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CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 1 - MCQExams.com
CBSE
Class 11 Engineering Maths
Complex Numbers And Quadratic Equations
Quiz 1
Express
1
(
1
−
c
o
s
θ
+
2
i
s
i
n
θ
)
in the form
x
+
i
y
Report Question
0%
(
1
5
+
3
c
o
s
θ
)
+
(
2
c
o
t
θ
/
2
5
+
3
c
o
s
θ
)
i
0%
(
1
5
−
3
c
o
s
θ
)
+
(
−
2
c
o
t
θ
/
2
5
−
3
c
o
s
θ
)
i
0%
(
1
5
+
3
c
o
s
θ
)
+
(
−
2
c
o
t
θ
/
2
5
+
3
c
o
s
θ
)
i
0%
(
1
5
−
3
c
o
s
θ
)
+
(
2
c
o
t
θ
/
2
5
−
3
c
o
s
θ
)
i
Explanation
To Express
1
(
1
−
c
o
s
θ
+
2
i
s
i
n
θ
)
in the form
x
+
i
y
We have ,
1
(
1
−
c
o
s
θ
+
2
i
s
i
n
θ
)
=
1
(
1
−
c
o
s
θ
)
+
2
i
s
i
n
θ
×
(
1
−
c
o
s
θ
)
−
2
i
s
i
n
θ
(
1
−
c
o
s
θ
)
−
2
i
s
i
n
θ
=
(
1
−
c
o
s
θ
−
2
i
s
i
n
θ
)
(
1
−
c
o
s
θ
)
2
+
(
2
s
i
n
θ
)
2
=
(
1
−
c
o
s
θ
−
2
i
s
i
n
θ
)
1
+
c
o
s
2
θ
−
2
c
o
s
θ
+
4
s
i
n
2
θ
=
(
1
−
c
o
s
θ
)
(
1
−
c
o
s
θ
)
(
3
c
o
s
θ
+
5
)
−
2
i
s
i
n
θ
(
1
−
c
o
s
θ
)
(
3
c
o
s
θ
+
5
)
=
1
(
3
c
o
s
θ
+
5
)
−
2
i
s
i
n
θ
2
c
o
s
θ
2
2
s
i
n
2
θ
2
(
3
c
o
s
θ
+
5
)
=
1
(
3
c
o
s
θ
+
5
)
−
2
c
o
t
θ
2
(
3
c
o
s
θ
+
5
)
i
=
1
(
5
+
3
c
o
s
θ
)
+
(
−
2
c
o
t
θ
/
2
)
(
5
+
3
c
o
s
θ
)
i
Hence , Option C
If
z
=
x
+
i
y
and
ω
=
(
1
−
i
z
)
(
z
−
i
)
, then
|
ω
|
=
1
implies that in the complex plane
Report Question
0%
z lies on the imaginary axis
0%
z lies on the real axis
0%
z lies on the unit circle
0%
none of these
Explanation
Given
w
=
1
−
i
z
z
−
i
and
|
w
|
=
1
⇒
|
1
−
i
z
z
−
i
|
=
1
...(1)
Substitute
z
=
x
+
i
y
in equation (1)
⇒
|
1
−
i
(
x
+
i
y
)
(
x
+
i
y
)
−
i
|
=
1
⇒
|
1
+
y
−
i
x
|
=
|
x
+
i
(
y
−
1
)
|
⇒
(
1
+
y
)
2
+
x
2
=
x
2
+
(
y
−
1
)
2
⇒
y
=
0
Therefore z lies on the real axis.
