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## Sunday, 21 February 2021

### CBSE Class 11 Maths - MCQ and Online Tests - Unit 5 - Complex Numbers and Quadratic Equations

#### CBSE Class 11 Maths  – MCQ and Online Tests – Unit 5 – Complex Numbers and Quadratic Equations

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#### CBSE Class 11 Maths  – MCQ and Online Tests – Unit 5 – Complex Numbers and Quadratic Equations

Question 1.
If arg (z) < 0, then arg (-z) – arg (z) =
(a) p
(b) -p
(c) -p/2
(d) p/2

Hint:
Given, arg (z) < 0
Now, arg (-z) – arg (z) = arg(-z/z)
? arg (-z) – arg (z) = arg(-1)
? arg (-z) – arg (z) = p {since sin p + i cos p = -1, So arg(-1) = p}

Question 2.
if x + 1/x = 1 find the value of x2000 + 1/x2000 is
(a) 0
(b) 1
(c) -1
(d) None of these

Hint:
Given x + 1/x = 1
? (xВ + 1) = x
? xВ – x + 1 = 0
? x = {-(-1) Б v(1В – 4 з 1 з 1)}/(2 з 1)
? x = {1 Б v(1 – 4)}/2
? x = {1 Б v(-3)}/2
? x = {1 Б v(-1)зv3}/2
? x = {1 Б iv3}/2 {since i = v(-1)}
? x = -w, -wВ
Now, put x = -w, we get
x2000 + 1/x2000 = (-w)2000 + 1/(-w)2000
= w2000 + 1/w2000
= w2000 + 1/w2000
= {(wГ)666 з wВ} + 1/{(wГ)666 з wВ}
= wВ + 1/wВ {since wГ = 1}
= wВ + wГ /wВ
= wВ + w
= -1 {since 1 + w + wВ = 0}
So, x2000 + 1/x2000 = -1

Question 3.
If the cube roots of unity are 1, ?, ?В, then the roots of the equation (x – 1)Г + 8 = 0 are
(a) -1, -1 + 2?, – 1 – 2?В
(b) – 1, -1, – 1
(c) – 1, 1 – 2?, 1 – 2?В
(d) – 1, 1 + 2?, 1 + 2?В

Answer: (c) – 1, 1 – 2?, 1 – 2?В
Hint:
Note that since 1, ?, and ?В are the cube roots of unity (the three cube roots of 1), they are the three solutions to xГ = 1 (note: ? and ?В are the two complex solutions to this)
If we let u = x – 1, then the equation becomes
uГ + 8 = (u + 2)(uВ – 2u + 4) = 0.
So, the solutions occur when u = -2 (giving -2 = x – 1 ? x = -1), or when:
uВ – 2u + 4 = 0,
which has roots, by the Quadratic Formula, to be u = 1 Б iv3
So, x – 1 = 1 Б iv3
? x = 2 Б iv3
Now, xГ = 1 when xГ – 1 = (x – 1)(xВ + x + 1) = 0, giving x = 1 and
xВ + x + 1 = 0
? x = (-1 Б iv3)/2
If we let ? = (-1 – iv3)/2 and ?2 = (-1 + iv3)/2
then 1 – 2? and 1 – 2?В yield the two complex solutions to (x – 1)Г + 8 = 0
So, the roots of (x – 1)Г + 8 are -1, 1 – 2?, and 1 – 2?В

Question 4:
The value of v(-25) + 3v(-4) + 2v(-9) is
(a) 13i
(b) -13i
(c) 17i
(d) -17i

Hint:
Given, v(-25) + 3v(-4) + 2v(-9)
= v{(-1) з (25)} + 3v{(-1) з 4} + 2v{(-1) з 9}
= v(-1) з v(25) + 3{v(-1) з v4} + 2{v(-1) з v9}
= 5i + 3з2i + 2з3i {since v(-1) = i}
= 5i + 6i + 6i
= 17i
So, v(-25) + 3v(-4) + 2v(-9) = 17i

Question 5.
if z lies on |z| = 1, then 2/z lies on
(a) a circle
(b) an ellipse
(c) a straight line
(d) a parabola

Answer: (a) a circle
Hint:
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 6.
If ? is an imaginary cube root of unity, then (1 + ? – ?В)7 equals
(a) 128 ?
(b) -128 ?
(c) 128 ?В
(d) -128 ?В

Answer: (d) -128 ?В
Hint:
Given ? is an imaginary cube root of unity.
So 1 + ? + ?В = 0 and ?Г = 1
Now, (1 + ? – ?В)7 = (-?В – ?В)7
? (1 + ? – ?2)7 = (-2?2)7
? (1 + ? – ?2)7 = -128 ?14
? (1 + ? – ?2)7 = -128 ?12 з ?2
? (1 + ? – ?2)7 = -128 (?3)4 ?2
? (1 + ? – ?2)7 = -128 ?2

Question 7.
The least value of n for which {(1 + i)/(1 – i)}n is real, is
(a) 1
(b) 2
(c) 3
(d) 4

Hint:
Given, {(1 + i)/(1 – i)}n
= [{(1 + i) з (1 + i)}/{(1 – i) з (1 + i)}]n
= [{(1 + i)В}/{(1 – iВ)}]n
= [(1 + iВ + 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 8.
Let z be a complex number such that |z| = 4 and arg(z) = 5p/6, then z =
(a) -2v3 + 2i
(b) 2v3 + 2i
(c) 2v3 – 2i
(d) -v3 + i

