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Byju's Answer
Standard XII
Mathematics
How to Find the Inverse of a Function
Evaluate : ...
Question
Evaluate :
∫
1
√
(
x
−
1
)
(
x
−
2
)
d
x
.
A
=
log
∣
∣
∣
(
x
−
3
2
)
+
√
x
2
−
3
x
+
2
∣
∣
∣
+
C
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B
=
log
(
(
x
−
3
2
)
+
√
x
2
−
3
x
+
2
)
+
C
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C
=
log
∣
∣
∣
(
x
−
3
2
)
+
√
x
2
−
3
x
+
2
∣
∣
∣
+
5
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D
=
log
∣
∣
∣
(
x
−
3
2
)
+
√
x
2
+
3
x
+
2
∣
∣
∣
+
C
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Solution
The correct option is
A
=
log
∣
∣
∣
(
x
−
3
2
)
+
√
x
2
−
3
x
+
2
∣
∣
∣
+
C
I
=
∫
1
√
x
2
−
3
x
+
2
d
x
=
∫
1
√
x
2
−
3
x
+
9
4
−
9
4
+
2
d
x
=
∫
1
√
(
x
−
3
2
)
2
−
(
1
2
)
2
d
x
=
log
∣
∣ ∣
∣
(
x
−
3
2
)
+
√
(
x
−
3
2
)
2
−
(
1
2
)
2
∣
∣ ∣
∣
+
C
=
log
∣
∣
∣
(
x
−
3
2
)
+
√
x
2
−
3
x
+
2
∣
∣
∣
+
C
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0
Similar questions
Q.
∫
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x
+
4
√
x
2
+
3
x
+
2
d
x
=
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√
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x
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−
B
ln
[
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)
+
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x
2
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x
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+
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Then
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B
=
?
Q.
Find
a
,
b
in
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x
+
2
(
x
2
+
3
x
+
3
)
√
x
+
1
d
x
=
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√
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{
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√
3
(
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)
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+
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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)