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AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper
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© examsnet.com
Question : 20
Total: 160
if
x
4
(
x
−
a
)
(
x
−
b
)
(
x
−
c
)
= P (x) +
A
x
−
a
+
B
x
−
b
+
C
x
−
c
, then P (0) + A (a - b) (a - c) =
a
4
+
b
4
+
c
4
+ a
a + b + c
a
4
- a - b - c
a + b + c +
a
4
Validate
Solution:
Consider the equation.
x
4
(
x
−
a
)
(
x
−
b
)
(
x
−
c
)
=
P
(
x
)
+
A
x
−
a
+
B
x
−
b
+
C
x
−
c
The above equation is solved as,
x
4
=
[
P
(
x
)
(
x
−
a
)
(
x
−
b
)
(
x
−
c
)
+
A
(
x
−
b
)
(
x
−
c
)
+
B
(
x
−
a
)
(
x
−
c
)
+
C
(
x
−
a
)
(
x
−
b
)
]
Substitute 0 for
x
in the above expression.
P
(
0
)
a
b
c
=
[
A
b
c
+
B
a
c
+
C
a
b
]
P
(
0
)
=
A
a
+
B
b
+
C
c
Now at
x
=
a
A
=
a
4
(
a
−
b
)
(
a
−
c
)
At
x
=
b
B
=
b
4
(
b
−
c
)
(
b
−
a
)
At
x
=
c
C
=
c
4
(
c
−
a
)
(
c
−
b
)
Thus
P
(
0
)
=
a
3
(
a
−
b
)
(
a
−
c
)
+
b
3
(
b
−
c
)
(
b
−
a
)
+
c
3
(
c
−
a
)
(
c
−
a
)
=
a
3
(
c
−
b
)
+
b
c
(
c
2
−
b
2
)
+
a
(
b
3
−
c
3
)
(
a
−
b
)
(
b
−
c
)
(
c
−
a
)
=
a
3
+
b
c
(
c
+
b
)
−
a
(
b
2
+
c
2
+
b
c
)
(
a
−
b
)
(
a
−
c
)
=
a
(
a
2
−
b
2
)
−
c
2
(
a
−
b
)
−
b
c
(
a
−
b
)
(
a
−
b
)
(
a
−
c
)
Solve further
P
(
0
)
=
(
a
2
+
b
a
)
−
c
2
−
b
c
(
a
−
c
)
=a + b + c
Therefore,
P
(
0
)
+
(
a
−
b
)
(
a
−
c
)
A
=
a
+
b
+
c
+
a
4
© examsnet.com
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