CBSE Class 11 Mathematics Chapter 5 Complex Numbers Multiple Choice Questions with Answers. MCQ Class 11 Mathematics Complex Numbers with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Mathematics Complex Numbers MCQs with Answers to know their preparation level.
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MCQ Class 11 Mathematics Complex Numbers with Answers - Set - 1
Question 1:
The complex numbers sin x + i cos 2x are conjugate to each other for
(a) x = nπ
(b) x = 0
(c) x = (n + 1/2) π
(d) no value of x
Correct Answer – (D)
Given complex number = sin x + i cos 2x
Conjugate of this number = sin x – i cos 2x
Now, sin x + i cos 2x = sin x – i cos 2x
⇒ sin x = cos x and sin 2x = cos 2x {comparing real and imaginary part}
⇒ tan x = 1 and tan 2x = 1
Now both of them are not possible for the same value of x.
So, there exist no value of x
Question 2 :
The value of i-999 is
(a) 1
(b) -1
(c) i
(d) -i
Correct Answer – (C)
Given, i-999
= 1/i999
= 1/(i996 × i³)
= 1/{(i4)249 × i3}
= 1/{1249 × i3} {since i4 = 1}
= 1/i3
= i4/i3 {since i4 = 1}
= i
So, i-999 = i
Question 3 :
The least value of n for which {(1 + i)/(1 – i)}n is real, is
(a) 1
(b) 2
(c) 3
(d) 4
Correct Answer – (B)
Given, {(1 + i)/(1 – i)}n
= [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]n
= [{(1 + i)²}/{(1 – i²)}]n
= [(1 + i² + 2i)/{1 – (-1)}]n
= [(1 – 1 + 2i)/{1 + 1}]n
= [2i/2]n
= in
Now, in is real when n = 2 {since i2 = -1 }
So, the least value of n is 2
Question 4 :
if z lies on |z| = 1, then 2/z lies on
(a) a circle
(b) an ellipse
(c) a straight line
(d) a parabola
Correct Answer – (A)
Let w = 2/z
Now, |w| = |2/z|
=> |w| = 2/|z|
=> |w| = 2
This shows that w lies on a circle with center at the origin and radius 2 units.
Question 5 :
The value of √(-144) is
(a) 12i
(b) -12i
(c) ±12i
(d) None of these
Correct Answer – (A)
Given, √(-144) = √{(-1) × 144}
= √(-1) × √(144)
= i × 12 {Since √(-1) = i}
= 12i
So, √(-144) = 12i
MCQ Class 11 Mathematics Complex Numbers with Answer
Question 6 :
Let z1 and z2 be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z1 and z1 form an equilateral triangle. Then
(a) a² = b
(b) a² = 2b
(c) a² = 3b
(d) a² = 4b
Correct Answer – (C)
Given, z1 and z2 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
Question 7 :
Let z be a complex number such that |z| = 4 and arg(z) = 5π/6, then z =
(a) -2√3 + 2i
(b) 2√3 + 2i
(c) 2√3 – 2i
(d) -√3 + i
Correct Answer – (A)
Let z = r(cos θ + i × sin θ)
Then r = 4 and θ = 5π/6
So, z = 4(cos 5π/6 + i × sin 5π/6)
⇒ z = 4(-√3/2 + i/2)
⇒ z = -2√3 + 2i
Question 8 :
If ω is an imaginary cube root of unity, then (1 + ω – ω²)7 equals
(a) 128 ω
(b) -128 ω
(c) 128 ω²
(d) -128 ω²
Correct Answer – (D)
Given ω is an imaginary cube root of unity.
So 1 + ω + ω² = 0 and ω³ = 1
Now, (1 + ω – ω²)7 = (-ω² – ω²)7
⇒ (1 + ω – ω2)7 = (-2ω2)7
⇒ (1 + ω – ω2)7 = -128 ω14
⇒ (1 + ω – ω2)7 = -128 ω12 × ω2
⇒ (1 + ω – ω2)7 = -128 (ω3)4 ω2
⇒ (1 + ω – ω2)7 = -128 ω2
Question 9 :
The value of √(-25) + 3√(-4) + 2√(-9) is
(a) 13i
(b) -13i
(c) 17i
(d) -17i
Correct Answer – (C)
= √{(-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
= 17i
So, √(-25) + 3√(-4) + 2√(-9) = 17i
Question 10 :
The value of √(-16) is
(a) -4i
(b) 4i
(c) -2i
(d) 2i
Correct Answer – (B)
Given, √(-16) = √(16) × √(-1)
= 4i {since i = √(-1) }
- NCERT Solutions Class 11 Mathematics Complex Numbers with Answers : Exercise 5.1
- NCERT Solutions Class 11 Mathematics Complex Numbers with Answers : Exercise 5.2
- NCERT Solutions Class 11 Mathematics Complex Numbers with Answers : Exercise 5.3
- NCERT Solutions Class 11 Mathematics Complex Numbers with Answers : Exercise 5 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
