MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers

Complex Numbers and Quadratic Equations Class 11 Maths MCQs Questions with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Complex Numbers and Quadratic Equations Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 5 Quiz

Class 11 Maths Chapter 5 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Complex Numbers and Quadratic Equations Class 11 Maths MCQ online test

Q1.	Let z₁ and z₂ be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z₁ and z₂ form an equilateral triangle. Then
	A.a² = b
	B.a² = 2b
	C.a² = 3b
	D.a² = 4b

Ans: a² = 3b
Given, z₁ and z₁ be two roots of the equation z²+ az + b = 0
Now, z₁ + z₂ = -a and z₁ × z₂ = b
Since z₁ and z₂ and z₃ from an equilateral triangle.
⇒ z₁² + z₂² + z₃² = z₁ × z₂ + z₂ × z₃ + z₁ × z₃
⇒ z₁² + z₂² = z₁ × z₂ {since z₃ = 0}
⇒ (z₁ + z₂)² – 2z₁ × z₂ = z₁ × z₂
⇒ (z₁ + z₂)² = 2z₁ × z₂ + z₁ × z₂
⇒ (z₁ + z₂)² = 3z₁ × z₂
⇒ (-a)² = 3b
⇒ a² = 3b

Q2.	The value of iⁱ is
	A.0
	B.e^-π
	C.2e^-π/2
	D.e^-π/2

Ans: e^-π/2
Let A = iⁱ
⇒ log A = i log i
⇒ log A = i log(0 + i)
⇒ log A = i [log 1 + i tan^-1 ∞]
⇒ log A = i [0 + i π/2]
⇒ log A = -π/2
⇒ A = e^-π/2

Q3.	The value of √(-25) + 3√(-4) + 2√(-9) is
	A.13 i
	B.-13 i
	C.17 i
	D.-17 i

Ans: 17 i
Given, √(-25) + 3√(-4) + 2√(-9)
= √{(-1) × (25)} + 3√{(-1) × 4} + 2√{(-1) × 9}
= √(-1) × √(25) + 3{√(-1) × √4} + 2{√(-1) × √9}
= 5i + 3 × 2i + 2 × 3i {since √(-1) = i}
= 5i + 6i + 6i
= 17 i
So, √(-25) + 3√(-4) + 2√(-9) = 17 i

Q4.	If the cube roots of unity are 1, ω and ω², then the value of (1 + ω / ω²)³ is
	A.1
	B.-1
	C.ω
	D.ω²

Ans: -1
Given, the cube roots of unity are 1, ω and ω²
So, 1 + ω + ω² = 0
and ω³ = 1
Now, {(1 + ω)/ ω²}³ = {-ω²/ ω²}³ = {-1}³ = -1

Q5.	If {(1 + i)/(1 – i)}ⁿ = 1 then the least value of n is
	A.1
	B.2
	C.3
	D.4

Ans: 4
Given, {(1 + i)/(1 – i)}ⁿ = 1
⇒ [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]ⁿ = 1
⇒ [{(1 + i)²}/{(1 – i²)}]ⁿ = 1
⇒ [(1 + i² + 2i)/{1 – (-1)}]ⁿ = 1
⇒ [(1 – 1 + 2i)/{1 + 1}]ⁿ = 1
⇒ [2i/2]ⁿ = 1
⇒ iⁿ = 1
Now, iⁿ is 1 when n = 4
So, the least value of n is 4

Q6.	The value of [i¹⁹ + (1/i)²⁵]² is
	A.-1
	B.-2
	C.-3
	D.-4

Ans: -4
Given, [i¹⁹ + (1/i)²⁵]²
= [i¹⁹ + 1/i²⁵]²
= [i¹⁶ × i³ + 1/(i²⁴ × i)]²
= [1 × i³ + 1/(1 × i)]² {since i⁴ = 1}
= [i³ + 1/i]²
= [i² × i + 1/i]²
= [(-1) × i + 1/i]² {since i² = -1}
= [-i + 1/i]²
= [-i + i⁴ /i]²
= [-i + i³]²
= [-i + i² × i]²
= [-i + (-1) × i]²
= [-i – i]²
= [-2i]²
= 4i²
= 4 × (-1)
= -4
So, [i¹⁹ + (1/i)²⁵]² = -4