Ans: B
If
a
,
b
and
c
are real numbers then the roots of the equation
(
x
−
a
)
(
x
−
b
)
+
(
x
−
b
)
(
x
−
c
)
+
(
x
−
c
)
(
x
−
a
)
=
0
are always
Report Question
0%
Real
0%
Imaginary
0%
Positive
0%
Negative
Explanation
Given equation is
(
x
−
a
)
(
x
−
b
)
+
(
x
−
b
)
(
x
−
c
)
+
(
x
−
c
)
(
x
−
a
)
=
0
⇒
3
x
2
−
2
(
b
+
a
+
c
)
x
+
a
b
+
b
c
+
c
a
=
0
Now, here
A
=
3
,
B
=
−
2
(
a
+
b
+
c
)
C
=
a
b
+
b
c
+
c
a
Therefore,
D
=
√
B
2
−
4
A
C
=
√
(
−
2
(
a
+
b
+
c
)
)
2
−
4
(
3
)
(
a
b
+
b
c
+
c
a
)
=
√
4
(
a
+
b
+
c
)
2
−
12
(
a
b
+
b
c
+
c
a
)
=
2
√
a
2
+
b
2
+
c
2
−
a
b
−
b
c
−
c
a
=
2
√
1
2
{
(
a
−
b
)
2
−
(
b
−
c
)
2
+
(
c
−
a
)
2
}
≥
0
This is always
≥
0
, we have real roots for the equation
(
x
−
a
)
(
x
−
b
)
+
(
x
−
b
)
(
x
−
c
)
+
(
x
−
c
)
(
x
−
a
)
=
0
If
(
x
+
i
y
)
(
2
−
3
i
)
=
4
+
i
then (x, y) =
Report Question
0%
(
1
,
1
13
)
0%
(
−
5
13
,
14
13
)
0%
(
5
13
,
14
13
)
0%
(
−
5
13
,
−
14
13
)
Explanation
(
x
+
i
y
)
(
2
−
3
i
)
=
4
+
i
2
x
−
(
3
x
)
i
+
(
2
y
)
i
−
3
y
i
2
=
4
+
i
2
x
+
3
y
⏟
R
e
a
l
+
(
2
y
−
3
x
)
⏟
I
m
a
g
i
n
a
r
y
i
=
4
+
i
Comparing the real & imaginary parts,
2
x
+
3
y
=
4
--------------------------(1)
2
y
−
3
x
=
1
----------------------------(2)
Solving eq(1) & eq(2),
4
x
+
6
y
=
8
−
9
x
+
6
y
=
3
13
x
=
5
⇒
x
=
5
13
y
=
14
13
∴
(
x
,
y
)
=
(
5
13
,
14
13
)
If
z
=
3
+
5
i
, then
z
3
+
z
+
198
=
Report Question
0%
3
−
15
i
0%
−
3
−
15
i
0%
−
3
+
15
i
0%
3
+
15
i
Explanation
z
=
3
+
5
i
z
3
=
(
3
+
5
i
)
3
=
3
3
+
3.3
2
(
5
i
)
+
3.3
(
5
i
)
2
+
(
5
i
)
3
=
27
−
125
i
+
135
i
−
225
=
−
225
+
27
+
(
135
−
125
)
i
=
−
198
+
10
i
∴
z
3
+
z
+
198
=
−
198
+
10
i
+
3
+
5
i
+
198
=
3
+
15
i
If
z
=
2
−
3
i
then
z
2
−
4
z
+
13
=
Report Question
0%
0
0%
1
0%
2
0%
3
Explanation
z
=
2
−
3
i
z
2
=
2
2
−
3
2
−
12
i
=
−
5
−
12
i
∴
z
2
−
4
z
+
13
=
(
−
5
−
12
i
)
−
4
(
2
−
3
i
)
+
13
=
−
5
−
12
i
−
8
+
12
i
+
13
=
−
13
+
13
=
0
The complex number
1
+
2
i
1
−
i
lies in the quadrant :
Report Question
0%
I
0%
II
0%
III
0%
IV
Explanation
Let
z
=
1
+
2
i
1
−
i
⇒
z
=
(
1
+
2
i
)
1
−
i
×
1
+
i
1
+
i
=
1
+
2
i
+
i
+
2
i
2
1
−
i
2
=
1
+
3
i
−
2
1
+
1
............
