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Joint Entrance Examination
Mathematics
Definite Integration
Question 58
Question 58 of 246
Medium
MCQ
l
i
m
x
→
π
2
(
(
∣
1
(
x
−
π
2
)
2
∫
x
3
(
π
2
)
3
cos
(
t
1
3
)
d
t
)
)
∣
\underset{x\to \frac{\pi }{2}}{lim}((|\frac{1}{{(x-\frac{\pi }{2})}^{2}}\int _{{x}^{3}}^{{(\frac{\pi }{2})}^{3}}\cos ({t}^{\frac{1}{3}})dt))|
x
→
2
π
l
im
((
∣
(
x
−
2
π
)
2
1
∫
x
3
(
2
π
)
3
cos
(
t
3
1
)
d
t
))
∣
is equal to
A
3
π
2
4
\frac{3{\pi }^{2}}{4}
4
3
π
2
B
3
π
2
8
\frac{3{\pi }^{2}}{8}
8
3
π
2
C
3
π
4
\frac{3\pi }{4}
4
3
π
D
3
π
8
\frac{3\pi }{8}
8
3
π
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Official solution
Using L'hospital rule
=
l
i
m
x
→
π
−
2
0
−
cos
x
×
3
x
2
2
(
x
−
π
2
)
=
l
i
m
x
→
π
−
2
sin
(
x
−
π
2
)
2
(
x
−
π
2
)
×
3
π
2
4
=
3
π
2
8
\begin{gathered} =\underset{x\to \frac{{\pi }^{-}}{2}}{lim}\frac{0-\cos x\times 3{x}^{2}}{2(x-\frac{\pi }{2})} \\ =\underset{x\to \frac{{\pi }^{-}}{2}}{lim}\frac{\sin (x-\frac{\pi }{2})}{2(x-\frac{\pi }{2})}\times \frac{3{\pi }^{2}}{4} \\ =\frac{3{\pi }^{2}}{8} \end{gathered}
=
x
→
2
π
−
l
im
2
(
x
−
2
π
)
0
−
cos
x
×
3
x
2
=
x
→
2
π
−
l
im
2
(
x
−
2
π
)
sin
(
x
−
2
π
)
×
4
3
π
2
=
8
3
π
2
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