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Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 213
Question 213 of 416
Medium
MCQ
If
I
=
∫
1
2
d
x
2
x
3
−
9
x
2
+
12
x
+
4
I=\displaystyle\int_1^2\frac{dx}{\sqrt{2x^3-9x^2+12x+4}}
I
=
∫
1
2
2
x
3
−
9
x
2
+
12
x
+
4
d
x
, then:
A
1
8
<
I
2
<
1
4
\dfrac18<I^2<\dfrac14
8
1
<
I
2
<
4
1
B
1
9
<
I
2
<
1
8
\dfrac19<I^2<\dfrac18
9
1
<
I
2
<
8
1
C
1
16
<
I
2
<
1
9
\dfrac1{16}<I^2<\dfrac19
16
1
<
I
2
<
9
1
D
1
6
<
I
2
<
1
2
\dfrac16<I^2<\dfrac12
6
1
<
I
2
<
2
1
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Solutions · 1
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Official solution
Let
f
(
x
)
=
1
2
x
3
−
9
x
2
+
12
x
+
4
f(x)=\dfrac1{\sqrt{2x^3-9x^2+12x+4}}
f
(
x
)
=
2
x
3
−
9
x
2
+
12
x
+
4
1
. Then
f
′
(
x
)
=
−
6
(
x
−
1
)
(
x
−
2
)
2
(
2
x
3
−
9
x
2
+
12
x
+
4
)
3
/
2
f'(x)=\dfrac{-6(x-1)(x-2)}{2(2x^3-9x^2+12x+4)^{3/2}}
f
′
(
x
)
=
2
(
2
x
3
−
9
x
2
+
12
x
+
4
)
3/2
−
6
(
x
−
1
)
(
x
−
2
)
, which is positive on
(
1
,
2
)
(1,2)
(
1
,
2
)
, so
f
f
f
is increasing there. \[f(1)=\frac13,\qquad f(2)=\frac1{\sqrt{8}}\] Since
f
f
f
is increasing,
1
3
<
I
<
1
8
⇒
1
9
<
I
2
<
1
8
\dfrac13<I<\dfrac1{\sqrt{8}}\Rightarrow\dfrac19<I^2<\dfrac18
3
1
<
I
<
8
1
⇒
9
1
<
I
2
<
8
1
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