[
∵
i
2
=
−
1
]
⇒
z
=
−
1
+
3
i
2
=
−
1
2
+
3
2
i
=
x
+
i
y
Clearly
x
<
0
and
y
>
0
Hence
z
lies in
II
quadrant
√
−
3
√
−
75
=
Report Question
0%
15
0%
15
i
0%
−
15
0%
−
15
i
Explanation
√
−
3
×
√
−
75
=
√
3
×
(
−
1
)
√
75
×
(
−
1
)
=
√
3
i
×
√
75
i
=
√
225
i
2
=
−
15
The sum of two complex numbers
a
+
i
b
and
c
+
i
d
is a real number if
Report Question
0%
a
+
c
=
0
0%
b
+
d
=
0
0%
a
+
b
=
0
0%
b
+
c
=
0
Explanation
It is given that
z
1
=
a
+
i
b
and
z
2
=
c
+
i
d
Then
z
1
+
z
2
=
(
a
+
c
)
+
i
(
b
+
d
)
Now
(
z
1
+
z
2
)
is purely real.
Then the imaginary part has to be
0
.
Hence
b
+
d
=
0
.
The locus of complex number z such that z is purely real and real part is equal to - 2 is
Report Question
0%
Negative y-axis
0%
Negative x-axis
0%
The point (-2, 0)
0%
The point (2, 0)
Explanation
z
=
x
+
i
y
(
x
,
y
)
z
is purely real and the real part equals
−
2
∴
y
=
0
&
x
=
−
2
z
=
−
2
Hence, this would be represented by the point (-2,0) on the Argand Plane.
1
i
−
1
+
1
i
+
1
is
Report Question
0%
positive rational number
0%
purely imaginary
0%
positive Integer
0%
negative integer
Explanation
Let
Z
=
1
i
−
1
+
1
i
+
1
=
i
+
1
+
i
−
1
(
i
−
1
)
(
i
+
1
)
=
i
+
i
(
i
2
−
1
2
)
=
2
i
−
2
∴
Z
=
−
i
The argument of every complex number is
Report Question
0%
Double valued
0%
Single valued
0%
Many valued
0%
Triple valued
Explanation
z
=
x
+
i
y
amplitude
=
t
a
n
−
1
y
x
⇒
amplitude
=
θ
±
2
k
π
where
θ
ϵ
[
−
π
,
π
]
∀
k
ϵ
R
since
k
ϵ
R
⇒
Amplitude of any complex number is many valued.
The sum of two complex numbers
a
+
i
b
and
c
+
i
d
is purely imaginary if
Report Question
0%
a
+
c
=
0
0%
a
+
d
=
0
0%
b
+
d
=
0
0%
b
+
c
=
0
Explanation
It is given that
z
1
=
a
+
i
b
and
z
2
=
c
+
i
d
z
1
+
z
2
=
(
a
+
c
)
+
i
(
b
+
d
)
z
1
+
z
2
is purely imaginary. (Given)
Then the real part has to be
0
.
Hence
a
+
c
=
0
.
For
a
<
0
, arg
(
i
a
)
=
Report Question
0%
π
2
0%
−
π
2
0%
π
0%
−
π
Explanation
Let
z
=
i
a
a
<
0
. Hence
z
must lie on the negative imaginary axis ,
Hence,
a
r
g
(
z
)
=
−
π
2
The principal value of the argument of
−
√
3
+
i
is :
Report Question
0%
π
6
0%
3
π
6
0%
5
π
6
0%
7
π
6
Explanation
Given,
z
=
−
√
3
+
i
We have
|
z
|
=
√
√
3
2
+
1
2
=
2
Therefore,
cos
θ
=
−
√
3
2
and
sin
θ
=
1
2
Hence,
θ
=
5
π
6
Amplitude of
1
+
i
1
−
i
is :
Report Question
0%
0
0%
π
0%
π
2
0%
−
π
Explanation
1
+
i
1
−
i
=
(
1
+
i
)
(
1
+
i
)
(
1
−
i
)
(
1
+
i
)
=
(
1
+
i
)
2
1
2
−
i
2
=
1
−
1
+
2
i
2
=
i
therefore amplitude
=
π
2
Which of the following equations has two distinct real roots ?
