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All questions of Complex Numbers & Quadratic Equations for A Level Exam
Imaginary part of −i(3i + 2) isa)−2b)2c)3d)−3Correct answer is option 'A'. Can you explain this answer?
|
Geetika Shah answered |
(-i)(3i) +2(-i) =-3(i^2)-2i =-3(-1)-2i =3-2i since i=√-1 =3+(-2)i comparing with a+bi,we get b=(-2)
Write in the simplest form: (i)-997- a)-i
- b)1
- c)-1
- d)i
Correct answer is 'A'. Can you explain this answer?
Write in the simplest form: (i)-997
a)
-i
b)
1
c)
-1
d)
i
|
Muskaan Mishra answered |
I^-997= 1/i^997, 1/(i^4)^249 × i, since i^4 = 1, i^4/i= i^3= -i
= Find the reciprocal (or multiplicative inverse) of -2 + 5i - a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Find the reciprocal (or multiplicative inverse) of -2 + 5i
a)
b)
c)
d)
|
|
Gaurav Kumar answered |
-2 + 5i
multiplicative inverse of -2 + 5i is
1/(-2+5i)
= 1/(-2+5i) * ((-2-5i)/(-2-5i))
= -2-5i/(-2)^2 -(5i)^2
= -2-5i/4-(-25)
= -2-5i/4+25
= -2-5i/29
= -2/29 -5i/29
multiplicative inverse of -2 + 5i is
1/(-2+5i)
= 1/(-2+5i) * ((-2-5i)/(-2-5i))
= -2-5i/(-2)^2 -(5i)^2
= -2-5i/4-(-25)
= -2-5i/4+25
= -2-5i/29
= -2/29 -5i/29
The argument of the complex number -1 – √3- a)π/3
- b)4π/3
- c)-π/3
- d)-2π/3
Correct answer is option 'D'. Can you explain this answer?
The argument of the complex number -1 – √3
a)
π/3
b)
4π/3
c)
-π/3
d)
-2π/3
|
Om Desai answered |
z = a + ib
a = -1, b = -√3
(-1, -√3) lies in third quadrant.
Arg(z) = -π + tan-1(b/a)
= - π + tan-1(√3)
= - π + tan-1(tan π/3)
= - π + π/3
= -2π/3
a = -1, b = -√3
(-1, -√3) lies in third quadrant.
Arg(z) = -π + tan-1(b/a)
= - π + tan-1(√3)
= - π + tan-1(tan π/3)
= - π + π/3
= -2π/3
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
a)
b)
c)
d)
|
Lohit Matani answered |
(x+iy)(x−iy) = (a+ib)(a−ib)/(c+id)(c−id)
⇒x2−iy2 = √[(a2−i2b2)/(c2−i2d2)]
⇒x2+y2 = √[(a2+b2)/(c2+d2)] [1i2 = -1]
(x2+y2)2 = (a2+b2)/(c2+d2)
⇒x2−iy2 = √[(a2−i2b2)/(c2−i2d2)]
⇒x2+y2 = √[(a2+b2)/(c2+d2)] [1i2 = -1]
(x2+y2)2 = (a2+b2)/(c2+d2)
Find the real numbers x and y such that : (x + iy)(3 + 2i) = 1 + i- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Find the real numbers x and y such that : (x + iy)(3 + 2i) = 1 + i
a)
b)
c)
d)
|
Suresh Iyer answered |
(x + iy)(3 + 2i) = (1 + i)
x + iy = (1 + i)/(3 + 2i)
x + iy = [(1 + i) * (3 - 2i)] / [(3 + 2i)*(3 - 2i)]
x + iy = (3 + 3i - 2i + 2) / [(3)2 + (2)2]
x + iy = (5 + i)/[ 9 + 4]
= (5 + i) / 13
=> 13x + 13iy = 5+i
13x = 5 13y = 1
x = 5/13 y = 1/13
x + iy = (1 + i)/(3 + 2i)
x + iy = [(1 + i) * (3 - 2i)] / [(3 + 2i)*(3 - 2i)]
x + iy = (3 + 3i - 2i + 2) / [(3)2 + (2)2]
x + iy = (5 + i)/[ 9 + 4]
= (5 + i) / 13
=> 13x + 13iy = 5+i
13x = 5 13y = 1
x = 5/13 y = 1/13
Express the following in standard form : 
- a)3+3i
- b)2 + 2i
- c)1 + 2i
- d)0 + 2i
Correct answer is option 'D'. Can you explain this answer?
