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JEE Main 10 Apr 2019 Paper 1 Solved Paper
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Section:
Mathematics
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© examsnet.com
Question : 64
Total: 90
If the coefficients of
x
2
‌
and
‌
x
3
are both zero, in the expansion of the expression
(
1
+
a
x
+
b
x
2
)
(
1
−
3
x
)
15
in powers of x, then the ordered pair (a, b) is equal to
[10 Apr 2019 Shift 1]
(28, 315)
(-54, 315)
(28, 861)
(-21, 714)
Validate
Solution:
(
1
+
a
x
+
b
x
2
)
(
1
−
3
x
)
15
=
(
1
+
a
x
+
b
x
2
)
(
1
−
15
C
1
(
3
x
)
+
15
C
2
(
3
x
)
2
−
15
C
3
(
3
x
)
3
+
.
.
.
.
)
Coefficient of
x
2
is
15
C
2
3
2
−
a
.
15
C
1
.
3
+
b
=
0
⇒
45
a
−
b
=
945
Coefficient of
x
3
is
−
15
C
3
.
3
3
+
a
15
C
2
.
3
2
−
b
.5
C
1
.3
=
0
⇒
21
a
−
b
=
273
Solving above equations, we get
a
=
28
,
b
=
315
© examsnet.com
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