Report Question
0%
2
x
2
−
3
√
2
x
+
9
4
=
0
0%
x
2
+
x
−
5
=
0
0%
x
2
+
3
x
+
2
√
2
=
0
0%
5
x
2
−
3
x
+
1
=
0
Explanation
An equation is said to have two distinct and real roots if the discriminant
b
2
−
4
a
c
>
0
Case (i): For equation:
2
x
2
−
3
√
2
x
+
9
4
=
0
.
Here
a
=
2
,
b
=
−
3
√
2
,
c
=
9
4
The discrimant is
(
−
3
√
2
)
2
−
4
(
2
)
(
9
4
)
=
18
−
18
=
0
Hence this equation has equal real roots
Case (ii): For equation:
x
2
+
x
−
5
=
0
.
Here
a
=
1
,
b
=
1
,
c
=
−
5
The discrimant is
1
2
−
4
(
1
)
(
−
5
)
=
1
+
20
=
21
>
0
Hence this equation has two distinct real roots
Case (iii): For equation:
x
2
+
3
x
+
2
√
2
=
0
.
Here
a
=
1
,
b
=
3
,
c
=
2
√
2
The discrimant is
3
2
−
4
(
1
)
(
2
√
2
)
=
9
−
8
√
2
<
0
Hence this equation has no real roots
Case (iv): For equation:
5
x
2
−
3
x
+
1
=
0
.
Here
a
=
5
,
b
=
−
3
,
c
=
1
The discrimant is
(
−
3
)
2
−
4
(
5
)
(
1
)
=
9
−
20
<
0
Hence this equation has no real roots
A quadratic equation
a
x
2
+
b
x
+
c
=
0
has two distinct real roots, if
Report Question
0%
a
=
0
0%
b
2
−
4
a
c
=
0
0%
b
2
−
4
a
c
<
0
0%
b
2
−
4
a
c
>
0
Explanation
If
a
=
0
, it becomes linear equation.
If
b
2
−
4
a
c
=
0
, then there will be real and equal roots.
If
b
2
−
4
a
c
<
0
, then the roots will be unreal.
Only if
b
2
−
4
a
c
>
0
, we will get two real distinct roots.
Option D is correct!
For
a
>
0
, arg
(
i
a
)
=
Report Question
0%
π
2
0%
−
π
2
0%
π
0%
−
π
Explanation
z
=
0
+
i
a
∴
a
r
g
(
z
)
=
a
r
g
(
i
a
)
=
tan
−
1
a
0
=
π
2
...(since a is greater than or equal to zero)
The modulus of
√
2
i
−
√
−
2
i
is:
Report Question
0%
2
0%
√
2
0%
0
0%
2
√
2
Explanation
Let
z
=
√
2
i
−
√
−
2
i
⇒
z
=
√
2
i
−
√
2
i
2
⇒
z
=
√
2
+
√
2
i
Now,
|
z
|
=
√
2
+
2
=
2
The roots of the equation
3
x
2
−
4
x
+
3
=
0
are :
Report Question
0%
real and unequal
0%
real and equal
0%
imaginary
0%
none of these
Explanation
Given equation is
3
x
2
−
4
x
+
3
=
0
To find, the nature of the roots of the equation
An equation is said to have
(i) two distinct and real roots if the discriminant
b
2
−
4
a
c
>
0
(ii) equal real roots if
b
2
−
4
a
c
=
0
(iii) no real roots or imaginary roots if
b
2
−
4
a
c
<
0
In the given equation
a
=
3
,
b
=
−
4
,
c
=
3
Hence the discriminant is
(
−
4
)
2
−
4
(
3
)
(
3
)
=
16
−
36
=
−
20
<
0
Therefore the roots of the given equation are imaginary in nature.
For
a
<
0
, arg
a
=
Report Question
0%
π
2
0%
−
π
2
0%
π
0%
−
π
Explanation
It is given that
a
<
0
Thus, 'a' is purely real.
Hence
a
r
g
(
a
)
will be
0
or,
π
But it is given that
a
<
0
Hence
c
o
s
(
a
r
g
(
a
)
)
=
−
1
Or
a
r
g
(
a
)
=
π
.