Express the following in standard form :
a)
3+3i
b)
2 + 2i
c)
1 + 2i
d)
0 + 2i
|
|
Gaurav Kumar answered |
(8 - 4i) - (-2 - 3i) + (-10 + 3i)
=> 8 - 4i + 2 + 3i-10 + 3i
=> 8 + 2 - 10 - 4i + 3i + 3i =>0 + 2i
=> 8 - 4i + 2 + 3i-10 + 3i
=> 8 + 2 - 10 - 4i + 3i + 3i =>0 + 2i
Find the real numbers x and y such that :a)
b)
c)
d)
Correct answer is option 'c'. Can you explain this answer?
|
Hansa Sharma answered |
(x + iy) (3 + 2i)
= 3x + 2xi + 3iy + 3i*y = 1+i
= 3x-2y + i(2x+3y) = 1+i
= 3x-2y-1 = 0 ; 2x + 3y -1 = 0
on equating real and imaginary parts on both sides
on solving two equations
x= 5/13 ; y = 1/13
= 3x + 2xi + 3iy + 3i*y = 1+i
= 3x-2y + i(2x+3y) = 1+i
= 3x-2y-1 = 0 ; 2x + 3y -1 = 0
on equating real and imaginary parts on both sides
on solving two equations
x= 5/13 ; y = 1/13
If If ω is a non real cube root of unity and (1+ω)9 = a+bω;a,b ∈ R, then a and b are respectively the numbers :- a)1, 0
- b)-1, 0
- c)0, 1
- d)0, -1
Correct answer is option 'B'. Can you explain this answer?
If If ω is a non real cube root of unity and (1+ω)9 = a+bω;a,b ∈ R, then a and b are respectively the numbers :
a)
1, 0
b)
-1, 0
c)
0, 1
d)
0, -1
|
Aryan Khanna answered |
Since w is the cube root of unity.
(1+w)9 =A+Bw
⇒(−w2)9 =A+Bw
⇒−(w3)6 = A+Bw
⇒−1=A+Bw
∴ A= -1 & B=0
(1+w)9 =A+Bw
⇒(−w2)9 =A+Bw
⇒−(w3)6 = A+Bw
⇒−1=A+Bw
∴ A= -1 & B=0
For a complex number x + iy, if x > 0 and y < 0 then the number lies in the.- a)Second quadrant
- b)Third quadrant
- c)First quadrant
- d)Fourth quadrant
Correct answer is option 'D'. Can you explain this answer?
For a complex number x + iy, if x > 0 and y < 0 then the number lies in the.
a)
Second quadrant
b)
Third quadrant
c)
First quadrant
d)
Fourth quadrant
|
Naina Sharma answered |
When x > 0 & y < 0, the quadrant contains positive X-axis and negative Y-axis
Therefore, the point lies in the fourth quadrant.
Therefore, the point lies in the fourth quadrant.
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
a)
b)
c)
d)
|
Manish Aggarwal answered |
(a + ib)1/2 = (x + iy)
Squaring both sides,
a + ib = (x + iy)2
a + ib = x2 - y2 + 2ixy
Equating real and imaginary
a = x2 - y2 b = 2xy............(1)
Using (x2 + y2)2 = (x2 - y2)2 + 4xy
(x2 + y2)2 = a2 + b2
(x2 + y2) = (a2 + b2)1/2.......(2)
Adding (1) and (2)
2x2 = (a2 + b2)1/2 + a
x = +-{1/2(a2 + b2)1/2 + a}1/2
Subtract (2) from (1)
2y2 = (a2 + b2)1/2 - a
y = x = +-{1/2(a2 + b2)1/2 - a}1/2
Therefore, (a+ib)1/2 = x+iy
=> +-{1/2(a2 + b2)1/2 + a}1/2 + i+-{1/2(a2 + b2)1/2 - a}1/2
Squaring both sides,
a + ib = (x + iy)2
a + ib = x2 - y2 + 2ixy
Equating real and imaginary
a = x2 - y2 b = 2xy............(1)
Using (x2 + y2)2 = (x2 - y2)2 + 4xy
(x2 + y2)2 = a2 + b2
(x2 + y2) = (a2 + b2)1/2.......(2)
Adding (1) and (2)
2x2 = (a2 + b2)1/2 + a
x = +-{1/2(a2 + b2)1/2 + a}1/2
Subtract (2) from (1)
2y2 = (a2 + b2)1/2 - a
y = x = +-{1/2(a2 + b2)1/2 - a}1/2
Therefore, (a+ib)1/2 = x+iy
=> +-{1/2(a2 + b2)1/2 + a}1/2 + i+-{1/2(a2 + b2)1/2 - a}1/2
- a)10
- b)4
- c)8
- d)6
Correct answer is option 'D'. Can you explain this answer?
a)
10
b)
4
c)
8
d)
6
|
Ciel Knowledge answered |
Let √(5 – 12i) = x + iy
Squaring both sides, we get
5 – 12i = x2 + 2ixy +(iy)2 = x2 – y2 + 2xyi.
Comparing real and imaginary parts , we get
5 = x2 – y2 ———– (1) and xy = – 6 ———— (2)
Squaring (1), we get
25 = (x2 – y2)2 = (x2 + y2)2 – 4x2y2
⇒ 25 = (x2 + y2)2 – 4(– 6)2
⇒ (x2 + y2)2 = 169
⇒ x2 + y2 = 13 ———- (3)
Adding (1) and (3) we get
2x2 = 18
⇒ x = ± 3.
