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National and Regional Contests
India Contests
ISI B.Stat Entrance Exam
2009 ISI B.Stat Entrance Exam
2009 ISI B.Stat Entrance Exam
Part of
ISI B.Stat Entrance Exam
Subcontests
(10)
5
1
Hide problems
rectangular parallelopiped in cylinder.... may be calculus
A cardboard box in the shape of a rectangular parallelopiped is to be enclosed in a cylindrical container with a hemispherical lid. If the total height of the container from the base to the top of the lid is
60
60
60
centimetres and its base has radius
30
30
30
centimetres, find the volume of the largest box that can be completely enclosed inside the container with the lid on.
1
1
Hide problems
Two trains intersect at an angle theta
Two train lines intersect each other at a junction at an acute angle
θ
\theta
θ
. A train is passing along one of the two lines. When the front of the train is at the junction, the train subtends an angle
α
\alpha
α
at a station on the other line. It subtends an angle
β
(
<
α
)
\beta (<\alpha)
β
(
<
α
)
at the same station, when its rear is at the junction. Show that
tan
θ
=
2
sin
α
sin
β
sin
(
α
−
β
)
\tan\theta=\frac{2\sin\alpha\sin\beta}{\sin(\alpha-\beta)}
tan
θ
=
sin
(
α
−
β
)
2
sin
α
sin
β
3
1
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M=max of BPR, APQ, PQCR. Minimize M
Let
A
B
C
ABC
A
BC
be a right-angled triangle with
B
C
=
A
C
=
1
BC=AC=1
BC
=
A
C
=
1
. Let
P
P
P
be any point on
A
B
AB
A
B
. Draw perpendiculars
P
Q
PQ
PQ
and
P
R
PR
PR
on
A
C
AC
A
C
and
B
C
BC
BC
respectively from
P
P
P
. Define
M
M
M
to be the maximum of the areas of
B
P
R
,
A
P
Q
BPR, APQ
BPR
,
A
PQ
and
P
Q
C
R
PQCR
PQCR
. Find the minimum possible value of
M
M
M
.
7
1
Hide problems
Easy geometry: Radius of pentagon is (xcsc36)/2
Show that the vertices of a regular pentagon are concyclic. If the length of each side of the pentagon is
x
x
x
, show that the radius of the circumcircle is
x
2
sin
3
6
∘
\frac{x}{2\sin 36^\circ}
2
s
i
n
3
6
∘
x
.
8
1
Hide problems
Select 3 frm {1,2,..,4n}, 4 divides their sum
Find the number of ways in which three numbers can be selected from the set
{
1
,
2
,
⋯
,
4
n
}
\{1,2,\cdots ,4n\}
{
1
,
2
,
⋯
,
4
n
}
, such that the sum of the three selected numbers is divisible by
4
4
4
.
4
1
Hide problems
K-th order arithmetic progression... [isi(bs)'09#4]
A sequence is called an arithmetic progression of the first order if the differences of the successive terms are constant. It is called an arithmetic progression of the second order if the differences of the successive terms form an arithmetic progression of the first order. In general, for
k
≥
2
k\geq 2
k
≥
2
, a sequence is called an arithmetic progression of the
k
k
k
-th order if the differences of the successive terms form an arithmetic progression of the
(
k
−
1
)
(k-1)
(
k
−
1
)
-th order. The numbers
4
,
6
,
13
,
27
,
50
,
84
4,6,13,27,50,84
4
,
6
,
13
,
27
,
50
,
84
are the first six terms of an arithmetic progression of some order. What is its least possible order? Find a formula for the
n
n
n
-th term of this progression.
