Practice AP Inter 1st Year Maths Study Material Chapter 4 Complex Numbers and Quadratic Equations MCQ to identify your strengths and weak areas.
AP Inter 1st Year Maths Complex Numbers and Quadratic Equations MCQ
Question 1.
The value of i-999 is
(1) 1
(2) -1
(3) i
(4) -i
Answer:
(3) i
Explanation:
i-999 = \(\frac{1}{i^{999}}=\frac{1}{i^3 i^{996}}=\frac{1}{-i \cdot\left(i^4\right)^{249}}=\frac{-1}{i}\)
= i
Question 2.
The multiplicative inverse of 1 + i is
(1) \(\frac{1}{2}\)(1 – i)
(2) \(\frac{1}{2}\)(1 + i)
(3) 1 – i
(4) i
Answer:
(1) \(\frac{1}{2}\)(1 – i)
Explanation:
Multiplicative inverse of 1 + i = \(\frac{1}{1+i}=\frac{1}{1+i} \times \frac{1-i}{1-i}=\frac{1-i}{1+1}=\frac{1}{2}\)
= \(\frac{1}{2}\)(1 – i)
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Question 3.
The modulus of 5 + 4i is
(1) 41
(2) -41
(3) \(\sqrt{41}\)
(4) \(\sqrt{-41}\)
Answer:
(3) \(\sqrt{41}\)
Explanation:
|5 + 4i| = \(\sqrt{5^2+4^2}=\sqrt{25+16}=\sqrt{41}\)
Question 4.
The value of \(\sqrt{-25}\) \(\sqrt{-9}\) is
(1) 15
(2) -15
(3) 15i
(4) None of these
Answer:
(2) -15
Explanation:
\(\frac{1}{2}\)
= 5i × 3i
= 15i2
= – 15
Question 5.
The Conjugate of \(\frac{2-i}{1-2 i}\) is
(1) \(\frac{4+3 i}{5}\)
(2) 4 – 3i
(3) \(\frac{4-3 i}{5}\)
(4) 1
Answer:
(3) \(\frac{4-3 i}{5}\)
Explanation:
Conjugate of z = z̄; z = \(\frac{2-i}{1-2 i}\)
z̄ = \(\frac{2-i}{1-2 i}=\frac{2+i}{1+2 i}=\frac{2+i}{1+2 i} \times \frac{1-2 i}{1-2 i}\)
= \(\frac{2-4 i+i-2 i^2}{1+4}=\frac{4-3 i}{5}\)
Question 6.
The value of (z + 3)(z̄ + 3) is equivalent to
(1) |z + 3|2
(2) |z – 3|
(3) z2 + 3
(4) None of these
Answer:
(1) |z + 3|2
Explanation:
(z + 3)\((\overline{z+3})\) = |z + 3|2 [∵ z.z̄ = |z|2]
Question 7.
Let x, y ∈ R, then x + iy is a non-real complex number if
(1) x = 0
(2) y = 0
(3) x ≠ 0
(4) None of these
Answer:
(4) None of these
Explanation:
y ≠ 0
Question 8.
If a + ib = c + id then
(1) a2 + c2 = 0
(2) b2 + c2 = 0
(3) b2 + d2 = 0
(4) a2 + b2 = c2 + d2
Answer:
(4) a2 + b2 = c2 + d2
Explanation:
a + ib = c + id
⇒ a = c, b = d
⇒ a2 = c2, b2 = d2
⇒ a2 + b2 = c2 + d2
Question 9.
The sum of the series i + i2 + i3 + ……………. upto 1000 terms is
(1) 1
(2) 0
(3) -1
(4) -2
Answer:
(2) 0
Explanation:
i + i2 + i3 + ……………. upto 1000 terms
= \(\frac{i\left(i^{1000}-1\right)}{i-1}=\frac{i\left(\left(i^4\right)^{250}-1\right)}{i-1}\)
= \(\frac{i(1-1)}{i-1}\) = 0
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Question 10.
The complex number z which satisfies the condition \(\left|\frac{\mathrm{i}+\mathrm{z}}{\mathrm{i}-\mathrm{z}}\right|\) = 1 lies on
1) Circle x2 + y2 = 1
2) The X-axis
3) The Y-axis
4) The line x + y = 1
Answer:
2) The X-axis
Explanation:
Given \(\left|\frac{\mathrm{i}+\mathrm{z}}{\mathrm{i}-\mathrm{z}}\right|\) = 1
⇒ |i + z| = |i – z|
⇒ |x + iy + i| = |i – (x + iy)|
⇒ x2 + (1 + y)2 = x2 + (1 – y)2
⇒ 1 + y2 + 2y = 1 + y2 – 2y
⇒ 4y = 0
⇒ y = 0
Which is the X-axis.
∴ Locus of z lies on the X-axis