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CBSE Questions for Class 9 Maths Polynomials Quiz 12 - MCQExams.com
CBSE
Class 9 Maths
Polynomials
Quiz 12
If
f
(
x
)
=
8
then
f
(
x
)
is called
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0%
Constant polynomials
0%
linear polynomials
0%
quadratic polynomials
0%
Cubic polynomials
Explanation
If
f
(
x
)
=
8
then
f
(
x
)
is called Constant polynomials
The degree of polynomial is
x
+
2
is:
Report Question
0%
2
0%
1
0%
3
0%
4
Explanation
Step 1: Apply the property of polynomial.
We have, x + 2
We know that the highest power of the variable in a polynomial in one variable is
called the degree of the polynomial.
Here, the highest power of variable x is 1.
Thus, degree of polynomial is 1, option - (B).
Write the correct alternative answer for the following question:
Which is the degree of the polynomial
2
x
2
+
5
x
3
+
7
?
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0%
3
0%
2
0%
5
0%
7
Explanation
The given polynomial is
2
x
2
+
5
x
3
+
7
.
We know,
in a polynomial in one variable,
the highest power of the polynomial term with a non-zero coefficient, is called the
degree
of the polynomial.
Here, the highest power of the variable
x
is
3
.
∴
The degree of the given polynomial is
3
.
Therefore, option
A
is correct.
The degree of the polynomial
x
4
+
x
3
is:
Report Question
0%
2
0%
3
0%
5
0%
4
Explanation
The highest power of the variable in a polynomial in one variable is called the
degree
of the polynomial.
Here, in the given polynomial
x
4
+
x
3
, the highest power of the variable
x
is
4
.
∴
The degree of the given polynomial is
4
.
Therefore, option
D
is correct.
The number of zeroes of linear polynomial at most is
Report Question
0%
0
0%
1
0%
2
0%
3
Explanation
Step-1: Explanation for number of zeroes of linear polynomial
The number of zeroes of a polynomial is equal to the degree of the polynomial
As the degree of a linear polynomial is 1, the number of zeroes is 1
Therefore, the number of zeroes of the linear polynomial at most is 1
Zero of the polynomial
p
(
x
)
=
√
3
x
+
3
is
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0%
−
√
3
0%
−
√
3
3
0%
3
√
3
0%
3
√
3
One zero of
p
(
x
)
=
2
x
+
1
will be
Report Question
0%
1
2
0%
3
0%
−
1
2
0%
1
Explanation
p
(
x
)
=
2
x
+
1
For zeroes
p
(
x
)
=
0
0
=
2
x
+
1
⇒
x
=
−
1
2
State True/False:-
One of the zeroes of the polynomial
x
2
+
2
x
−
15
is
11
Report Question
0%
True
0%
False
Explanation
To find the zeroes of the polynomial, we have to proceed in the following way:-
Let f(x)=
x
2
+
2
x
−
15
For zeroes
f
(
x
)
=
0
⇒
x
2
+
2
x
−
15
=
0
⇒
x
2
+
5
x
−
3
x
−
15
=
0
⇒
x
(
x
+
5
)
−
3
(
x
+
5
)
=
0
⇒
(
x
−
3
)
(
x
+
5
)
=
0
x
−
3
=
0
or
x
+
5
=
0
x
=
3
or
x
=
−
5
Hence the zeroes of the polynomial
x
2
+
2
x
−
15
are
3
,
−
5
.
Hence, given statement is
false.
Identify the zeroes of the given polynomial.
p
(
z
)
=
4
z
2
−
15
z
π
−
4
π
3
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0%
4
π
,
−
π
4
0%
−
4
π
,
π
4
0%
4
π
,
π
4
0%
−
4
π
,
−
π
4
For
p
(
x
)
=
5
x
2
−
2
3
x
+
8
, then value of
p
(
3
)
=
Report Question
0%
50
0%
51
0%
55
0%
35
Explanation
The polynomial given is
p
(
x
)
=
5
x
2
−
2
3
x
+
8
after substituting the value
x
=
3
in the polynomial we get:
p
(
3
)
=
5
x
2
−
2
3
x
+
8
=
5
(
3
)
2
−
(
2
3
×
3
)
+
8
=
(
5
×
9
)
−
2
+
8
=
45
−
2
+
8
=
53
−
2
=
51
Hence,
p
(
3
)
=
51
.
