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Class 11 Maths MCQ Questions of Complex Numbers and Quadratic Equations with Answers?

Answer 1

Students are advised to resolve the Class 11 Maths MCQ Questions of Complex Numbers and Quadratic Equations to understand different concepts. Practicing the MCQ Questions class 11 Maths chapter-wise with answers will boost your confidence thereby helping you score well within the exam.

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Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The value of \(\sqrt{(-16)}\) is

(a) -4i
(b) 4i
(c) -2i
(d) 2i

2. The value of \(\sqrt{(-144)}\) is

(a) -12i
(b) 12i
(c) -10i
(d) 2i

3.  The value of \(\sqrt{-25}+3\sqrt{-4}+2\sqrt{-9}\) is

(a) -17i
(b) 15i
(c) 17i
(d) 8i

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

5. 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

6. The value of i¹³⁵ is

(a) 1
(b) -1
(c) i
(d) -i

7. 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

8. The curve represented by Im(z²) = k, where k is a non-zero real number, is

(a) a pair of striaght line
(b) an ellipse
(c) a parabola
(d) a hyperbola 

9. The value of x and y if (3y – 2) + i(7 – 2x) = 0

(a) x = 7/2, y = 2/3
(b) x = 2/7, y = 2/3
(c) x = 7/2, y = 3/2
(d) x = 2/7, y = 3/2

10. Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary

(a) θ = nπ ± π/2 where n is an integer
(b) θ = nπ ± π/3 where n is an integer
(c) θ = nπ ± π/4 where n is an integer
(d) None of these

11. If arg (z) < 0, then arg (-z) – arg (z) =

(a) π
(b) -π
(c) -π/2
(d) π/2

12. The modulus of 3 + 4i is

(a) 5
(b) 25
(c) 15
(d) -5

13. The modulus of 5 + 3i is

(a) \(\sqrt{34}\)
(b) 25
(c) 15
(d) -5

14. The value of 1+ i² + i⁴ + i⁶ + i⁸ + ........+ i²⁰

(a) 1
(b) 2
(c) -1
(d) 3

15. Number of solutions of the equation z²+|z|² =0  is

(a) 1
(b) 2
(c) 3
(d) infinitely many

16. If ∣z−4∣<∣z−2∣, its solution is given by

(a) Re(z)>0
(b) Re(z)<0
(c) Re(z)>3
(d) Re(z)>2

17. If a + ib = c + id, then

(a) a² + c² = 0
(b) b² + c² = 0
(c) b² + d² = 0
(d) a² + b² = c² + d²

18.  If a complex number z lies in the interior or on the boundary of a circle of radius 3 units and centre (– 4, 0), the greatest value of |z +1| is

(a) 4
(b) 6
(c) 3
(d) 10

19. If 1 – i, is a root of the equation x2 + ax + b = 0, where a, b ∈ R, then the value of a – b is

(a) -4
(b) 0
(c) 2
(d) 1

20. The simplified value of (1 – i)³/(1 – i³) is

(a) 1
(b) -2
(c) -i
(d) 2i

Answer 2

1. Answer: (b) 4i

Explanation: Given,

  \(=\sqrt{(16)}\times\sqrt{(-1)}\)

= 4i {since i = \(\sqrt{(-1)}\)}

2. Answer: (b) 12i

Explanation: Given,

  \(=\sqrt{(144)}\times\sqrt{(-1)}\)

= 12i {since i = \(\sqrt{(-1)}\)}

3.  Answer: (c) 17i

Explanation: \(\sqrt{-25}+3\sqrt{-4}+2\sqrt{-9}\)

\(=5\sqrt{-1}+6\sqrt{-1}+6\sqrt{-1}\) \(\therefore \sqrt{-1}=i\)

=5i+6i+6i

= 17i

4. Answer: (a) a circle

Explanation: Let \(w=\frac{2}{z}\)

 \(w=2\overset{-}{z}\) .....(Multiplying and dividing by \(\overset{-}{z}\) and ∣z∣² =1

Now locus of w represents a circle.

5. Answer: (d) -128 ω²

Explanation: ​​​​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

6. Answer: (d) -i

Explanation: 135 leaves remainder as 3 when it is divided by 4. 

