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

Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.

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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

Answer

Answer: (a) p
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

Answer

Answer: (c) -1
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

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

Answer

Answer: (c) 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

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

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

Answer

Answer: (b) 2
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

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

Answer

Answer: (c) 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

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

Answer

Answer: (b) 4i
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

Answer

Answer: (a) 12i
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

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

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

Answer

Answer: (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

Answer

Answer: (b) 22n
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

Answer

Answer: (c) 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

Answer

Answer: (a) 12i
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

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

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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