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JEE Main Math Class 12 Limits and Derivatives Part 2 Questions
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© examsnet.com
Question : 5
Total: 56
Let
f
(
x
)
=
a
x
3
+
b
x
2
+
e
x
+
41
be such that
f
(
1
)
=
40
,
f
′
(
1
)
=
2
and
f
′
(
1
)
=
4
.
Then
a
2
+
b
2
+
c
2
is equal to :
[9 Apr 2024 Shift 1]
62
73
54
51
Validate
Solution:
👈: Video Solution
f
(
x
)
=
ax
3
+
b
x
2
+
c
x
+
41
f
′
(
x
)
=
3
ax
2
+
2
bx
+
cx
⇒
f
′
(
1
)
=
3
a
+
2
b
+
c
=
2
.
.
.
(
1
)
f
′
′
(
n
)
=
6
ax
+
2
b
⇒
f
′
′
(
1
)
=
6
a
+
2
b
=
4
3
a
+
b
=
2
.
.
.
.
.
.
.
.
.
(
2
)
(
1
)
−
(
2
)
b
+
c
=
0
.
.
.
.
.
.
.
.
.
(
3
)
f
(
1
)
=
40
a
+
b
+
c
+
41
=
40
use
(
3
)
a
+
41
=
40
by
(
2
)
−
3
+
b
=
2
⇒
b
=
5
&
c
=
−
5
a
2
+
b
2
+
c
2
=
1
+
25
+
25
=
51
© examsnet.com
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