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CBSE Questions for Class 9 Maths Polynomials Quiz 8 - MCQExams.com
CBSE
Class 9 Maths
Polynomials
Quiz 8
If
g
(
x
)
=
3
x
2
−
2
x
−
5
, what is the value of
g
(
−
1
)
?
Report Question
0%
−
4
0%
−
10
0%
−
6
0%
0
Explanation
Given,
g
(
x
)
=
x
2
−
2
x
−
5
Then
g
(
−
1
)
=
3
(
−
1
)
2
−
2
(
−
1
)
−
5
=
3
+
2
−
5
=
0
For what value of
k
,
−
2
is a zero of the polynomial
3
x
2
+
4
x
+
2
k
?
Report Question
0%
5
0%
3
0%
−
8
0%
−
2
Explanation
If
−
2
is a zero of
f
(
x
)
=
3
x
2
+
4
x
+
2
k
Thus,
f
(
−
2
)
=
0
∴
3
(
−
2
)
2
+
4
(
−
2
)
+
2
k
=
0
=>
12
−
8
+
2
k
=
0
=>
2
k
=
−
4
=>
k
=
−
2
The value of the polynomial
x
3
−
27
when
x
=
3
is
Report Question
0%
26
0%
25
0%
0
0%
1
Explanation
Let
f
(
x
)
=
x
3
−
27.
Therefore, at
x
=
3
,
f
(
3
)
=
3
3
−
27
=
27
−
27
=
0
Hence, option
C
is correct.
(
a
−
b
)
2
−
(
a
+
b
)
2
is equal to:
Report Question
0%
2
a
b
0%
4
a
b
0%
6
a
b
0%
−
4
a
b
Explanation
Given,
(
a
−
b
)
2
−
(
a
+
b
)
2
.
We know,
(
x
+
y
)
2
=
x
2
+
2
x
y
+
y
2
and
(
x
−
y
)
2
=
x
2
−
2
x
y
+
y
2
.
Then,
(
a
−
b
)
2
−
(
a
+
b
)
2
=
(
a
2
+
b
2
−
2
a
b
)
−
(
a
2
+
b
2
+
2
a
b
)
=
a
2
+
b
2
−
2
a
b
−
a
2
−
b
2
−
2
a
b
)
=
−
4
a
b
.
Therefore, option
D
is correct.
Say true or false:
The zeros of the polynomial
x
2
−
14
x
+
49
is equal to
7
.
Report Question
0%
True
0%
False
Explanation
The given polynomial
x
2
−
14
x
+
49
=
0
can be factorized as follows:
x
2
−
14
x
+
49
=
0
⇒
x
2
−
7
x
−
7
x
+
49
=
0
⇒
x
(
x
−
7
)
−
7
(
x
−
7
)
=
0
⇒
(
x
−
7
)
(
x
−
7
)
=
0
⇒
(
x
−
7
)
=
0
,
(
x
−
7
)
=
0
⇒
x
=
7
,
x
=
7
Hence, the zeroes of
x
2
−
14
x
+
49
is
7
.
Say true or false:
The zero of
x
−
5
is
5
.
Report Question
0%
True
0%
False
Explanation
To find the zeroes of any polynomial, we need to equate the polynomial expression to zero.
Here, the polynomial is
x
−
5
that is
x
−
5
=
0
⟹
x
=
5
.
Hence, the zero of
x
−
5
is
5
.
The value of the polynomial
7
x
4
+
6
x
2
+
x
when
x
=
−
2
is
Report Question
0%
131
0%
132
0%
134
0%
136
Explanation
given,
f
(
x
)
=
7
x
4
+
6
x
2
+
x
f
(
−
2
)
=
7
(
−
2
)
4
+
6
(
−
2
)
2
+
(
−
2
)
f
(
−
2
)
=
7
(
16
)
+
6
(
4
)
−
2
f
(
−
2
)
=
112
+
22
=
134
The value of the polynomial
4
x
2
+
3
x
−
7
at
x
=
1
is:
Report Question
0%
1
0%
2
0%
0
0%
None of the above
Explanation
f
(
x
)
=
4
x
2
+
3
x
−
7
So at
x
=
1
,
f
(
1
)
=
4
(
1
)
2
+
3
(
1
)
−
7
f
(
1
)
=
4
+
3
−
7
=
0
State the following statement is True or False
The zero of the polynomial
x
+
2
+
4
is
−
2
.