α
,
β
,
γ
be the zeroes of the expression
a
x
3
+
b
x
2
+
4
x
+
7
, then the value of
α
β
+
β
γ
+
γ
α
is:
Report Question
0%
-4
0%
4
α
0%
−
4
α
0%
None of these
If
p
(
x
)
=
1
+
7
x
−
9
x
2
+
2
3
x
3
, then
p
(
−
3
)
=
Report Question
0%
−
81
0%
−
137
0%
−
119
0%
−
100
Explanation
Consider the polynomial
p
(
x
)
=
1
+
7
x
−
9
x
2
+
2
3
x
3
and substitute
x
=
−
3
in the polynomial:
p
(
−
3
)
=
1
+
(
7
×
−
3
)
−
9
(
−
3
)
2
+
2
3
(
−
3
)
3
=
1
−
21
−
(
9
×
9
)
+
(
2
3
×
−
27
)
=
1
−
21
−
81
−
18
=
−
119
Hence,
p
(
−
3
)
=
−
119
.
Calculate the number of real numbers
k
such that
f
(
k
)
=
2
if
f
(
x
)
=
x
4
−
3
x
3
−
9
x
2
+
4
.
Report Question
0%
None
0%
One
0%
Two
0%
Three
0%
Four
Explanation
Given,
f
(
x
)
=
x
4
−
3
x
3
−
9
x
2
+
4
,
f
(
k
)
=
2
Therefore,
y
=
2
=
f
(
x
)
⇒
2
=
x
4
−
3
x
3
−
9
x
2
+
4
⇒
x
4
−
3
x
3
−
9
x
2
+
2
=
0
Then plug in consecutive numbers for
x
and if the sign of
f
(
x
)
changes,
f
(
−
2
)
=
6
f
(
−
1
)
=
−
3
so there's one
f
(
0
)
=
2
so that's two
f
(
1
)
=
−
9
so that's three
And eventually it has to become positive again so there are four.
If
f
(
x
)
=
x
6
−
10
x
5
−
10
x
4
−
10
x
3
−
10
x
2
−
10
x
+
10
, the value of
f
(
11
)
is
Report Question
0%
1
0%
10
0%
11
0%
21
Explanation
The given expression
f
(
x
)
=
x
6
−
10
x
5
−
10
x
4
−
10
x
3
−
10
x
2
−
10
x
+
10
can be rewritten as
f
(
x
)
=
x
6
−
10
(
x
5
+
x
4
+
x
3
+
x
2
+
x
−
1
)
.
Now, substitute
x
=
11
in the above expression as shown below:
⇒
f
(
x
)
=
x
6
−
10
(
x
5
+
x
4
+
x
3
+
x
2
+
x
−
1
)
⇒
f
(
11
)
=
11
6
−
10
(
11
5
+
11
4
+
11
3
+
11
2
+
11
−
1
)
⇒
f
(
11
)
=
1771561
−
10
(
161051
+
14641
+
1331
+
121
+
11
−
1
)
⇒
f
(
11
)
=
1771561
−
(
10
×
177154
)
⇒
f
(
11
)
=
1771561
−
1771540
⇒
f
(
11
)
=
21
Hence,
f
(
11
)
=
21
.
Two roots of the equation
4
x
3
−
p
x
2
+
q
x
−
2
p
=
0
are 4 andWhat is the third root?
Report Question
0%
11
27
0%
11
13
0%
11
0%
11
15
0%
−
22
27
The equation
8
x
6
+
72
x
5
+
b
x
4
+
c
x
3
−
687
x
2
−
2160
x
−
1700
=
0
, as shown in the figure, has two complex roots. The product of these complex roots is
Report Question
0%
−
4
0%
17
2
0%
9
0%
−
687
2
0%
427
2
L
e
t
x
=
√
3
−
√
5
a
n
d
y
=
√
3
+
√
5
.