If the square of
(
a
+
i
b
)
is real, then
a
b
=
Report Question
0%
0
0%
1
0%
−
1
0%
2
Explanation
(
a
+
i
b
)
2
=
a
2
−
b
2
+
2
i
a
b
is given to be real
⇒
a
b
=
0
Find the argument of
−
1
−
i
√
3
Report Question
0%
θ
=
−
2
π
/
3
0%
θ
=
2
π
/
3
0%
θ
=
−
4
π
/
3
0%
θ
=
4
π
/
3
Explanation
Let
z
=
−
1
−
i
√
3
Then
α
=
tan
−
1
|
b
/
a
|
=
tan
−
1
|
√
3
/
1
|
=
π
/
3
Here,
z
is in the third quadrant.
Therefore, argument is
θ
=
−
(
π
−
α
)
=
−
(
π
−
π
/
3
)
=
−
2
π
/
3
Ans: A
The roots of
x
2
−
x
+
1
=
0
are:
Report Question
0%
Real and equal
0%
Real and not equal
0%
Imaginary
0%
Reciprocals
Explanation
Given equation is
x
2
−
x
+
1
=
0
We know
D
=
b
2
−
4
a
c
Here
a
=
1
,
b
=
−
1
,
c
=
1
Therefore,
D
=
(
−
1
)
2
−
4
(
1
)
(
1
)
=
1
−
4
=
−
3
<
0
Thus roots are imaginary.
Nature of the roots of the quadratic equation
2
x
2
−
2
√
6
x
+
3
=
0
is:
Report Question
0%
Real, irrational, unequal
0%
Real, rational, equal
0%
Real, rational, unequal
0%
Complex
Explanation
(B)
b
2
−
4
a
c
=
(
−
2
√
6
)
2
−
4
(
2
)
(
3
)
=
24
−
24
=
0
∴
Real, rational, equal
Determine the nature of roots of the equation
x
2
+
2
x
√
3
+
3
=
0
.
Report Question
0%
Real and distinct
0%
Non-real and distinct
0%
Real and equal
0%
Non-real and equal
Explanation
The nature of the roots can be determined from the discriminant
b
2
−
4
a
c
∴
b
2
−
4
a
c
=
(
2
√
3
)
2
−
(
4
×
1
×
3
)
⇒
b
2
−
4
a
c
=
12
−
12
⇒
b
2
−
4
a
c
=
0
∵
b
2
−
4
a
c
=
0
There are two real and equal roots.
Find the value of
x
of the equation
(
1
−
i
)
x
=
2
x
Report Question
0%
1
0%
2
0%
0
0%
none of these
Explanation
Then
(
1
−
i
)
x
=
2
x
⇒
|
(
1
−
i
)
|
x
=
|
2
|
x
⇒
(
√
1
+
1
)
x
=
2
x
⇒
(
√
2
)
x
=
2
x
⇒
x
2
=
x
⇒
2
x
=
x
⇒
x
=
0
Hence, option C is correct.
If the discriminant of a quadratic equation is negative, then its roots are:
Report Question
0%
unequal
0%
equal
0%
inverse
0%
imaginary
Explanation
If the discriminant
b
2
−
4
a
c
of a quadratic equation is negative, then its roots are imaginary.
Solve
(
1
−
i
)
x
+
(
1
+
i
)
y
=
1
−
3
i
,
Report Question
0%
x
=
−
1
,
y
=
2.
0%
x
=
2
,
y
=
−
1.
0%
x
=
2
,
y
=
1.
0%
x
=
1
,
y
=
2.
Explanation
(
1
−
i
)
x
+
(
1
+
i
)
y
=
1
−
3
i
.
Equating real and imaginary parts, we get
x
+
y
=
1
and
−
x
+
y
=
−
3.
Adding both equations we get
y
=
−
1
Substituting this value of
y
in any of the 2 equations, we get
x
=
2
∴
x
=
2
,
y
=
−
1.
