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Test Index
JEE Main Math Class 12 Application of Derivatives Part 2 Questions
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© examsnet.com
Question : 7
Total: 100
The curve
y
(
x
)
=
a
x
3
+
b
x
2
+
c
x
+
5
touches the
x
-axis at the point
P
(
−
2
,
0
)
and cuts the
y
axis at the point
Q
, where
y
′
is equal to 3 . Then the local maximum value of
y
(
x
)
is:
[25-Jul-2022-Shift-1]
27
4
29
4
37
4
9
2
Validate
Solution:
f
(
x
)
=
y
=
a
x
3
+
b
x
2
+
c
x
+
5
.
.
.
(i)
d
y
d
x
=
3
a
x
2
+
2
b
x
+
c
.
.
.
(
i
i
)
Touches
x
-axis at
P
(
−
2
,
0
)
.
⇒
y
|
x
=
−
2
=
0
⇒
−
8
a
+
4
b
−
2
c
+
5
=
0
.
.
.
.
.
.
(iii)
Touches
x
-axis at
P
(
−
2
,
0
)
also implies
.
d
y
d
x
|
x
=
−
2
=
0
⇒
12
a
−
4
b
+
c
=
0
.
.
.
.
.
.
(
i
v
)
y
=
f
(
x
)
cuts
y
-axis at
(
0
,
5
)
Given,
.
d
y
d
x
|
x
=
0
=
c
=
3
From (iii), (iv) and (v)
a
=
−
1
2
,
b
=
−
3
4
,
c
=
3
⇒
f
(
x
)
=
−
x
2
2
−
3
4
x
2
+
3
x
+
5
f
′
(
x
)
=
−
3
2
x
2
−
3
2
x
+
3
=
−
3
2
(
x
+
2
)
(
x
−
1
)
f
′
(
x
)
=
0
at
x
=
−
2
and
x
=
1
By first derivative test
x
=
1
in point of local maximum Hence local maximum value of
f
(
x
)
is
f
(
1
)
i.e.,
27
4
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