Report Question
0%
True
0%
False
Explanation
Rewrite the polynomial
x
+
2
+
4
as
x
+
6
and equate it to
0
as follows:
x
+
6
=
0
⇒
x
=
−
6
Hence, the only zero of the polynomial
x
+
2
+
4
is
−
6
.
Using standard identity, find the value of
102
2
.
Report Question
0%
10402
0%
10408
0%
10204
0%
10404
Explanation
Given,
102
2
=
(
100
+
2
)
2
.
We know,
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
.
Then,
102
2
=
(
100
+
2
)
2
=
100
2
+
2
2
+
2
(
2
)
(
100
)
=
10000
+
4
+
400
=
10404
.
Therefore, option
D
is correct.
Using standard identity, find
(
2
x
−
y
)
2
.
Report Question
0%
4
x
2
+
y
2
−
4
x
y
0%
4
x
2
+
y
2
+
4
x
y
0%
2
x
2
+
y
2
−
4
x
y
0%
4
x
2
+
y
2
−
2
x
y
Explanation
Given,
(
2
x
−
y
)
2
We know,
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
.
Then,
(
2
x
−
y
)
2
=
(
2
x
)
2
+
y
2
−
2
×
2
x
×
y
=
4
x
2
+
y
2
−
4
x
y
.
Therefore, option
A
is correct.
Using standard identity, find the value of
(
a
+
2
b
)
2
.
Report Question
0%
a
2
+
2
b
2
+
4
a
b
0%
a
2
+
4
b
2
+
2
a
b
0%
a
2
+
4
b
2
+
4
a
b
0%
2
a
2
+
4
b
2
+
4
a
b
Explanation
Given,
(
a
+
2
b
)
2
.
We know,
(
x
+
y
)
2
=
x
2
+
2
x
y
+
y
2
.
Then,
(
a
+
2
b
)
2
=
a
2
+
(
2
b
)
2
+
2
×
a
×
2
b
=
a
2
+
4
b
2
+
4
a
b
.
Therefore, option
C
is correct.
The value of the polynomial
2
x
5
−
5
x
3
−
10
x
+
9
when
x
=
−
1
Report Question
0%
21
0%
22
0%
23
0%
None of the above
Explanation
f
(
x
)
=
2
x
5
−
5
x
3
−
10
x
+
9
f
(
−
1
)
=
2
(
−
1
)
5
−
5
(
−
1
)
3
−
10
(
−
1
)
+
9
f
(
−
1
)
=
2
(
−
1
)
−
5
(
−
1
)
+
10
+
9
f
(
−
1
)
=
−
2
+
5
+
19
f
(
−
1
)
=
22
Using standard identity, find the value of
99
2
.
Report Question
0%
9901
0%
9801
0%
1001
0%
9701
Explanation
Given,
=
99
2
=
(
100
−
1
)
2
We know,
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
Then,
99
2
=
(
100
−
1
)
2
=
100
2
+
1
2
−
2
×
100
×
1
=
10000
+
1
−
200
=
10001
−
200
=
9801
Therefore, option
B
is correct.
Which of the following is correct?
Report Question
0%
(
x
−
y
)
2
=
x
2
+
2
x
y
−
y
2
0%
(
x
−
y
)
2
=
x
2
−
2
x
y
+
y
2
0%
(
x
−
y
)
2
=
x
2
−
y
2
0%
(
x
+
y
)
2
=
x
2
+
2
x
y
−
y
2
Explanation
We know,
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
.
Then,
(
x
−
y
)
2
=
x
2
−
2
(
x
)
(
y
)
+
y
2
=
x
2
−
2
x
y
+
y
2
.
To verify this, let us consider:
(
x
−
y
)
2
=
(
x
−
y
)
(
x
−
y
)
=
x
(
x
−
y
)
−
y
(
x
−
y
)
=
x
2
−
x
y
−
y
x
+
y
2
=
x
2
−
x
y
−
x
y
+
y
2
=
x
2
−
2
x
y
+
y
2
.