Ans: B
The roots of
4
x
2
−
2
x
+
8
=
0
are:
Report Question
0%
Real and equal
0%
Rational and not equal
0%
Irrational
0%
Not real
Explanation
Given equation is
4
x
2
−
2
x
+
8
=
0
We know value of discriminant
D
=
b
2
−
4
a
c
Here
a
=
4
,
b
=
−
2
,
c
=
8
Therefore,
D
=
(
−
2
)
2
−
4
(
4
)
(
8
)
=
4
−
128
=
−
124
<
0
Hence, roots are not real.
Evaluate :
√
−
25
+
3
√
−
4
+
2
√
−
9
Report Question
0%
−
17
i
0%
5
i
0%
17
i
0%
6
i
Explanation
√
−
25
+
3
√
−
4
+
2
√
−
9
=
5
√
−
1
+
6
√
−
1
+
6
√
−
1
we know,
√
−
1
=
i
=
5
i
+
6
i
+
6
i
=
17
i
If
x
2
−
2
p
x
+
8
p
−
15
=
0
has equal roots, then
p
=
Report Question
0%
3
or
−
5
0%
3
or
5
0%
−
3
or
5
0%
−
3
or
−
5
Explanation
Given equation is
x
2
−
2
p
x
+
8
p
−
15
=
0
Here
a
=
1
,
b
=
−
2
p
,
c
=
8
p
−
15
We know for equal roots,
b
2
−
4
a
c
=
0
Therefore,
(
−
2
p
)
2
−
4
(
1
)
(
8
p
−
15
)
=
0
⇒
4
p
2
−
32
p
+
60
=
0
⇒
p
2
−
8
p
+
15
=
0
⇒
(
p
−
5
)
(
p
−
3
)
=
0
i.e.,
p
=
5
or
3
Determine the values of
p
for which the quadratic equation
2
x
2
+
p
x
+
8
=
0
has
equal roots.
Report Question
0%
p
=
±
64
0%
p
=
±
8
0%
p
=
±
4
0%
p
=
±
16
Explanation
In
a
x
2
+
b
x
+
c
=
0
,
Discriminant
D
=
b
2
−
4
a
c
,
D
=
0
, for the roots to be real and equal
In
2
x
2
+
p
x
+
8
=
0
Thus, we have
a
=
2
,
b
=
p
and
c
=
8
Then
D
=
b
2
−
4
a
c
=
0
∴
p
2
−
4
×
2
×
8
=
0
⇒
p
2
−
64
=
0
⇒
p
=
±
8
.
Find the values of
k
for the following quadratic equation, so that they have two real and equal roots:
2
x
2
+
k
x
+
3
=
0
Report Question
0%
k
=
±
2
√
3
0%
k
=
±
2
√
6
0%
k
=
±
√
6
0%
k
=
±
√
3
Explanation
In
a
x
2
+
b
x
+
c
=
0
Discriminant
D
=
b
2
−
4
a
c
If
D
=
0
, so they have equal and real roots.
Then,
2
x
2
+
k
x
+
3
=
0
a
=
2
,
b
=
k
and
c
=
3
D
=
k
2
−
4
×
2
×
3
=
0
⇒
k
2
−
24
=
0
⇒
k
=
±
√
24
⇒
k
=
±
2
√
6
i
4
n
+
3
+
(
−
i
)
8
n
−
3
(
i
)
12
n
−
1
−
i
2
−
16
n
,
n
ε
N
is equal to
Report Question
0%
1 + i
0%
2i
0%
-2i
0%
-1 - i
Explanation
Given expression
=
i
3
+
(
−
1
)
(
i
)
−
3
(
−
i
)
−
1
−
(
i
)
2
=
−
i
−
i
i
+
1
=
−
2
i
1
+
i
×
1
−
i
1
−
i
=
−
2
i
−
2
1
+
1
=
−
1
−
i
1+
i
2
+
i
4
+
i
6
+
.
.
.
.
.
.
.
.
+
i
2
n
is
Report Question
0%
Positive
0%
Negative
0%
Zero
0%
Cannot be determined
Explanation
1
+
i
2
+
i
4
+
.
.
.
.
