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Byju's Answer
Standard XII
Mathematics
Integration by Partial Fractions
x4/x-1x-2=x2+...
Question
x
4
(
x
−
1
)
(
x
−
2
)
=
x
2
+
3
x
+
k
+
16
x
−
2
−
1
x
−
1
⇒
k
=
A
8
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B
1
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C
9
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D
7
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Solution
The correct option is
D
7
The given equation
x
4
(
x
−
1
)
(
x
−
2
)
=
x
2
+
3
x
+
k
+
16
x
−
2
−
1
x
−
1
..........
(
1
)
⟹
x
4
(
x
−
1
)
(
x
−
2
)
=
(
x
2
+
3
x
+
k
)
(
x
−
2
)
(
x
−
1
)
+
16
(
x
−
1
)
−
(
x
−
2
)
(
x
−
2
)
(
x
−
1
)
⟹
x
4
=
(
x
2
+
3
x
+
k
)
(
x
−
2
)
(
x
−
1
)
+
16
(
x
−
1
)
−
(
x
−
2
)
⟹
x
4
=
(
x
2
+
3
x
+
k
)
(
x
2
−
3
x
+
2
)
+
16
x
−
16
−
x
+
2
⟹
x
4
=
(
x
4
−
3
x
2
+
2
x
2
+
3
x
3
−
9
x
2
+
6
x
+
k
x
2
−
3
k
x
+
2
k
)
+
16
x
−
16
−
x
+
2
⟹
0
=
−
7
x
2
+
k
x
2
+
21
x
−
3
k
x
−
14
+
2
k
⟹
0
=
x
2
(
k
−
7
)
+
3
x
(
7
−
k
)
+
2
(
k
−
7
)
⟹
0
=
(
x
2
−
3
x
+
2
)
(
k
−
7
)
⟹
(
x
2
−
3
x
+
2
)
(
k
−
7
)
=
0
⟹
(
x
−
2
)
(
x
−
1
)
(
k
−
7
)
=
0
⟹
x
−
2
=
0
o
r
x
−
1
=
0
o
r
k
−
7
=
0
Now,
x
≠
2
and
x
≠
1
........ (Since, the denominator in equation
(
1
)
becomes zero)
∴
k
−
7
=
0
Hence,
k
=
7
Suggest Corrections
0
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Q.
Which of the following are quadratic equations?
(i) x
2
+ 6x − 4 = 0
(ii)
3
x
2
-
2
x
+
1
2
=
0
(iii)
x
2
+
1
x
2
=
5
(iv)
x
-
3
x
=
x
2
(v)
2
x
2
-
3
x
+
9
=
0
(vi)
x
2
-
2
x
-
x
-
5
=
0
(vii) 3x
2
− 5x + 9 = x
2
− 7x + 3
(viii)
x
+
1
x
=
1
(ix) x
2
− 3x = 0
(x)
x
+
1
x
2
=
3
x
+
1
x
+
4
(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)
x
+
1
x
=
x
2
,
x
≠
0
(xiii) 16x
2
− 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)
3
= x
3
− 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)