Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations MCQs are provided here for students to help them in their preparation for the exams. The MCQs of Class 11 Maths Chapter 5 given here contain the right answers and explanations. These objective type questions cover all the concepts of the NCERT curriculum, which are important from the examination point of view.
Get MCQs for all chapters of Class 11 Maths here.
Students can solve the MCQs provided here for Chapter 5 of Class 11 Maths to improve their skills in identifying complex numbers and solving quadratic equations, whose roots are complex numbers.
Download PDF – Complex Numbers and Quadratic Equations MCQs
Also check:
- Complex numbers Class 11 Notes
- Quadratic equations for Class 11 Notes
- Important Questions for Class 11 Maths Chapter 5
MCQs for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers
1. The value of 1 + i2 + i4 + i6 + … + i2n is
(a) positive
(b) negative
(c) 0
(d) cannot be evaluated
Correct option: (d) cannot be evaluated
Solution:
1 + i2 + i4 + i6 + … + i2n = 1 – 1 + 1 – 1 + … (–1)n
This cannot be evaluated unless the value of n is known.
2. If a + ib = c + id, then
(a) a2 + c2 = 0
(b) b2 + c2 = 0
(c) b2 + d2 = 0
(d) a2 + b2 = c2 + d2
Correct option: (d) a2 + b2 = c2 + d2
Solution:
Given,
a + ib = c + id
⇒ |a + ib| = |c + id|
⇒ √(a2 + b2) = √(c2 + d2)
Squaring on both sides, we get;
a2 + b2 = c2 + d2
3. 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
Correct option: (b) 6
Solution:
The distance of the point representing z from the centre of the circle is |z – (-4 + i0)| = |z + 4|
According to the given,
|z + 4| ≤ 3
Now,
|z + 1| = |z + 4 – 3| ≤ |z + 4| + |-3| ≤ 3 + 3 ≤ 6
Hence, the greatest value of |z + 1| is 6.
4. The value of arg (x) when x < 0 is
(a) 0
(b) π/2
(c) π
(d) none of these
Correct option: (c) π
Solution:
Let z = x + 0i and x < 0
Since the point (-x, 0) lies on the negative side of the real axis,
|z| = |x + oi| = √[(-1)2 + 0)] = 1
∴ Principal argument (z) = π
Alternative method:
Let x = cos θ + i sin θ
For θ = π, x should be negative.
Thus, x < 0 for θ = π.
5. 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
Correct option: (a) -4
Solution:
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
6. Number of solutions of the equation z2 + |z|2 = 0 is
(a) 1
(b) 2
(c) 3
(d) infinitely many
Correct option: (d) infinitely many
Solution:
Given,
z2 + |z|2 = 0, z ≠0
⇒ (x + iy)2 + [√(x2 + y2)]2 = 0
⇒ x2 – y2 + i2xy + x2 + y2 = 0
⇒ 2x2 + i2xy = 0
⇒2x (x + iy) = 0
⇒ x = 0 or x + iy = 0 (not possible)
Therefore, x = 0 and z ≠0.
Thus, y can have any real value.
Hence, there exist infinitely many solutions.
7. If [(1 + i)/(1 – i)]x = 1, then
(a) x = 2n + 1, where n ∈ N
(b) x = 4n, where n ∈ N
(c) x = 2n, where n ∈ N
(d) x = 4n + 1, where n ∈ N
Correct option: (b) x = 4n, where n ∈ N
Solution:
Given,
[(1 + i)/(1 – i)]x = 1By rationalising the denominator,
[(1 + i)(1 + i)/ (1 – i)(1 + i)]x = 1 [(1 + i)2/ (1 – i + i – i2)]x = 1 [(1 + i2 + 2i)/(1 + 1)]x = 1 [(1 – 1 + 2i)/ 2]x = 1ix = 1
Thus, ix = i4n, where n is any positive integer.
8. If the complex number z = x + iy satisfies the condition |z + 1| = 1, then z lies on
(a) x-axis
(b) circle with centre (1, 0) and radius 1
(c) circle with centre (–1, 0) and radius 1
(d) y-axis
Correct option: (c) circle with centre (–1, 0) and radius 1
Solution:
Given,
z = x + iy
and
|z + 1| = 1
|x + iy + 1| = 1
⇒ |(x + 1) + iy| = 1
⇒ √[(x +1)2 + y2] = 1
Squaring on both sides,
(x + 1)2 + y2 = 1
This is the equation of a circle with centre (–1, 0) and radius 1.
9. The simplified value of (1 – i)3/(1 – i3) is
(a) 1
(b) -2
(c) -i
(d) 2i
Correct option: (b) -2
Solution:
(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
10. sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for:
(a) x = nπ
(b) x = [n + (1/2)] (Ï€/2)
(c) x = 0
(d) No value of x
Correct option: (d) No value of x
Solution:
Consider sin x + i cos 2x and cos x – i sin 2x are conjugate to each other.
So, sin x – i cos 2x = cos x – i sin 2x
On comparing real and imaginary parts of both sides, we get
⇒ sin x = cos x and cos 2x = sin 2x
⇒ sin x/cos x = 1 and (cos 2x/sin 2x) = 1
⇒ tan x = 1 and tan 2x = 1
Now, consider tan 2x = 1
Using the formula tan 2A = 2 tan A/(1 – tan2A),
(2 tan x)/(1 – tan2x) = 1
However, this is not possible for tan x = 1.
Therefore, for no value of x, sinx + i cos 2x and cos x – i sin 2x are conjugate to each other
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