.
.
.
.
+
i
2
n
=
1
−
1
+
1
−
1
+
1
−
1......
Hence it can b positive or negative depending on the value of
2
n
Hence It cannot be determined from the given data.
Find the modulus and the principal value of the argument of the number
1
−
i
Report Question
0%
√
2
,
π
/
4
0%
√
2
,
−
π
/
4
0%
√
2
,
−
π
/
3
0%
√
2
,
3
π
/
4
Explanation
Let
1
=
r
cos
θ
,
−
1
=
r
sin
θ
Squaring and adding these relations, we get
r
2
(
cos
2
θ
+
sin
2
θ
)
=
1
2
+
(
−
1
)
2
=
2
.
Since,
c
o
s
2
θ
+
s
i
n
2
θ
=
1
,
⇒
r
=
±
√
2
r
cannot be negative
Hence.
r
=
√
2
Then
cos
θ
=
1
√
2
,
sin
θ
=
−
1
√
2
,
The value of
θ
between-
π
and
π
which satisfies these equations is
−
(
π
/
4
)
Thus
|
1
−
i
|
=
r
=
√
2
and
a
r
g
(
1
−
i
)
=
−
(
π
/
4
)
.
Ans: B
If
i
2
=
−
1
, then the value of
200
∑
n
=
1
i
n
is
Report Question
0%
50
0%
-50
0%
0
0%
100
Explanation
Given,
i
2
=
−
1
200
∑
n
=
1
i
n
=
i
+
i
2
+
i
3
+
.
.
.
.
.
.
.
.
.
+
i
200
=
i
(
1
−
i
200
)
1
−
i
(since G. P.)
=
i
(
1
−
1
)
1
−
i
=
0
If i =
√
−
1
,
t
h
e
n
1
+
i
2
+
i
3
−
i
6
+
i
8
is equal to -
Report Question
0%
2- i
0%
1
0%
-3
0%
-1
Explanation
Given,
i
=
√
−
1
∴
1
+
i
2
+
i
3
−
i
6
+
i
8
=
1
−
1
−
i
+
1
+
1
=
2
−
i
Check whether
2
x
2
−
3
x
+
5
=
0
has real roots or no.
Report Question
0%
The equation has real roots.
0%
The equation has no real roots.
0%
Data insufficient
0%
None of these
Explanation
Here the quadratic equation is
2
x
2
−
3
x
+
5
=
0
Comparing it with
a
x
2
+
b
x
+
c
=
0
, we get
a
=
2
,
b
=
−
3
,
c
=
5
Therefore, discriminant,
D
=
b
2
−
4
a
c
(
−
3
)
2
−
4
×
2
×
5
=
9
−
40
=
−
31
Here,
D
<
0
Therefore the equation has no real roots.
(
i
10
+
1
)
(
i
9
+
1
)
(
i
8
+
1
)
.
.
.
.
.
.
.
(
i
+
1
)
equal to
Report Question
0%
-1
0%
1
0%
0
0%
i
Explanation
(
i
10
+
1
)
(
i
9
+
1
)
(
i
8
+
1
)
.
.
.
.
.
.
.
(
i
+
1
)
=
(
i
10
+
1
)
(
i
9
+
1
)
(
i
8
+
1
)
.
.
.
.
.
.
.
(
i
3
+
1
)
(
i
2
+
1
)
(
i
+
1
)
=
(
i
10
+
1
)
(
i
9
+
1
)
(
i
8
+
1
)
.
.
.
.
.
.
.
(
i
3
+
1
)
(
−
1
+
1
)
(
i
+
1
)
[
∵
i
2
=
−
1
]
=
(
i
10
+
1
)
(
i
9
+
1
)
(
i
8
+
1
)
.
.
.
.
.
.
.
(
i
3
+
1
)
(
0
)
(
i
+
1
)
=
0
If
i
2
=
−
1
, then find the odd one out of the following expressions.
Report Question
0%
−
i
2
0%
(
−
i
)
2
0%
i
4
0%
(
−
i
)
4
0%
−
i
6
Explanation
i
is an imagiary number which has a value of
i
=
√
−
1
. Then
A.