Subtracting (1) from (3) we get
2y2 = 8
⇒ y = ± 2.
Hence, square root of √(5 – 12i) is (3 – 2i)
Similarly, √(5 + 12i) is (3 + 2i)
√(5 + 12i) + √(5 – 12i)
⇒ (3 + 2i) + (3 - 2i)
⇒ 6
Squaring both sides, we get
5 – 12i = x2 + 2ixy +(iy)2 = x2 – y2 + 2xyi.
Comparing real and imaginary parts , we get
5 = x2 – y2 ———– (1) and xy = – 6 ———— (2)
Squaring (1), we get
25 = (x2 – y2)2 = (x2 + y2)2 – 4x2y2
⇒ 25 = (x2 + y2)2 – 4(– 6)2
⇒ (x2 + y2)2 = 169
⇒ x2 + y2 = 13 ———- (3)
Adding (1) and (3) we get
2x2 = 18
⇒ x = ± 3.
Subtracting (1) from (3) we get
2y2 = 8
⇒ y = ± 2.
Hence, square root of √(5 – 12i) is (3 – 2i)
Similarly, √(5 + 12i) is (3 + 2i)
√(5 + 12i) + √(5 – 12i)
⇒ (3 + 2i) + (3 - 2i)
⇒ 6
Let 
Suppose α1 and B1are the roots of the equation x2 – 2x sec a + 1 = 0 and α2 and β2 are the roots of the equation x2 + 2x tanθ – 1 = 0. If α1 > b1 and α2 > β2, then α1 + β2 equals (JEE Adv. 2016)- a)2 (secθ – tanθ)
- b)2 secθ
- c)–2 tanθ
- d)0
Correct answer is option 'C'. Can you explain this answer?
Let Suppose α1 and B1are the roots of the equation x2 – 2x sec a + 1 = 0 and α2 and β2 are the roots of the equation x2 + 2x tanθ – 1 = 0. If α1 > b1 and α2 > β2, then α1 + β2 equals (JEE Adv. 2016)
Suppose α1 and B1are the roots of the equation x2 – 2x sec a + 1 = 0 and α2 and β2 are the roots of the equation x2 + 2x tanθ – 1 = 0. If α1 > b1 and α2 > β2, then α1 + β2 equals (JEE Adv. 2016)a)
2 (secθ – tanθ)
b)
2 secθ
c)
–2 tanθ
d)
0
|
Krishna Iyer answered |
x2 – 2x secθ + 1 = 0 ⇒ x = secθ ± tanθ and x2 + 2x tanθ – 1 = 0 ⇒ x = –tanθ ± secθ
α1,β1 are roots of x2 – 2x secθ + 1 = 0 and α1> b1
∴ α1 = secθ – tanθ and b1 = secθ + tanθ α2,
β2 are roots of x2 + 2x tanθ – 1 = 0 and α2 > β2
∴ α2 = -tanθ + secθ, β2 = – tanθ – secθ
∴ α1 + β2 = secθ – tanθ – tanθ – secθ = – 2tanθ
β2 are roots of x2 + 2x tanθ – 1 = 0 and α2 > β2
∴ α2 = -tanθ + secθ, β2 = – tanθ – secθ
∴ α1 + β2 = secθ – tanθ – tanθ – secθ = – 2tanθ
Express the following in standard form :
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Express the following in standard form :
a)
b)
c)
d)
|
Pooja Shah answered |
first write above equation in complex number format , ie using iota
(3-4i) / (2-3i)*(2+3i) / (2+3i) = (6+9i-8i+12) / 13=(18/13)+(i/13)
(3-4i) / (2-3i)*(2+3i) / (2+3i) = (6+9i-8i+12) / 13=(18/13)+(i/13)
Multiplicative inverse of the non zero complex number x + iy (x,y ∈ R,)- a)
- b)
- c)
- 
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Multiplicative inverse of the non zero complex number x + iy (x,y ∈ R,)
a)
b)
c)
- d)
none of these
|
Impact Learning answered |
Let u be multiplicative inverse
zu = 1
u = 1/z
u = 1/(x+iy)
Rationalise it
[1/(x+iy)]*[(x-iy)/(x-iy)]
= (x-iy)/(x2+y2)
u = x/(x2+y2) +i(-y)/(x2+y2)
zu = 1
u = 1/z
u = 1/(x+iy)
Rationalise it
[1/(x+iy)]*[(x-iy)/(x-iy)]
= (x-iy)/(x2+y2)
u = x/(x2+y2) +i(-y)/(x2+y2)
Let (x0, y0) be the solution of the following equations
(2x)ℓn2 = (3y)ℓn3
3ℓnx =2ℓny
Then x0 is (2011)- a)
- b)
- c)
- d)6
Correct answer is option 'C'. Can you explain this answer?