Q7.	If z and w be two complex numbers such that \|z\| ≤ 1, \|w\| ≤ 1 and \|z + iw\| = \|z – iw\| = 2, then z equals {w is congugate of w}
	A.1 or i
	B.i or – i
	C.1 or – 1
	D.i or – 1

Ans: 1 or – 1
Given |z + iw| = |z – iw| = 2 {w is congugate of w}
⇒ |z – (-iw)| = |z – (iw)| = 2
⇒ |z – (-iw)| = |z – (-iw)|
So, z lies on the perpendicular bisector of the line joining -iw and -iw.
Since, -iw is the mirror in the x-axis, the locus of z is the x-axis.
Let z = x + iy and y = 0
⇒ |z| < 1 and x² + 0² < 0
⇒ -1 ≤ x ≤ 1
So, z may take value 1 or -1

Q8.	The value of {-√(-1)}⁴ⁿ⁺³, n ∈ N is
	A.i
	B.-i
	C.1
	D.-1

Ans: i
Given, {-√(-1)}⁴ⁿ⁺³
= {-i}⁴ⁿ⁺³ {since √(-1) = i}
= {-i}⁴ⁿ × {-i}³
= {(-i)⁴}ⁿ × (-i³) {since i⁴ = 1}
= 1ⁿ ×(-i × i²)
= -i × (-1) {since i² = -1}
= i

Q9.	Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is real
	A.π
	B.nπ
	C.nπ/2
	D.2nπ

Ans: nπ
Given,
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 + 2i × sin θ)×(1 – 2i × sin θ)}/(1 – 4i² × sin² θ)
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 – 4sin² θ) + 8i × sin θ}/(1 + 4sin² θ) …………. 1
Now, equation 1 is real if sin θ = 0
⇒ sin θ = sin nπ
⇒ θ = nπ

Q10.	If i = √(-1) then 4 + 5(-1/2 + i√3/2)³³⁴ + 3(-1/2 + i√3/2)³⁶⁵ is equals to
	A.1 – i√3
	B.-1 + i√3
	C.i√3
	D.-i√3

Ans: i√3
Given, 4 + 5(-1/2 + i√3/2)³³⁴ + 3(-1/2 + i√3/2)³⁶⁵
= 4 + 5w³³⁴ + 3w³⁶⁵ {since w = -1/2 + i√3/2}
= 4 + 5w + 3w² {since w³ = 1}
= 4 + 5(-1/2 + i√3/2) + 3(-1/2 – i√3/2) {since w² = (-1/2 – i√3/2)}
= i√3

Q11.	The real part of the complex number √9 + √(-16) is
	A.3
	B.-3
	C.4
	D.-4

Ans: 3
Given, √9 + √(-16) = √9 + √(16) × √(-1)
= 3 + 4i {since i = √(-1)}
So, the real part of the complex number is 3

Q12.	The modulus of 5 + 4i is
	A.41
	B.-41
	C.√41
	D.-√41

Ans: √41
Let Z = 5 + 4i
Now modulus of Z is calculated as
|Z| = √(5² + 4²)
⇒ |Z| = √(25 + 16)
⇒ |Z| = √41
So, the modulus of 5 + 4i is √41

Q13.	The modulus of 1 + i√3 is
	A.1
	B.2
	C.3
	D.None of these

Ans: 2
Let Z = 1 + i√3
Now modulus of Z is calculated as
|Z| = √{1² + (√3)²}
⇒ |Z| = √(1 + 3)
⇒ |Z| = √4
⇒ |Z| = 2
So, the modulus of 1 + i√3 is 2

Q14.	The value of {-√(-1)}⁴ⁿ⁺³, n ∈ N is
	A.i
	B.-i
	C.1
	D.-1

Ans: i
Given, {-√(-1)}⁴ⁿ⁺³
= {-i}⁴ⁿ⁺³ {since √(-1) = i}
= {-i}⁴ⁿ × {-i}³
= {(-i)⁴}ⁿ ×(-i³) {since i⁴ = 1}
= 1ⁿ × (-i × i²)
= -i × (-1) {since i² = -1}
= i

Q15.	If ω is cube root of unity (ω ≠ 1) , then the least value of n where n is a positive integer such that (1 + ω²)ⁿ = (1 + ω⁴)ⁿ is
	A.2
	B.3
	C.5
	D.6

