Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
AP EAPCET 24-Aug-2021 Shift 1 Paper
Show Para
Hide Para
Share question:
© examsnet.com
Question : 14
Total: 160
Let
Q
(
x
)
be a polynomial of degree
n
. If
Q
(
1
)
=
1
and
Q
(
2
x
)
Q
(
x
+
1
)
+
56
x
+
7
−
8
=
0
, then the value of
n
C
0
+
n
C
1
+
.
.
.
.
.
.
+
n
C
n
=
32
64
8
16
Validate
Solution:
Q
(
x
)
be a polynomial of degree
n
Q
(
1
)
=
1
and
Q
(
2
x
)
Q
(
x
+
1
)
+
56
x
+
7
−
8
=
0
...(i)
Put
x
=
0
in Eq. (i), we get
Q
(
0
)
Q
(
1
)
+
56
7
−
8
=
0
⇒
Q
(
0
)
1
+
8
−
8
=
0
Q
(
0
)
=
0
Solving Eq. (i), we get
Q
(
2
x
)
Q
(
x
+
1
)
+
56
−
8
x
−
56
x
+
7
=
0
⇒
Q
(
2
x
)
Q
(
x
+
1
)
=
8
x
x
+
7
=
0
⇒
Q
(
2
x
)
Q
(
x
+
1
)
=
8
x
x
+
7
...(ii)
∵
Q
(
0
)
=
0
∴
x
is a factor of
Q
(
x
)
.
Let
Q
(
x
)
=
x
P
(
x
)
∴
Q
(
2
x
)
Q
(
x
+
1
)
=
8
x
x
+
7
2
x
.
P
(
2
x
)
(
x
+
1
)
P
(
x
+
1
)
=
8
x
x
+
7
P
(
2
x
)
P
(
x
+
1
=
4
(
x
+
1
)
x
+
7
...(ii)
Put
x
=
−
1
P
(
−
2
)
P
(
0
)
=
0
⇒
P
(
−
2
)
=
0
⇒
x
+
2
is factor of
P
(
x
)
P
(
x
)
=
(
x
+
2
)
R
(
x
)
Now, in Eq, (ii), we get
4
(
x
+
1
)
x
+
7
=
P
(
2
x
)
P
(
x
+
1
)
=
(
2
x
+
2
)
R
(
2
x
)
(
x
+
1
+
2
)
R
(
x
+
1
)
⇒
4
(
x
+
1
)
x
+
7
=
2
x
+
2
x
+
3
R
(
2
x
)
R
(
x
+
1
)
2
(
x
+
3
)
x
+
7
=
R
(
2
x
)
R
(
x
+
1
)
∵
x
=
−
3
⇒
0
=
R
(
−
6
)
R
(
−
2
)
⇒
R
(
−
6
)
=
0
=
x
+
6
is a factor of
R
(
x
)
.
So,
R
(
x
)
=
(
x
+
6
)
S
(
x
)
2
(
x
+
3
)
x
+
7
=
R
(
2
x
)
R
(
x
+
1
)
=
(
2
x
+
6
)
S
(
2
x
)
(
x
+
6
+
1
)
S
(
x
+
1
)
S
(
2
x
)
S
(
x
+
1
)
=
1
⇒
S
(
2
x
)
=
S
(
x
+
1
)
for all
x
S
(
x
)
is constant function.
∴
Q
(
x
)
=
x
P
(
x
)
=
x
(
x
+
2
)
R
(
x
)
=
x
(
x
+
2
)
(
x
+
3
)
S
(
x
)
∴
Q
(
x
)
is cubic polynomial.
∴
n
=
3
Required value
=
3
C
0
+
3
C
1
+
3
C
2
+
3
C
3
=
1
+
3
+
3
+
1
=
8
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
Prev Question
Next Question
Examsnet