Let (x0, y0) be the solution of the following equations
(2x)ℓn2 = (3y)ℓn3
3ℓnx =2ℓny
Then x0 is (2011)
(2x)ℓn2 = (3y)ℓn3
3ℓnx =2ℓny
Then x0 is (2011)
a)
b)
c)
d)
6
|
|
Manish Aggarwal answered |
We have (2x)ℓn2 = (3y)ℓn3
⇒ ℓn2. ℓn2x = ℓn3. ℓn3y
⇒ ℓn2. ℓn2x = ℓn3. (ℓn3 + ℓny) ...(1)
Also given 3ℓnx = 2lny
⇒ ℓn2. ℓn2x = ℓn3. ℓn3y
⇒ ℓn2. ℓn2x = ℓn3. (ℓn3 + ℓny) ...(1)
Also given 3ℓnx = 2lny
⇒ ℓnx. ℓn3 = ℓny. ℓn2 ⇒ ℓny = 
Substituting this value of ℓny in equation (1), we get
ℓn2. ℓn2x = ℓn3 
⇒ (ℓn2)2 ℓn2x = (ℓn3)2 ℓn2 + (ℓn3)2 ℓnx
⇒ (ℓn2)2 ℓn2x = (ℓn3)2 (ℓn2 + ℓnx)
⇒ (ℓn2)2 ℓn2x – (ℓn3)2 ℓn2x = 0
⇒ [(ℓn2)2 – (ℓn3)2] ℓn2x = 0 ⇒ ℓn2x = 0
⇒ 2x = 1 or x = 
If α is a complex a number such that α2+α+1 = 0 then α31 is- a)0
- b)1
- c)α
- d)α2
Correct answer is option 'C'. Can you explain this answer?
If α is a complex a number such that α2+α+1 = 0 then α31 is
a)
0
b)
1
c)
α
d)
α2
|
Knowledge Hub answered |
Since α3 +1=(α+1)(α2 −α+1), using the given information, we get α3 +1 = 0
If z1 = 2 + i, z2 = 1 + 3i, then Re ( z1 - z2) =- a)ι
- b)1
- c)2 ι
- d)2
Correct answer is option 'B'. Can you explain this answer?
If z1 = 2 + i, z2 = 1 + 3i, then Re ( z1 - z2) =
a)
ι
b)
1
c)
2 ι
d)
2
|
|
Devika Roy answered |
The numbers in the questions are not very clear.
z1 = 2 + i
z2 = 1+3i
z1 – z2 = (2 – 1) + i (1 – 3)
= 1 – 2i
z1 = 2 + i
z2 = 1+3i
z1 – z2 = (2 – 1) + i (1 – 3)
= 1 – 2i
is positive, then S contains
Which of the following is a finite set?- a)Set of roots of the equation x2 – 25 = 0
- b)The set of numbers which are multiples of 5
- c)The set of lines parallel to y – axis
- d)Set of natural numbers
Correct answer is option 'A'. Can you explain this answer?
Which of the following is a finite set?
a)
Set of roots of the equation x2 – 25 = 0
b)
The set of numbers which are multiples of 5
c)
The set of lines parallel to y – axis
d)
Set of natural numbers
|
Jatin Mehra answered |
Understanding Finite Sets
In set theory, a finite set is one that contains a limited number of elements, while an infinite set has unlimited elements. Let’s analyze the options given:
Option A: Set of roots of the equation x² - 25 = 0
- The roots of this equation can be found by solving it:
- x² - 25 = 0
- This factors to (x - 5)(x + 5) = 0
- The solutions are x = 5 and x = -5.
- Hence, the set of roots is {5, -5}, which contains exactly two elements. This qualifies as a finite set.
Option B: The set of numbers which are multiples of 5
- This set includes numbers like 0, 5, 10, 15, and so on, extending infinitely in both directions (negative and positive).
- Therefore, this is an infinite set.
Option C: The set of lines parallel to the y-axis
- Lines parallel to the y-axis can be represented by the equation x = k, where k can take any real value.
- Since there are infinitely many real numbers, this set is also infinite.
Option D: Set of natural numbers
- The natural numbers include 1, 2, 3, 4, and so forth, extending infinitely.
- Thus, this set is infinite as well.
Conclusion
Among the options analyzed, only Option A represents a finite set. The other options describe sets that are infinite in nature.
In set theory, a finite set is one that contains a limited number of elements, while an infinite set has unlimited elements. Let’s analyze the options given:
Option A: Set of roots of the equation x² - 25 = 0
- The roots of this equation can be found by solving it:
- x² - 25 = 0
- This factors to (x - 5)(x + 5) = 0
- The solutions are x = 5 and x = -5.
- Hence, the set of roots is {5, -5}, which contains exactly two elements. This qualifies as a finite set.
Option B: The set of numbers which are multiples of 5
- This set includes numbers like 0, 5, 10, 15, and so on, extending infinitely in both directions (negative and positive).
