Y=ax2+bx+c;PT ¶=a2/2 at x=1/2a (√a2-1-b
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Answer:
Explanation:
The graph of the function (fun.)
y
=
a
x
2
+
b
x
+
c
passes through
the point (pt.)
(
3
,
1
)
, so, the co-ordinates must satisfy the
equation.
∴
1
=
9
a
+
3
b
+
c
...
...
...
...
...
...
.
(
1
)
y
is maximum at
x
=
1
∴
[
d
y
d
x
]
x
=
1
=
0
∴
2
a
+
b
=
0
...
...
(
2
)
Next,
y
max
=
a
2
+
4
occurs at
x
=
1
∴
a
+
b
+
c
=
a
2
+
4
...
...
.
(
3
)
(
1
)
,
(
2
)
,
&
,
(
3
)
⇒
(
a
,
b
,
c
)
=
(
−
3
,
6
,
10
)
,
or
,
(
−
1
,
2
,
4
)
.
For
y
max
at x=1, we must have,
[
d
2
y
d
x
2
]
x
=
1
<
0
d
y
d
x
=
2
a
x
+
b
⇒
d
2
y
d
x
2
=
2
a
=
−
6
or
−
2
,
readily
<
0
.
It is easy to verify that the pair of triads so derived satisfies the given
conditions.
Therefore,
(
a
,
b
,
c
)
=
(
−
3
,
6
,
10
)
,
or
,
(
−
1
,
2
,
4
)
.
Enjoy Maths.!
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