Hence,
(
x
−
y
)
2
=
x
2
−
2
x
y
+
y
2
Therefore, option
B
is correct.
Find the value of
49
2
using standard identity.
Report Question
0%
2391
0%
2401
0%
2411
0%
2421
Explanation
As we know that,
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
Substitute
50
for
a
and
1
for
b
in above formula,
(
50
−
1
)
2
=
50
2
−
2
×
50
×
1
+
1
2
49
2
=
2500
−
100
+
1
=
2401
Therefore,
B
is correct.
Find the value of
99
×
101
using standard identity.
Report Question
0%
9999
0%
9989
0%
9979
0%
1009
Explanation
Given,
99
×
101
=
(
100
−
1
)
(
100
+
1
)
.
We know,
(
a
+
b
)
(
a
−
b
)
=
a
2
−
b
2
.
Then,
99
×
101
=
(
100
−
1
)
(
100
+
1
)
=
(
100
)
2
−
1
2
=
10000
−
1
=
9999
.
Therefore, option
A
is correct.
Square of
3
a
−
4
b
is:
Report Question
0%
9
a
2
+
16
b
2
−
24
a
b
0%
6
a
2
−
8
b
2
0%
9
a
2
−
16
b
2
0%
9
a
2
+
16
b
2
+
24
a
b
Explanation
Given, square of
3
a
−
4
b
i.e.
(
3
a
−
4
b
)
2
.
We know,
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
.
Then,
(
3
a
−
4
b
)
2
=
(
3
a
)
2
−
2
(
3
a
)
(
4
b
)
+
(
4
b
)
2
=
9
a
2
−
24
a
b
+
16
b
2
.
Therefore, option
A
is correct.
Find the value of
47
×
53
.
Report Question
0%
2451
0%
2471
0%
2491
0%
2501
Explanation
Given,
47
×
53
=
(
50
−
3
)
(
50
+
3
)
.
We know,
(
a
+
b
)
(
a
−
b
)
=
a
2
−
b
2
.
Then,
47
×
53
=
(
50
−
3
)
(
50
+
3
)
=
(
50
)
2
−
3
2
=
2500
−
9
=
2491
.
Therefore, option
C
is correct.
Find the value of
87
2
−
13
2
.
Report Question
0%
7300
0%
7350
0%
7400
0%
7450
Explanation
Given,
87
2
−
13
2
.
We know,
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
.
Then,
=
87
2
−
13
2
=
(
87
+
13
)
(
87
−
13
)
=
100
×
74
=
7400
.
Therefore, option
C
is correct.
Find the value of
52
2
using standard identity.
Report Question
0%
2604
0%
2704
0%
2804
0%
2904
Explanation
Given,
52
2
=
(
50
+
2
)
2
.
We know,
(
a
+
b
)
2
=
a
2
+
b
2
+
2
a
b
.
Then,
52
2
=
(
50
+
2
)
2
=
50
2
+
2
2
+
2
×
50
×
2
=
2500
+
4
+
200
=
2704
.
Therefore, option
B
is correct.
Find the value of
199
×
201
.
Report Question
0%
39989
0%
40001
0%
39999
0%
400011
Explanation
Given,
199
×
201
=
(
200
−
1
)
(
200
+
1
)
.
We know,
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
.
Then,
199
×
201
=
(
200
−
1
)
(
200
+
1
)
=
200
2
−
1
2
=
40000
−
1
=
39999
.
Therefore, option
C
is correct.
Find the value of
995
2
−
5
2
.
Report Question
0%
99000
0%
9900
0%
99005
0%
990000
Explanation
Given,
995
2
−
5
2
.
We know,
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
.
Then,
995
2
−
5
2
=
(
995
+
5
)
(
995
−
5
)
=
1000
×
990
=
990000
.
Therefore, option
D
is correct.
Find the value of
(
a
+
b
)
2
−
(
a
−
b
)
2
.
Report Question
0%
a
b
0%
2
a
b
0%
3
a
b
0%
4
a
b
Explanation
Given,
(
a
+
b
)
2
−
(
a
−
b
)
2
.