- Therefore, this is an infinite set.
Option C: The set of lines parallel to the y-axis
- Lines parallel to the y-axis can be represented by the equation x = k, where k can take any real value.
- Since there are infinitely many real numbers, this set is also infinite.
Option D: Set of natural numbers
- The natural numbers include 1, 2, 3, 4, and so forth, extending infinitely.
- Thus, this set is infinite as well.
Conclusion
Among the options analyzed, only Option A represents a finite set. The other options describe sets that are infinite in nature.
The inequality | z − 6 | < | z − 2 | represents the region given by- a)Re(z) < 2
- b)Re(z) > 3
- c)Re(z) > 4
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The inequality | z − 6 | < | z − 2 | represents the region given by
a)
Re(z) < 2
b)
Re(z) > 3
c)
Re(z) > 4
d)
none of these
|
Rajiv Reddy answered |
Let, z = x + iy
Putting in the given inequality,
∣(x−4) − iy∣ < ∣(x−2) + iy∣
squaring both sides,
⇒ (x−4)2 + y2 < (x−2)2 + y2
⇒ −8x + 16 < −4x + 4
⇒ 4x > 12
⇒ x > 3
⇒ Re(z) > 3
⇒ −8x + 16 < −4x + 4
⇒ 4x > 12
⇒ x > 3
⇒ Re(z) > 3
then a and b are respectively :- a)64 and - 64√3
- b)128 and 128√3
- c)512 and - 512√3
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
then a and b are respectively :a)
64 and - 64√3
b)
128 and 128√3
c)
512 and - 512√3
d)
none of these
|
|
Om Desai answered |
(√3 + i)10 = a + ib
Z = √3 + i = rcosθ + i rsinθ
⇒ √3 = rcosθ i = rsinθ
⇒ (√3)2 + (1)2 = r2cos2θ + r2sin2θ
⇒ 4 = r2
⇒ r = 2
tan = 1/√3
⇒ tan π/6
Therefore, Z = √3 + i = 2(cos π/6 + i sin π/6)
(Z)10 = √3 + i = (2cos π/6 + 2i sin π/6)10
= 210 (cos π/6 + i sin π/6)10
210 (cos 10π/6 + i sin 10π/6)
= 210 (cos(2π - π/3) + i sin(2π - π/3))
= 210 (cos π/3 - i sin π/3)
= 210 (1/2 - i√3/2)
29(1 - i√3)
a = 29 = 512
b = - 29(√3) = -512√3
Z = √3 + i = rcosθ + i rsinθ
⇒ √3 = rcosθ i = rsinθ
⇒ (√3)2 + (1)2 = r2cos2θ + r2sin2θ
⇒ 4 = r2
⇒ r = 2
tan = 1/√3
⇒ tan π/6
Therefore, Z = √3 + i = 2(cos π/6 + i sin π/6)
(Z)10 = √3 + i = (2cos π/6 + 2i sin π/6)10
= 210 (cos π/6 + i sin π/6)10
210 (cos 10π/6 + i sin 10π/6)
= 210 (cos(2π - π/3) + i sin(2π - π/3))
= 210 (cos π/3 - i sin π/3)
= 210 (1/2 - i√3/2)
29(1 - i√3)
a = 29 = 512
b = - 29(√3) = -512√3
- a)9p/10
- b)5p/6
- c)6p/5
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
a)
9p/10
b)
5p/6
c)
6p/5
d)
none of these
|
|
Lohit Matani answered |
sin(6π/5) + i(1+cos(6π/5)) or −sin(π/5) + i(1−cos(π/5)) lies in the second quadrant of complex plane hence its argument is given as
arg(x+iy) = π − tan-1 |y/x| (∀ x<0,y≥0)
= π−tan-1 |1−cos(π/5)/sin(π/5)|
= π−tan-1 |2sin2(π/10)/2sin(π/10)cos(π/10)|
= π−tan-1 |sin(π/10)/cos(π/10)|
= π − tan-1 |tan(π/10)|
= π−tan-1 (tan(π/10)) (∵ tan(π/10)>0)
= π−π/10(∵−π/2≤tan−1(x)≤π/2)
= 9π/10
arg(x+iy) = π − tan-1 |y/x| (∀ x<0,y≥0)
= π−tan-1 |1−cos(π/5)/sin(π/5)|
= π−tan-1 |2sin2(π/10)/2sin(π/10)cos(π/10)|
= π−tan-1 |sin(π/10)/cos(π/10)|
= π − tan-1 |tan(π/10)|
= π−tan-1 (tan(π/10)) (∵ tan(π/10)>0)
= π−π/10(∵−π/2≤tan−1(x)≤π/2)
= 9π/10
then z69 is equal to :
A man walks a distance of 3 units from the origin towards the north-east (N 45° E) direction. From there, he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point P. Then the position of P in the Argand plane is (2007 -3 marks)- a)3eip/4 + 4i
- b)(3 – 4i)eip/4
- c)(4 + 3i)eip/4
- d)(3 + 4i)eip/4
Correct answer is option 'D'. Can you explain this answer?