Answer: (a) -2v3 + 2i
Hint:
Let z = r(cos ? + i з sin ?)
Then r = 4 and ? = 5p/6
So, z = 4(cos 5p/6 + i з sin 5p/6)
? z = 4(-v3/2 + i/2)
? z = -2v3 + 2i

Question 9:
The value of i-999 is
(a) 1
(b) -1
(c) i
(d) -i

Hint:
Given, i-999
= 1/i999
= 1/(i996 з iГ)
= 1/{(i4)249 з i3}
= 1/{1249 з i3} {since i4 = 1}
= 1/i3
= i4/i3 {since i4 = 1}
= i
So, i-999 = i

Question 10.
Let z1 and z2 be two roots of the equation zВ + az + b = 0, z being complex. Further assume that the origin, z1 and z1 form an equilateral triangle. Then
(a) aВ = b
(b) aВ = 2b
(c) aВ = 3b
(d) aВ = 4b

Answer: (c) aВ = 3b
Hint:
Given, z1 and z2 be two roots of the equation zВ + az + b = 0
Now, z1 + z2 = -a and z1 з z2 = b
Since z1 and z2 and z3 from an equilateral triangle.
? z12 + z22 + z32 = z1 з z2 + z2 з z3 + z1 з z3
? z12+ z22 = z1 з z2 {since z3 = 0}
? (z1 + z2)В – 2z1 з z2 = z1 з z2
? (z1 + z2)В = 2z1 з z2 + z1 з z2
? (z1 + z2)В = 3z1 з z2
? (-a)В = 3b
? aВ = 3b

Question 11.
The value of v(-16) is
(a) -4i
(b) 4i
(c) -2i
(d) 2i

Hint:
Given, v(-16) = v(16) з v(-1)
= 4i {since i = v(-1) }

Question 12.
The value of v(-144) is
(a) 12i
(b) -12i
(c) Б12i
(d) None of these

Hint:
Given, v(-144) = v{(-1) з 144}
= v(-1) з v(144)
= i з 12 {Since v(-1) = i}
= 12i
So, v(-144) = 12i

Question 13.
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

Answer: (a) x = 7/2, y = 2/3
Hint:
Given, (3y – 2) + i(7 – 2x) = 0
Compare real and imaginary part, we get
3y – 2 = 0
? y = 2/3
and 7 – 2x = 0
? x = 7/2
So, the value of x = 7/2 and y = 2/3

Question 14.
Find real ? such that (3 + 2i з sin ?)/(1 – 2i з sin ?) is imaginary
(a) ? = np Б p/2 where n is an integer
(b) ? = np Б p/3 where n is an integer
(c) ? = np Б p/4 where n is an integer
(d) None of these

Answer: (b) ? = np Б p/3 where n is an integer
Hint:
Given,
(3 + 2i з sin ?)/(1 – 2i з sin ?) = {(3 + 2i з sin ?)з(1 – 2i з sin ?)}/(1 – 4iВ з sinВ ?)
(3 + 2i з sin ?)/(1 – 2i з sin ?) = {(3 – 4sinВ ?) + 8i з sin ?}/(1 + 4sinВ ?) …………. 1
Now, equation 1 is imaginary if
3 – 4sinВ ? = 0
? 4sinВ ? = 3
? sinВ ? = 3/4
? sin ? = Бv3/2
? ? = np Б p/3 where n is an integer

Question 15.
If {(1 + i)/(1 – i)}n = 1 then the least value of n is
(a) 1
(b) 2
(c) 3
(d) 4

Hint:
Given, {(1 + i)/(1 – i)}n = 1
? [{(1 + i) з (1 + i)}/{(1 – i) з (1 + i)}]n = 1
? [{(1 + i)В}/{(1 – iВ)}]n = 1
? [(1 + iВ + 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 16.
(1 – w + wВ)з(1 – wВ + w4)з(1 – w4 + w8) з …………… to 2n factors is equal to
(a) 2n
(b) 22n
(c) 23n
(d) 24n

Hint:
Given, (1 – w + wВ)з(1 – wВ + w4)з(1 – w4 + w8) з …………… to 2n factors
= (1 – w + w2)з(1 – w2 + w )з(1 – w + w2) з …………… to 2n factors
{Since w4 = w, w8 = w2}
= (-2w) з (-2wВ) з (-2w) з (-2wВ)з …………… to 2n factors
= (2В wГ)з(2В wГ)з(2В wГ) …………… to 2n factors
= (2В)n {since wГ = 1}
= 22n

Question 17.
The modulus of 5 + 4i is
(a) 41
(b) -41
(c) v41
(d) -v41

Hint:
Let Z = 5 + 4i
Now modulus of Z is calculated as
|Z| = v(5В + 4В)
? |Z| = v(25 + 16)
? |Z| = v41
So, the modulus of 5 + 4i is v41

Question 18.
The value of v(-144) is
(a) 12i
(b) -12i
(c) Б12i
(d) None of these

Hint:
Given, v(-144) = v{(-1)з144}
= v(-1) з v(144)
= i з 12 {Since v(-1) = i}
= 12i
So, v(-144) = 12i

Question 19:
The complex numbers sin x + i cos 2x are conjugate to each other for
(a) x = np
(b) x = 0
(c) x = (n + 1/2) p
(d) no value of x

Answer: (d) no value of x
Hint:
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 20.
The curve represented by Im(zВ) = 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

Answer: (d) a hyperbola
Hint:
Let z = x + iy
Now, zВ = (x + iy)В
? zВ = xВ – yВ + 2xy
Given, Im(zВ) = k
? 2xy = k
? xy = k/2 which is a hyperbola.

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