Ans: 3
Given ω is an imaginary cube root of unity.
So 1 + ω + ω² = 0 and ω³ = 1
Now, (1 + ω²)ⁿ = (1 + ω⁴)ⁿ
⇒ (-1)ⁿ ×(ω)ⁿ = (1 + ω × ω³)ⁿ
⇒ (-1)ⁿ × (ω)ⁿ = (1 + ω)ⁿ
⇒ (-1)ⁿ × (ω)ⁿ = (-ω²)ⁿ
⇒ (-1)ⁿ × (ω)ⁿ = (-1)ⁿ × ω²ⁿ
⇒ ωⁿ = ω²ⁿ
Since ω³ = 1, So the least value of n is 3

Q16.	The value of i⁹ + i¹⁰ + i¹¹ + i¹² is
	A.i
	B.2i
	C.0
	D.1

Ans: 0
Given, i⁹ + i¹⁰ + i¹¹ + i¹²
= i9 (1 + i + i2 + i3 )
= i9 (1 + i – 1 – i ) {since i2 = (-1) and i4 = 1}
= i9 × 0
= 0

Q17.	If a = cos α + i sin α and b = cos β + i sin β , then the value of 1/2(ab + 1/ ab) is
	A.sin (α + β)
	B.cos (α + β)
	C.sin (α – β)
	D.cos (α – β)

Ans: cos (α + β)
Given a = cos α + i sin α and b = cos β + i sin β
Now, 1/a = 1/(cos α + i sin α)
⇒ 1/a = {1 × (cos α – i sin α)/{(cos α + i sin α) × (cos α + i sin α)}
⇒ 1/a = (cos α – i sin α)/(cos² α + i sin² α)
⇒ 1/a = (cos α – i sin α)
Again, 1/b = 1/(cos β + i sin β)
⇒ 1/b = {1 × (cos β – i sin β)/{(cos β + i sin β) × (cos β + i sin β)}
⇒ 1/b = (cos β – i sin β)/(cos² β + i sin² β)
⇒ 1/b = (cos β – i sin β)
Now, ab = (cos α + i sin α) × (cos β + i sin β)
⇒ ab = cos α × cos β + i cos α × sin β + i sin α × cos β – sin α × sin β
Again, 1/ab = (cos α – i sin α) × (cos β – i sin β)
⇒ 1/ab = cos α × cos β – i cos α × sin β – i sin α × cos β – sin α × sin β
Now, ab + 1/ab = cos α × cos β + i cos α × sin β + i sin α × cos β – sin α × sin β + cos α × cos β – i cos α × sin β – i sin α × cos β – sin α × sin β
⇒ ab + 1/ab = 2(cos α × cos β – sin α × sin β)
⇒ 1/2(ab + 1/ ab) = 2(cos α × cos β – sin α × sin β)/2
⇒ 1/2(ab + 1/ ab) = cos α × cos β – sin α × sin β
⇒ 1/2(ab + 1/ ab) = cos(α + β)

Q18.	The polar form of -1 + i is
	A.√2(cos π/2 + i × sin π/2)
	B.√2(cos π/4 + i × sin π/4)
	C.√2(cos 3π/2 + i × sin 3π/2)
	D.√2(cos 3π/4 + i × sin 3π/4)

Ans: √2(cos 3π/4 + i × sin 3π/4)
The polar form of a com plex number = r(cos θ + i × sin θ)
Given, complex number = -1 + i
Let x + iy = -1 + i
Now, x = -1, y = 1
Now, r = √{(-1)² + 1²} = √(1 + 1) = √2
and tan θ = y/x
⇒ tan θ = 1/(-1)
⇒ tan θ = -1
⇒ θ = 3π/4
Now, polar form is √2(cos 3π/4 + i × sin 3π/4)

Q19.	For all complex numbers z₁, z₂ satisfying \|z₁\| = 12 and \|z₂ – 3 – 4i\| = 5, the minimum value of \|z₁ – z₂\| is
	A.0
	B.2
	C.7
	D.17

Ans: 2
Given For all complex numbers z₁, z₂ satisfying |z₁| = 12 and |z₂ – 3 – 4i| = 5
Now, mod(z₁) = 12 represents a circle centred at 0 and radius 12
mod(z₂ – 3 – 4i) = 5 represents a circle centred at (3, 4) and radius 5
This circle passes through the origin. Distance of diametrically opposite end is 10
So, the minimum value (z₁ – z₂) = 2