−
i
2
=
−
1
∗
−
1
=
1
B.
(
−
i
)
2
=
i
2
=
−
1
C.
i
4
=
(
i
2
)
2
=
(
−
1
)
2
=
1
D.
(
−
i
)
4
=
(
i
2
)
2
=
(
−
1
)
2
=
1
E.
−
i
6
=
−
(
i
2
)
3
=
−
1
×
(
−
1
)
3
=
−
1
×
(
−
1
)
=
1
All are
1
except option
B
which is
−
1
.
When
(
3
−
2
i
)
is subtracted from
(
4
+
7
i
)
, then the result is
Report Question
0%
1
+
5
i
0%
1
+
9
i
0%
7
+
5
i
0%
7
+
9
i
Explanation
We need to subtract
(
3
−
2
i
)
from
(
4
+
7
i
)
∴
4
+
7
i
−
(
3
−
2
i
)
=
4
+
7
i
−
3
+
2
i
=
1
+
9
i
The value of k for which polynomial
x
2
−
k
x
+
4
has equal zeroes is
Report Question
0%
4
0%
2
0%
−
4
0%
−
2
Explanation
x
2
−
k
x
+
4
has equal roots
⇒
D
=
0
⇒
k
2
−
4
×
4
×
1
⇒
k
2
=
16
⇒
k
=
±
4
If the discriminant of a quadratic equation is negative, then its roots are
Report Question
0%
Unequal
0%
Equal
0%
Inverses
0%
Imaginary
Explanation
As
x
=
−
b
±
√
D
2
a
if D is
−
ve
⇒
imaginary Roots
If the equation
(
1
+
m
2
)
x
2
+
2
m
c
x
+
(
c
2
−
a
2
)
=
0
has equal roots, then
c
2
=
Report Question
0%
a
2
(
1
+
m
2
)
0%
a
(
1
+
m
2
)
0%
a
4
(
1
−
m
2
)
0%
a
2
(
1
−
m
2
)
Explanation
As the eq
n
has equal roots
⇒
D
=
0
⇒
(
2
m
c
)
2
−
4
(
1
+
m
2
)
(
c
2
−
a
2
)
=
0
⇒
4
−
m
2
c
2
−
4
[
c
2
−
a
2
+
m
2
c
2
−
m
2
a
2
]
=
0
⇒
c
2
=
m
2
a
2
+
a
2
⇒
c
2
=
a
2
(
1
+
m
2
)
Amplitude of
1
+
√
3
i
√
3
+
i
is
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0%
π
3
0%
π
2
0%
0
0%
π
6
Explanation
Z
=
1
+
√
3
i
√
3
+
i
⇒
(
1
+
√
3
i
√
3
+
i
)
(
√
3
−
i
√
3
−
i
)
=
2
√
3
+
2
i
4
∴
Amplitude of
Z
is
tan
−
1
(
1
√
3
)
=
π
6
Hence, option D.
For
i
=
√
−
1
, what is the sum
(
7
+
3
i
)
+
(
−
8
+
9
i
)
?
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0%
−
1
+
12
i
0%
−
1
−
6
i
0%
15
+
12
i
0%
15
−
6
i
Explanation
Given:
(
7
+
3
i
)
+
(
−
8
+
9
i
)
=
(
7
+
(
−
8
)
)
+
(
3
i
+
9
i
)
=
(
7
−
8
)
+
12
i
=
−
1
+
12
i
Option A is correct.
If the equation
x
2
−
b
x
+
1
=
0
does not possess real roots then
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0%
−
3
<
b
<
3
0%
−
2
<
b
<
2
0%
b
>
2
0%
b
<
−
2
Explanation
Given equation is
x
2
−
b
x
+
1
=
0
If the equation doesnot possess real roots,then
b
2
−
4
a
x
<
0
=>
(
−
b
)
2
−
4
(
1
)
(
1
)
<
0
=>
b
2
<
4
=>
b
<
±
2
∴
−
2
<
b
<
2
0:0:1
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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