If the value of expression
x
−
y
+
2
x
2
y
+
2
x
y
2
−
x
4
y
+
x
y
4
can be expressed in the form
√
p
+
√
q
w
h
e
r
e
p
,
q
∈
N
,
then
(
p
+
q
)
has the value equal
Report Question
0%
448
0%
610
0%
510
0%
540
Explanation
The zeros of the polynomial
x
2
−
9
are
Report Question
0%
3,3
0%
√
3
,
−
√
3
0%
3,-3
0%
−
√
3
,
−
√
3
Number of root of equation
3|x|-|2-x|=1
is
Report Question
0%
0
0%
2
0%
4
0%
7
The number of polynomials having zeroes as
−
2
and
5
is?
Report Question
0%
0
0%
2
0%
3
0%
More than
3
If a + b + 2 c = 0,
c
≠
0
, then equation $$
Report Question
0%
At least one root in (0, 1)
0%
At least one root in (0, 2)
0%
At least one root in (-1, 1)
0%
None of these
The roots of the equation
(
x
−
1
)
3
+
8
=
0
are
Report Question
0%
−
1
,
1
+
2
ω
,
1
+
2
ω
2
0%
−
1
,
1
−
2
ω
,
1
−
2
ω
2
0%
2
,
2
ω
,
2
ω
2
0%
2
,
1
+
2
ω
,
1
+
2
ω
2
Number of common roots of the equation
x
4
+
x
2
+
1
=
0
and
x
8
−
1
=
0
is
Report Question
0%
zero
0%
1
0%
2
0%
4
For equation
x
3
−
6
x
2
+
9
x
+
k
=
0
to have exactly one root in (1, 3), the set of values of k is
Report Question
0%
(-4, 0)
0%
(1, 3)
0%
(0, 4)
0%
None of these
If the sum of two roots of the equation
x
3
−
p
x
2
+
q
x
−
r
=
0
is zero, then
Report Question
0%
pq=r
0%
qr=p
0%
pr=q
0%
pqr=1
Let
f
(
x
)
be a polynomial of degree 5 with leading coefficient unity, such that
f
(
1
)
=
5
,
f
(
2
)
=
4
,
f
(
3
)
=
3
,
f
(
4
)
=
2
and
f
(
5
)
=
1
,
then
Sum of the roots of
f
(
x
)
is equal to:
Report Question
0%
15
0%
−
15
0%
21
0%
Can't be determine
Let
f
(
x
)
be a polynomial of degree 5 with leading coefficient unity, such that
f
(
1
)
=
5
,
f
(
2
)
=
4
,
f
(
3
)
=
3
,
f
(
4
)
=
2
and
f
(
5
)
=
1
,
then
Product of the roots of
f
(
x
)
is equal to:
Report Question
0%
120
0%
−
120
0%
114
0%
−
114
Let for
a
≠
a
1
≠
0
,
f
(
x
)
=
a
x
2
+
b
x
+
c
,
g
(
x
)
=
a
1
x
2
+
b
1
x
+
c
1
and
p
(
x
)
=
f
(
x
)
−
g
(
x
)
. If
p
(
x
)
=
0
only for
x
=
−
1
and
p
(
−
2
)
=
2
then the value of
p
(
2
)
is
Report Question
0%
18
0%
3
0%
9
0%
6
Let
f
(
x
)
be a polynomial of degree 5 with leading coefficient unity, such that
f
(
1
)
=
5
,
f
(
2
)
=
4
,
f
(
3
)
=
3
,
f
(
4
)
=
2
and
f
(
5
)
=
1
,
then
f
(
6
)
is equal to:
Report Question
0%
120
0%
−
120
0%
0
0%
6
The product of the zero of the polynomal
x
2
−
4
x
+
3
is ....
Report Question
0%
4
0%
1
0%
-4
0%
3
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Incorrect : 0
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