9
1
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Perimeter of R: ISI(bs)'09 #9
Consider
6
6
6
points located at
P
0
=
(
0
,
0
)
,
P
1
=
(
0
,
4
)
,
P
2
=
(
4
,
0
)
,
P
3
=
(
−
2
,
−
2
)
,
P
4
=
(
3
,
3
)
,
P
5
=
(
5
,
5
)
P_0=(0,0), P_1=(0,4), P_2=(4,0), P_3=(-2,-2), P_4=(3,3), P_5=(5,5)
P
0
=
(
0
,
0
)
,
P
1
=
(
0
,
4
)
,
P
2
=
(
4
,
0
)
,
P
3
=
(
−
2
,
−
2
)
,
P
4
=
(
3
,
3
)
,
P
5
=
(
5
,
5
)
. Let
R
R
R
be the region consisting of all points in the plane whose distance from
P
0
P_0
P
0
is smaller than that from any other
P
i
P_i
P
i
,
i
=
1
,
2
,
3
,
4
,
5
i=1,2,3,4,5
i
=
1
,
2
,
3
,
4
,
5
. Find the perimeter of the region
R
R
R
.
6
1
Hide problems
n-th derivative of lnx/x
Let
f
(
x
)
f(x)
f
(
x
)
be a function satisfying
x
f
(
x
)
=
ln
x
for
x
>
0
xf(x)=\ln x \ \ \ \ \ \ \ \ \text{for} \ \ x>0
x
f
(
x
)
=
ln
x
for
x
>
0
Show that
f
(
n
)
(
1
)
=
(
−
1
)
n
+
1
n
!
(
1
+
1
2
+
⋯
+
1
n
)
f^{(n)}(1)=(-1)^{n+1}n!\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)
f
(
n
)
(
1
)
=
(
−
1
)
n
+
1
n
!
(
1
+
2
1
+
⋯
+
n
1
)
where
f
(
n
)
(
x
)
f^{(n)}(x)
f
(
n
)
(
x
)
denotes the
n
n
n
-th derivative evaluated at
x
x
x
.
10
1
Hide problems
Show that x_n=n+(n)
Let
x
n
x_n
x
n
be the
n
n
n
-th non-square positive integer. Thus
x
1
=
2
,
x
2
=
3
,
x
3
=
5
,
x
4
=
6
,
x_1=2, x_2=3, x_3=5, x_4=6,
x
1
=
2
,
x
2
=
3
,
x
3
=
5
,
x
4
=
6
,
etc. For a positive real number
x
x
x
, denotes the integer closest to it by
⟨
x
⟩
\langle x\rangle
⟨
x
⟩
. If
x
=
m
+
0.5
x=m+0.5
x
=
m
+
0.5
, where
m
m
m
is an integer, then define
⟨
x
⟩
=
m
\langle x\rangle=m
⟨
x
⟩
=
m
. For example,
⟨
1.2
⟩
=
1
,
⟨
2.8
⟩
=
3
,
⟨
3.5
⟩
=
3
\langle 1.2\rangle =1, \langle 2.8 \rangle =3, \langle 3.5\rangle =3
⟨
1.2
⟩
=
1
,
⟨
2.8
⟩
=
3
,
⟨
3.5
⟩
=
3
. Show that
x
n
=
n
+
⟨
n
⟩
x_n=n+\langle \sqrt{n}\rangle
x
n
=
n
+
⟨
n
⟩
2
1
Hide problems
f"(x)>0, show that int(f(x)cosx dx) >0
Let
f
(
x
)
f(x)
f
(
x
)
be a continuous function, whose first and second derivatives are continuous on
[
0
,
2
π
]
[0,2\pi]
[
0
,
2
π
]
and
f
′
′
(
x
)
≥
0
f''(x) \geq 0
f
′′
(
x
)
≥
0
for all
x
x
x
in
[
0
,
2
π
]
[0,2\pi]
[
0
,
2
π
]
. Show that
∫
0
2
π
f
(
x
)
cos
x
d
x
≥
0
\int_{0}^{2\pi} f(x)\cos x dx \geq 0
∫
0
2
π
f
(
x
)
cos
x
d
x
≥
0