∴ i¹³⁵ = i³ = -i

7. Answer: (c) a² = 3b

Explanation:  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

8. Answer: (d) a hyperbola 

Explanation: Let z = x + iy

Now, z² = (x + iy)²

⇒ z² = x² – y² + 2xy

Given, Im(z²) = k

⇒ 2xy = k

 ⇒ xy = k/2 which is a hyperbola.

9. Answer: (a) x = 7/2, y = 2/3

Explanation: Given, (3y – 2) + i(7 – 2x) = 0

3y – 2 = 0

⇒ y = 2/3 and 

7 – 2x = 0

⇒ x = 7/2

So, the value of x = 7/2 and y = 2/3

10. Answer: (a) x = 7/2, y = 2/3

Explanation: 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) …………. (i)

3 – 4sin² θ = 0

⇒ 4sin² θ = 3

⇒ sin² θ = 3/4

⇒ sin θ = \(\pm\frac{\sqrt3}{2}\)

⇒ θ = nπ ± π/3 where n is an integer

Answer 3

11. Answer: (a) π

Explanation: Given, arg (z) < 0

Now, arg (-z) – arg (z) = arg(-z/z)

⇒ arg (-z) – arg (z) = arg(-1)

⇒ arg (-z) – arg (z) = π {since sin π + i cos π = -1, So arg(-1) = π}

12. Answer: (a) 5

Explanation: \(\sqrt{3^2+4^2}\)

= 5

13. Answer: (a) 

Explanation: \(\sqrt{5^2+3^2}\)

= \(\sqrt{34}\)

14. Answer: (a) 1

Explanation: 1+ i² + i⁴ + i⁶ + i⁸ + i¹⁰ + i¹² + i¹⁴ + i¹⁶ + i¹⁸.+ i²⁰

= 1+(−1)+1+(−1)+1+(−1)+1+(−1)+1+(−1)+1

= 1

15. Answer: (d) infinitely many

Explanation: z²+|z|²=0 ,

z ≠ 0

⇒x²−y² +i²xy+x²+y² =0

⇒2x²+i²xy = 0

⇒2x(x+iy) = 0

⇒x=0 or x+iy=0 (not possible)

∴x=0 and z≠0

So y can have any real value.

Hence infinitely many solutions.

16. Answer: (c) Re(z)>3

Explanation: Given ∣z−4∣<∣z−2∣ all complex numbers lie farther from (2,0) than from (4,0) as

∣z−4∣<∣z−2∣

So they lie on the side of (z,0) which is closer to (4,0) than (2,0)

z = x+iy

⇒ x>3

∴Re(z)>3 

17. Answer: (d) a² + b² = c² + d²

Explanation: a + ib = c + id

⇒ |a + ib| = |c + id|

⇒ \(\sqrt{(a^2+b^2)}\) = \(\sqrt{(c^2+d^2)}\)

Squaring on both sides, we get;

a² + b² = c² + d²

18.  Answer: (b) 6

Explanation: The distance of the point representing z from the centre of the circle is |z – (-4 + i0)| = |z + 4|

|z + 4| ≤ 3

|z + 1| = |z + 4 – 3| ≤ |z + 4| + |-3| ≤ 3 + 3 ≤ 6

Hence, the greatest value of |z + 1| is 6.

19. Answer: (a) -4

Explanation: Given that 1 – i is the root of x² + ax + b = 0.

Thus, 1 + i is also the root of the given equation since non-real complex roots occur in conjugate pairs.

Sum of roots = −a/1 = (1 – i) + (1 + i)

⇒ a = – 2

Product of roots, b/1 = (1 – i)(1 + i)

b = 1 – i²

b = 1 + 1 {since i2 = -1}

⇒ b = 2

Now, a – b = -2 – 2 = -4

20. Answer: (b) -2

Explanation: (1 – i)³/(1 – i3)

= (1 – i)³/(1³ – i³)

= (1 – i)³/ [(1 – i)(1 + i + i²)]

= (1 – i)²/(1 + i – 1)

= (1 – i)²/i

= (1 + i² – 2i)/i

= (1 – 1 – 2i)/i

= -2i/i

= -2

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