A man walks a distance of 3 units from the origin towards the north-east (N 45° E) direction. From there, he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point P. Then the position of P in the Argand plane is (2007 -3 marks)
a)
3eip/4 + 4i
b)
(3 – 4i)eip/4
c)
(4 + 3i)eip/4
d)
(3 + 4i)eip/4
|
|
Tejas Verma answered |
If p, q, r are +ve and are in A.P., the roots of quadratic equation px2 + qx + r = 0 are all real for (1994)- a)
- b)
- c)all p and r
- d)no p an d r
Correct answer is option 'B'. Can you explain this answer?
If p, q, r are +ve and are in A.P., the roots of quadratic equation px2 + qx + r = 0 are all real for (1994)
a)
b)
c)
all p and r
d)
no p an d r
|
|
Geetika Shah answered |
For real roots q2 - 4 pr≥0
(∵ p, q, r are in A.P.)
If α and β (α < β) are the roots of the equation x2 + bx + c = 0, where c < 0 < β, then (2000S)- a)0 < α < β
- b)α < 0 < β < | α |
- c)α < β < 0
- d)α < 0 < | α | < β
Correct answer is option 'B'. Can you explain this answer?
If α and β (α < β) are the roots of the equation x2 + bx + c = 0, where c < 0 < β, then (2000S)
a)
0 < α < β
b)
α < 0 < β < | α |
c)
α < β < 0
d)
α < 0 < | α | < β
|
Chirag Verma answered |
Given c < 0 < b and α + β = – b ....(1) αβ = c
....(2)
....(2)
From (2), c < 0 ⇒ αβ < 0 ⇒ either a is -ve or β is - ve and second ;quantity is positive.
from (1), b > 0 ⇒ – b < 0 ⇒ α + β < 0 ⇒ the sum is negative
⇒ modules of nengative quantity is > modulus of positive quantity but α< β is given.
Therefore, it is clear that α is negative and β is positive and modulus of α is greater than modulus of β ⇒ α< 0 <β< |α|
Therefore, it is clear that α is negative and β is positive and modulus of α is greater than modulus of β ⇒ α< 0 <β< |α|
The value of the sum 
, where i = 
, equals (1998 - 2 Marks)- a)i
- b) i – 1
- c)–i
- d) 0
Correct answer is option 'B'. Can you explain this answer?
The value of the sum , where i = , equals (1998 - 2 Marks)
, where i = , equals (1998 - 2 Marks)a)
i
b)
i – 1
c)
–i
d)
0
|
|
Ananya Das answered |
This forms a G.P.
Sum of G.P.= i(1 + i)
For all complex numbers z1, z2 satisfying |z1|=12 and | z2-3-4i| = 5, the minimum value of |z1-z2| is (2002S)- a)0
- b)2
- c)7
- d)17
Correct answer is option 'B'. Can you explain this answer?
For all complex numbers z1, z2 satisfying |z1|=12 and | z2-3-4i| = 5, the minimum value of |z1-z2| is (2002S)
a)
0
b)
2
c)
7
d)
17
|
|
Sneha Singh answered |
Min.dis. is difference of their radii
A point z = P(x,y) lies on the negative direction of y axis, so amp(z) =- a)π
- b)-π
- c)3(π/2)
- d)π/2
Correct answer is option 'C'. Can you explain this answer?
A point z = P(x,y) lies on the negative direction of y axis, so amp(z) =
a)
π
b)
-π
c)
3(π/2)
d)
π/2
|
|
Geetika Shah answered |
P(-x,-y) lies on the 3rd or 4th quadrant
So, according to the question, points lies in 3rd quadrant
i.e. 3(π/2)
So, according to the question, points lies in 3rd quadrant
i.e. 3(π/2)
If P and Q are two sets such that n(P) = 120, n(Q) = 50 and n(P ∪ Q) = 140 then, n(P ∩ Q) is:- a)90
- b)30
- c)20
- d)70
Correct answer is option 'B'. Can you explain this answer?
If P and Q are two sets such that n(P) = 120, n(Q) = 50 and n(P ∪ Q) = 140 then, n(P ∩ Q) is:
a)
90
b)
30
c)
20
d)
70
|
Seelam Yogitha answered |
N(AUB)=N(A)+N(B)-N(A^B)=120 +50-140=30
If points corresponding to the complex numbers z1, z2, z3 and z4 are the vertices of a rhombus, taken in order, then for a non-zero real number k- a)z1 – z3 = i k( z2 –z4)
- b)z1 – z2 = i k( z3 –z4)
- c)z1 + z3 = k( z2 +z4)
- d)z1 + z2 = k( z3 +z4)
Correct answer is option 'A'. Can you explain this answer?
