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Mathematics
The sixth term in the expansion of [2 log 2 √9x-1+7+(1/2(1)5 log 2(3x-1+1))] is 84. Then the number of values of x is
Q. The sixth term in the expansion of
[
2
l
o
g
2
9
x
−
1
+
7
+
2
5
1
l
o
g
2
(
3
x
−
1
+
1
)
1
]
is 84.
Then the number of values of
x
is
3007
211
Binomial Theorem
Report Error
A
0
30%
B
1
20%
C
2
30%
D
3
20%
Solution:
2
l
o
g
2
9
x
−
1
+
7
=
9
x
−
1
+
7
2
2
1
l
o
g
2
(
3
x
−
1
+
1
)
=
(
3
x
−
1
+
1
)
1/5
Also, the sixth term in the expansion is
84
∴
7
C
5
(
9
x
−
1
+
7
)
2
⋅
3
x
−
1
+
1
1
=
84
⇒
3
x
−
1
+
1
9
x
−
1
+
7
=
21
84
⇒
(
3
x
−
1
)
2
−
4
(
3
x
−
1
)
+
3
=
0
⇒
(
3
x
−
1
−
1
)
(
3
x
−
1
−
3
)
=
0
∴
3
x
−
1
=
1
or
3
⇒
x
=
1
or
2
.