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 - 3
Question 1:
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
Correct Answer – (C)
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
Question 2 :
The value of {-√(-1)}4n+3, n ∈ N is
(a) i
(b) -i
(c) 1
(d) -1
Correct Answer – (A)
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
Question 3 :
The value of [i19 + (1/i)25]² is
(a) -1
(b) -2
(c) -3
(d) -4
Correct Answer – (D)
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
Question 4 :
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
Correct Answer – (B)
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
Question 5 :
The value of ii is
(a) 0
(b) e-π
(c) 2e-π/2
(d) e-π/2
Correct Answer – (D)
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
MCQ Class 11 Mathematics Complex Numbers with Answer
Question 6 :
Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is real
(a) π
(b) nπ
(c) nπ/2
(d) 2nπ
Correct Answer – (B)
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
Question 7 :
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
Correct Answer – (C)
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
Question 8 :
If {(1 + i)/(1 – i)}ⁿ = 1 then the least value of n is
(a) 1
(b) 2
(c) 3
(d) 4
Correct Answer – (D)
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
Question 9 :
The value of √(-25) + 3√(-4) + 2√(-9) is
(a) 13 i
(b) -13 i
(c) 17 i
(d) -17 i
Correct Answer – (C)
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
Question 10 :
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
Correct Answer – (C)
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
- 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