If points corresponding to the complex numbers z1, z2, z3 and z4 are the vertices of a rhombus, taken in order, then for a non-zero real number k
a)
z1 – z3 = i k( z2 –z4)
b)
z1 – z2 = i k( z3 –z4)
c)
z1 + z3 = k( z2 +z4)
d)
z1 + z2 = k( z3 +z4)
|
|
Riya Banerjee answered |
AC = z3 = z1 eiπ
= z1 (cosπ + i sinπ)
= z3 = z1(-1 + i(0))
= z3 = -z1
AC = z1 - z3
BC = z2 - z4
(z1 - z3)/(z2 - z4) = k
(z1 - z3) = eiπ/2(z2 - z4)
(z1 - z3) k(cosπ/2 + sinπ/2) (z2 - z4)
z1 - z3 = ki(z2 - z4)
z1 - z3 = ik(z2 - z4)
= z1 (cosπ + i sinπ)
= z3 = z1(-1 + i(0))
= z3 = -z1
AC = z1 - z3
BC = z2 - z4
(z1 - z3)/(z2 - z4) = k
(z1 - z3) = eiπ/2(z2 - z4)
(z1 - z3) k(cosπ/2 + sinπ/2) (z2 - z4)
z1 - z3 = ki(z2 - z4)
z1 - z3 = ik(z2 - z4)
Let a, b, c be the sides of a triangle where a ≠ b ≠ c and λ ∈ R. If the roots of the equation x2 + 2(a + b + c)x + 3λ (ab + bc + ca) = 0 are real, then (2006 - 3M, –1)- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Let a, b, c be the sides of a triangle where a ≠ b ≠ c and λ ∈ R. If the roots of the equation x2 + 2(a + b + c)x + 3λ (ab + bc + ca) = 0 are real, then (2006 - 3M, –1)
a)
b)
c)
d)
|
|
Chirag Verma answered |
∵ a, b, c are sides of a triangle and a ≠ b≠c
∴
Similarly, we have
b2 + c2 - 2bc < a2 ; c2 + a2 -2ca<b2
On adding, we get
a2 + b2 + c2 < 2(ab + bc+ ca)
....(1)∵ Roots of the given equation are real
∴
....(2)From (1) and (2), we get 
If the roots of the equation x2 – 2ax + a2 + a – 3 = 0 are real and less than 3, then (1999 - 2 Marks)- a)a < 2
- b)2 ≤ a ≤ 3
- c)3 < a ≤ 4
- d)a > 4
Correct answer is option 'A'. Can you explain this answer?
If the roots of the equation x2 – 2ax + a2 + a – 3 = 0 are real and less than 3, then (1999 - 2 Marks)
a)
a < 2
b)
2 ≤ a ≤ 3
c)
3 < a ≤ 4
d)
a > 4
|
|
Anand Kumar answered |
Use conditions
case(1):-D>0
case(2):-b/2a< />
case(3):-af(3)>0
case(1):-D>0
case(2):-b/2a< />
case(3):-af(3)>0
The amplitude of a complex number is called the principal value amplitude if it lies between.- a)-2π/3 , 2π/3
- b)π , 2π/3
- c)-π , π
- d)-π/2 , π/2
Correct answer is option 'C'. Can you explain this answer?
The amplitude of a complex number is called the principal value amplitude if it lies between.
a)
-2π/3 , 2π/3
b)
π , 2π/3
c)
-π , π
d)
-π/2 , π/2
|
Akshay Kumar Gupta answered |
We know complex number is a form of a straight line so maximum angle of the line is range of Π.
The set of all solutions of the inequality 
< 1/4 contains the set- a) (–¥, 0)
- b)(–¥, 1)
- c) (1, ¥)
- d) (3, ¥)
Correct answer is option 'D'. Can you explain this answer?
The set of all solutions of the inequality < 1/4 contains the set
< 1/4 contains the seta)
(–¥, 0)
b)
(–¥, 1)
c)
(1, ¥)
d)
(3, ¥)
|
|
Hansa Sharma answered |
(1/2)(x2 - 2x) < (1/4)
(1/2)(x2 - 2x) < (1/2)2
x2 − 2x > 2......(as after multiplicative inverse sign of inequality changes)
x2 − 2x − 2 > 0
x2 - 2x + 1 - 3 >
(x-1)2 - 3 > 0
(x-1)2 > 3
So for the above to hold good both the expression must be positive or both must be negative. After finding the solution the range of the solution will be
either x > 3
(3,¥)
(1/2)(x2 - 2x) < (1/2)2
x2 − 2x > 2......(as after multiplicative inverse sign of inequality changes)
x2 − 2x − 2 > 0
x2 - 2x + 1 - 3 >
(x-1)2 - 3 > 0
(x-1)2 > 3
So for the above to hold good both the expression must be positive or both must be negative. After finding the solution the range of the solution will be
either x > 3
(3,¥)
If (1 – p) is root of quadratic equation x2 + px + (1 – p) = 0, then its roots are- a)0, 1
- b)–1, 1
- c)0, –1
- d)–1, 2
Correct answer is option 'C'. Can you explain this answer?