We know,
(
a
+
b
)
2
=
a
2
+
b
2
+
2
a
b
and
(
a
−
b
)
2
a
2
+
b
2
−
2
a
b
.
Then,
(
a
+
b
)
2
−
(
a
−
b
)
2
=
a
2
+
b
2
+
2
a
b
−
(
a
2
+
b
2
−
2
a
b
)
=
a
2
+
b
2
+
2
a
b
−
a
2
−
b
2
+
2
a
b
=
4
a
b
.
Therefore, option
D
is correct.
The zero of the polynomial
2
x
+
1
is:
Report Question
0%
x
=
−
1
2
0%
x
=
−
1
0%
x
=
−
2
0%
None of the above
Explanation
Let
p
(
x
)
=
2
x
+
1
Consider
p
(
x
)
=
0
⇒
2
x
+
1
=
0
⇒
x
=
−
1
2
Therefore,
x
=
−
1
2
is the zero of the polynomial
2
x
+
1
=
0
A zero of a polynomial:
Report Question
0%
needs to be
0
.
0%
can be any number including
0
.
0%
needs to be an integer.
0%
None of the above
Explanation
Consider the polynomial,
p
(
x
)
=
a
x
2
+
b
x
+
c
.
To find the zero of a polynomial, we write
p
(
x
)
=
0
.
Hence, a real number
c
is said to be a zero of the polynomial
p
(
x
)
, if
p
(
c
)
=
0
.
E.g. : Suppose,
p
(
x
)
=
x
2
.
When
x
=
0
, we get
p
(
x
)
=
(
0
)
2
=
0
.
Clearly, a zero of the polynomial can be any number including
0
.
Therefore, option
B
is correct.
The zero of the polynomial
3
x
+
5
is :
Report Question
0%
−
5
3
0%
5
3
0%
1
2
0%
−
1
2
Explanation
We find the zeros of the given polynomial using:
3
x
+
5
=
0
3
x
=
−
5
x
=
−
5
3
.
(
3
−
√
7
)
(
3
+
√
7
)
=
?
Report Question
0%
4
0%
2
0%
6
0%
8
Explanation
Given,
(
3
−
√
7
)
(
3
+
√
7
)
.
We know,
(
x
−
y
)
(
x
+
y
)
=
x
2
−
y
2
.
Thus,
(
3
−
√
7
)
(
3
+
√
7
)
=
3
2
−
(
√
7
)
2
=
9
−
7
=
2
.
Therefore, option
B
is correct.
The value of
(
a
+
b
)
2
−
2
(
a
−
b
)
2
+
(
a
−
b
)
(
a
+
b
)
is:
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0%
4
a
b
−
b
2
0%
2
a
b
−
b
2
0%
3
a
b
−
b
2
0%
6
a
b
−
2
b
2
Explanation
Given,
(
a
+
b
)
2
−
2
(
a
−
b
)
2
+
(
a
−
b
)
(
a
+
b
)
.
We know,
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
,
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
and
(
a
+
b
)
(
a
−
b
)
=
a
2
−
b
2
.
Then,
(
a
+
b
)
2
−
2
(
a
−
b
)
2
+
(
a
−
b
)
(
a
+
b
)
=
(
a
2
+
b
2
+
2
a
b
)
−
2
(
a
2
+
b
2
−
2
a
b
)
+
(
a
2
−
b
2
)
=
a
2
+
b
2
+
2
a
b
−
2
a
2
−
2
b
2
+
4
a
b
+
a
2
−
b
2
=
2
a
2
+
b
2
+
6
a
b
−
2
a
2
−
3
b
2
=
6
a
b
−
2
b
2
.
Therefore, option
D
is correct.
Find the value of
1.05
×
0.95
using standard identity.
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0%
0.9985
0%
0.9975
0%
0.9875
0%
0.9995
Explanation
Given,
1.05
×
0.95
=
(
1
+
0.05
)
(
1
−
0.05
)
.
We know,
(
a
+
b
)
(
a
−
b
)
=
a
2
−
b
2
.
Then,
1.05
×
0.95
=
(
1
+
0.05
)
(
1
−
0.05
)
=
1
2
−
(
0.05
)
2
=
1
−
0.0025
=
0.9975
.
Therefore, option
B
is correct.
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