } \right) ^ { x } = 1 , {/tex} then

A

{tex} x = 2 n + 1 , {/tex} where {tex} n {/tex} is any positive integer

{tex} x = 4 n , {/tex} where {tex} n {/tex} is any positive integer

C

{tex} x = 2 n , {/tex} where {tex} n {/tex} is any positive integer

D

{tex} x = 4 n + 1 , {/tex} where {tex} n {/tex} is any positive integer

Explanation

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Q 2. Let {tex} z _ { 1 } {/tex} and {tex} z _ { 2 } {/tex} be the roots of the equation {tex} z ^ { 2 } + a z + b = 0 , z {/tex} being complex number. Further, assume that the origin, {tex} z _ { 1 } {/tex} and {tex} z _ { 2 } {/tex} form an equilateral triangle, then

A

{tex} a ^ { 2 } = 4 b {/tex}

B

{tex} a ^ { 2 } = b {/tex}

C

{tex} a ^ { 2 } = 2 b {/tex}

{tex} a ^ { 2 } = 3 b {/tex}

Explanation



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Q 3. If {tex} z , \omega {/tex} are two non-zero complex numbers such that {tex} | z \omega | = 1 {/tex} and {tex} \arg ( z ) - \arg ( \omega ) = \pi / 2 , {/tex} then {tex} \overline { z } \omega {/tex} is equal to

A

{tex}-1{/tex}

B

{tex}1{/tex}

{tex} -i {/tex}

D

{tex} i {/tex}

Explanation


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Q 4. If {tex} \left| z ^ { 2 } - 1 \right| = | z | ^ { 2 } + 1 , {/tex} then {tex} z {/tex} lies on

A

the real axis

the imaginary axis

C

a circle

D

an ellipse

Explanation


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Q 5. If the cube roots of unity are 1 ,{tex}\omega{/tex}, {tex} \omega ^ { 2 } , {/tex} then the roots of the equation {tex} ( x - 1 ) ^ { 3 } + 8 = 0 {/tex} are

A

{tex} - 1,1 + 2 \omega , 1 + 2 \omega ^ { 2 } {/tex}

{tex} - 1,1 - 2 \omega , 1 - 2 \omega ^ { 2 } {/tex}

C

{tex} - 1 , - 1 , - 1 {/tex}

D

{tex} - 1 , - 1 + 2 \omega, - 1 , - 2 \omega ^ { 2 } {/tex}

Explanation

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Q 6. If {tex} z _ { 1 } {/tex} and {tex} z _ { 2 } {/tex} are two non-zero complex numbers such that {tex} \left| z _ { 1 } + z _ { 2 } \right| = \left| z _ { 1 } \right| + \left| z _ { 2 } \right| , {/tex} then {tex} \arg \left( z _ { 1 } \right) - \arg \left( z _ { 2 } \right) {/tex} is equal to

A

{tex} - \frac { \pi } { 2 } {/tex}

{tex}0{/tex}

C

{tex} - \pi {/tex}

D

{tex} \frac { \pi } { 2 } {/tex}

Explanation


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Q 7. If {tex} \omega = \frac { z } { z - \frac { 1 } { 3 } i } {/tex} and {tex} | \omega | = 1 , {/tex} then {tex} z {/tex} lies on

A

a parabola

a straight line

C

a circle

D

an ellipse

Explanation

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Q 8. If {tex} z ^ { 2 } + z + 1 = 0 , {/tex} where {tex} z {/tex} is a complex number, then the value of {tex} \left( z + \frac { 1 } { z } \right) ^ { 2 } + \left( z ^ { 2 } + \frac { 1 } { z ^ { 2 } } \right) ^ { 2 } + \left( z ^ { 3 } + \frac { 1 } { z ^ { 3 } } \right) ^ { 2 } + \cdots + \left( z ^ { 6 } + \frac { 1 } { z ^ { 6 } } \right) ^ { 2 } {/tex} is

A

18

B

54

C

6

12

Explanation


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Q 9. If {tex} | z + 4 | \leq 3 , {/tex} then the maximum value of {tex} | z + 1 | {/tex} is

A

0

B

4

C

10

6

Explanation

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Q 10. The conjugate of a complex number is {tex} \frac { 1 } { i - 1 } {/tex} . Then that complex number is

A

{tex} \frac { - 1 } { i - 1 } {/tex}

B

{tex} \frac { 1 } { i + 1 } {/tex}

{tex} \frac { - 1 } { i + 1 } {/tex}

D

{tex} \frac { 1 } { i - 1 } {/tex}

Explanation

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Q 11. If {tex} \left| z - \frac { 4 } { z } \right| = 2 , {/tex} then the maximum value of {tex} | z | {/tex} is equal to

A

{tex} \sqrt { 3 } + 1 {/tex}

{tex} \sqrt { 5 } + 1 {/tex}

C

{tex}2{/tex}

D

{tex} 2 + \sqrt { 2 } {/tex}

Explanation

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Q 12. The number of complex number z such that {tex} | z - 1 | = | z + 1 | = | z - i | {/tex} equals