If (1 – p) is root of quadratic equation x2 + px + (1 – p) = 0, then its roots are
a)
0, 1
b)
–1, 1
c)
0, –1
d)
–1, 2
|
|
Anand Kumar answered |
Put (1-p) In the given equation and solve for p.
after solving it u will have two values of p
now, for the solution put values p one by one in (1-p)
after solving it u will have two values of p
now, for the solution put values p one by one in (1-p)
The value of (-1 + √-3)2 + (-1 - √-3)2 is
- a)8
- b)4
- c)-4
- d)-2
Correct answer is option 'C'. Can you explain this answer?
The value of (-1 + √-3)2 + (-1 - √-3)2 is
a)
8
b)
4
c)
-4
d)
-2
|
Learners Habitat answered |
The given expression can be simplified as follows:
-1 + √(-3) = -1 + i√3
-1 - √(-3) = -1 - i√3
Now, (-1 + i√3)2 = (-1)2 + 2*(-1)*(i√3) + (i√3)2
= 1 - 2i√3 - 3
= -2i√3 - 2
Similarly, (-1 - i√3)2 = (-1)2 + 2*(-1)*(-i√3) + (-i√3)2
= 1 + 2i√3 - 3
= 2i√3 - 2
Adding the two results together:
(-2i√3 - 2) + (2i√3 - 2) = -4
Therefore, the value of the expression is -4.
-1 + √(-3) = -1 + i√3
-1 - √(-3) = -1 - i√3
Now, (-1 + i√3)2 = (-1)2 + 2*(-1)*(i√3) + (i√3)2
= 1 - 2i√3 - 3
= -2i√3 - 2
Similarly, (-1 - i√3)2 = (-1)2 + 2*(-1)*(-i√3) + (-i√3)2
= 1 + 2i√3 - 3
= 2i√3 - 2
Adding the two results together:
(-2i√3 - 2) + (2i√3 - 2) = -4
Therefore, the value of the expression is -4.
Two towns A and B are 60 km apart. A school is to be built to serve 150 students in town A and 50 students in town B. If the total distance to be travelled by all 200 students is to be as small as possible, then the school should be built at (1982 - 2 Marks)- a)town B
- b)45 km from town A
- c)town A
- d)45 km from town B
Correct answer is option 'C'. Can you explain this answer?
Two towns A and B are 60 km apart. A school is to be built to serve 150 students in town A and 50 students in town B. If the total distance to be travelled by all 200 students is to be as small as possible, then the school should be built at (1982 - 2 Marks)
a)
town B
b)
45 km from town A
c)
town A
d)
45 km from town B
|
Lavanya Menon answered |
Let the distance of school from A = x
∴ The distance of the school form B = 60 – x
Total distance covered by 200 students = 2[150 x + 50 (60 – x) ] = 2[100 x + 3000] This is min., when x = 0
∴ school should be built at town A.
∴ The distance of the school form B = 60 – x
Total distance covered by 200 students = 2[150 x + 50 (60 – x) ] = 2[100 x + 3000] This is min., when x = 0
∴ school should be built at town A.
If α and β are the roots of x2 + px + q = 0 and α4, β4 are the roots of x2 – rx + s = 0, then the equation x2 – 4qx + 2q2 – r = 0 has always (1989 - 2 Marks)- a)two real roots
- b)two positive roots
- c)two negative roots
- d)one positive and one negative root
Correct answer is option 'D'. Can you explain this answer?
If α and β are the roots of x2 + px + q = 0 and α4, β4 are the roots of x2 – rx + s = 0, then the equation x2 – 4qx + 2q2 – r = 0 has always (1989 - 2 Marks)
a)
two real roots
b)
two positive roots
c)
two negative roots
d)
one positive and one negative root
|
|
Ram Mohith answered |
- a)- i
- b)1
- c)i
- d)-1
Correct answer is option 'C'. Can you explain this answer?
a)
- i
b)
1
c)
i
d)
-1
|
|
Raghav Bansal answered |
x = (√3+i)/2
x3 = 1/8(√3+i)3
Apply formula (a+b)3 = a3 + b3 + 3a2b + 3ab2
= (3√3 + i3 + 3*3*i + 3*√3*i2)/8
= (3√3 - i + 9i - 3√3)/8
= 8i/8
= i
x3 = 1/8(√3+i)3
Apply formula (a+b)3 = a3 + b3 + 3a2b + 3ab2
= (3√3 + i3 + 3*3*i + 3*√3*i2)/8
= (3√3 - i + 9i - 3√3)/8
= 8i/8
= i
- a)0
- b)-1
- c)1
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
a)
0
b)
-1
c)
1
d)
none of these
|
|
Parth Patel answered |
If it is -1 then it is given that the w is cube root of unity
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