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Definite Integrals Part 3
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Section:
Mathematics
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© examsnet.com
Question : 67
Total: 115
Let
f
:
R
→
R
be defined
f
(
x
)
=
a
e
2
x
+
b
e
x
+
c
x
. If
f
(
0
)
=
−
1
,
f
′
(
log
e
2
)
=
21
and
∫
0
log
e
4
(
f
(
x
)
−
c
x
)
dx
=
39
2
, then the value of
|
a
+
b
+
c
|
equals :
[30-Jan-2024 Shift 2]
16
10
12
8
Validate
Solution:
👈: Video Solution
f
(
x
)
=
ae
2
x
+
be
x
+
cx
f
(
0
)
=
−
1
a
+
b
=
−
1
f
′
(
x
)
=
2
ae
2
x
+
b
e
x
+
c
f
′
(
ln
2
)
=
21
8
a
+
2
b
+
c
=
21
ln
4
∫
0
(
ae
2
x
+
b
e
x
)
dx
=
39
2
[
ae
2
x
+
be
x
]
0
ln
4
=
39
2
⇒
8
a
+
4
b
−
a
2
−
b
=
39
2
15
a
+
6
b
=
39
15
a
−
6
a
−
6
=
39
9
a
=
45
⇒
a
=
5
b
=
−
6
c
=
21
−
40
+
12
=
−
7
a
+
b
+
c
=
−
8
|
a
+
b
+
c
∣
=
8
© examsnet.com
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