A

{tex} \infty {/tex}

B

0

C

1

2

Explanation

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Q 13. If {tex} \alpha {/tex} and {tex} \beta {/tex} are the roots of the equation {tex} x ^ { 2 } - x + 1 = 0 {/tex} , then {tex} \alpha ^ { 2009 } + \beta ^ { 2009 } = ? {/tex}

A

2

B

-2

C

-1

1

Explanation

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Q 14. If {tex} \omega ( \neq 1 ) {/tex} is a cube root of unity, and {tex} ( 1 + \omega ) ^ { 7 } = A + B \omega . {/tex} Then {tex} ( A , B ) {/tex} equals

{tex} ( 1,1 ) {/tex}

B

{tex} ( 1,0 ) {/tex}

C

{tex} ( - 1,1 ) {/tex}

D

{tex} ( 0,1 ) {/tex}

Explanation

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Q 15. Let {tex} \alpha , \beta {/tex} be real and {tex} z {/tex} be a complex number. If {tex} z ^ { 2 } + \alpha z + \beta = 0 {/tex} has two distinct roots on the line {tex} R z = 1 {/tex} , then it is necessary that

A

{tex} \beta \in ( - 1,0 ) {/tex}

B

{tex} | \beta | = 1 {/tex}

{tex} \beta \in ( 1 , \infty ) {/tex}

D

{tex} \beta \in ( 0,1 ) {/tex}

Explanation

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Q 16. If {tex} z {/tex} is a complex number of unit modulus and argument {tex} \theta {/tex} , then {tex} \arg \left( \frac { 1 + z } { 1 + \overline { z } } \right) {/tex} equals

A

{tex}0{/tex}

B

{tex} \frac { \pi } { 2 } - \theta {/tex}

{tex}\theta{/tex}

D

{tex} \pi - \theta {/tex}

Explanation


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Q 17. If , but , then the equation whose roots are and is

A

C

D

None of these

Explanation


, . α, β are roots of . Therefore ,

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Q 18. If one root of the equation is the square of the other, then

A

B

C

Explanation

Let α and be the roots then ,
Now

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Q 19. If the roots of the equation x2 - 2ax + a2 + a - 3 = 0 are less than 3 then

a < 2

B

2 ≤ a ≤ 3

C

3 < a ≤ 4

D

a > 4

Explanation

Since roots are less than a real number, roots must be real
⇒ 4a2 - 4(a2 + a - 3) ≥ 0 ⇒ a ≤ 3. ... (1)
Let f(x) = x2 - 2ax + a2 + a - 3. Since 3 lie outside the roots,
f(3) > 0 ⇒ a < 2 or a > 3. . . . .(2)
From (1), (2) we have a < 2

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Q 20. Sum of the real roots of the equation x2 + 5|x| + 6 = 0

A

Equals to 5

B

Equals to 10

C

Equals to -5

Does not exit

Explanation

Since x2, 5|x| and 6 are positive so x2 + 5|x| + 6 = 0 does not have any real root. Therefore sum does not exist

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Q 21. If = 7 then x =

A

B

log7(3/4)

D

log7(4/3)

Explanation

x = log(3/4) 7 = =

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Q 22. The value of 'a' for which the sum of the squares of the roots of the equation x2 - (a - 2)x - a - 1 = 0 assumes the least value is :

A

0

1

C

2

D

3

Explanation

Here α+β=a-2,αβ=-(a-1), Now,α22=(α+β)2- 2αβ
= (a- 2)2 + 2 (a +1) = a2 - 2a + 6 = (a-1)2 + 5
So α2 + β2 will be maximum at a = 1.

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Q 23. If a, b, c are in G.P. then the equation ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a common root if d/a , e/b, f/c are in :

AP

B

GP

C

HP

D

None of these

Explanation

Here b = (As a, b, c → GP)
So, ax2 + 2 + c = 0 ⇒ (x+)2 = 0
or x = -/, Now, substituting in eqn (2);
dc/a - 2e/ + f = 0 ⇒ d/a - 2e/ + f/c = 0
⇒ d/a -2e/b + f/c = 0

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Q 24. If the roots of the equation = are such that α + β = 0, then the value of λ is-

B

c

C

D

Explanation

(λ + 1) (x2 - bx) = (λ - 1) (ax - c)
⇒ λx2 + (-bλ - b - aλ + a) x + cλ - c = 0
Since α + β = 0, we have
= 0
⇒ -bλ - b -aλ + a = 0 ⇒ λ = .

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Q 25. The equation x + = 1 + has-

No real root

B

One real root

C

Two equal roots

D

Infinitely

Explanation

We have x + = 1 +
⇒ x = 1 provided x = 1 ≠ 0 i.e., x ≠ 1.
The given equation has no solution.