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Contests
National and Regional Contests
India Contests
ISI B.Stat Entrance Exam
2006 ISI B.Stat Entrance Exam
2006 ISI B.Stat Entrance Exam
Part of
ISI B.Stat Entrance Exam
Subcontests
(10)
4
1
Hide problems
Archimedes broken chord theorem
In the figure below,
E
E
E
is the midpoint of the arc
A
B
E
C
ABEC
A
BEC
and the segment
E
D
ED
E
D
is perpendicular to the chord
B
C
BC
BC
at
D
D
D
. If the length of the chord
A
B
AB
A
B
is
l
1
l_1
l
1
, and that of the segment
B
D
BD
B
D
is
l
2
l_2
l
2
, determine the length of
D
C
DC
D
C
in terms of
l
1
,
l
2
l_1, l_2
l
1
,
l
2
. [asy] unitsize(1 cm); pair A=2dir(240),B=2dir(190),C=2dir(30),E=2dir(135),D=foot(E,B,C); draw(circle((0,0),2)); draw(A--B--C); draw(E--D); draw(rightanglemark(C,D,E,8)); label("
A
A
A
",A,.5A); label("
B
B
B
",B,.5B); label("
C
C
C
",C,.5C); label("
E
E
E
",E,.5E); label("
D
D
D
",D,dir(-60)); [/asy]
10
1
Hide problems
f(0)=1,f(1)=0, f(n)+f(n-1)=nf(n-1)+(n-1)f(n-2)
Consider a function
f
f
f
on nonnegative integers such that
f
(
0
)
=
1
,
f
(
1
)
=
0
f(0)=1, f(1)=0
f
(
0
)
=
1
,
f
(
1
)
=
0
and
f
(
n
)
+
f
(
n
−
1
)
=
n
f
(
n
−
1
)
+
(
n
−
1
)
f
(
n
−
2
)
f(n)+f(n-1)=nf(n-1)+(n-1)f(n-2)
f
(
n
)
+
f
(
n
−
1
)
=
n
f
(
n
−
1
)
+
(
n
−
1
)
f
(
n
−
2
)
for
n
≥
2
n \ge 2
n
≥
2
. Show that
f
(
n
)
n
!
=
∑
k
=
0
n
(
−
1
)
k
k
!
\frac{f(n)}{n!}=\sum_{k=0}^n \frac{(-1)^k}{k!}
n
!
f
(
n
)
=
k
=
0
∑
n
k
!
(
−
1
)
k
5
1
Hide problems
Easy geometry [ISI (BS) 2006 #5]
Let
A
,
B
A,B
A
,
B
and
C
C
C
be three points on a circle of radius
1
1
1
.(a) Show that the area of the triangle
A
B
C
ABC
A
BC
equals
1
2
(
sin
(
2
∠
A
B
C
)
+
sin
(
2
∠
B
C
A
)
+
sin
(
2
∠
C
A
B
)
)
\frac12(\sin(2\angle ABC)+\sin(2\angle BCA)+\sin(2\angle CAB))
2
1
(
sin
(
2∠
A
BC
)
+
sin
(
2∠
BC
A
)
+
sin
(
2∠
C
A
B
))
(b) Suppose that the magnitude of
∠
A
B
C
\angle ABC
∠
A
BC
is fixed. Then show that the area of the triangle
A
B
C
ABC
A
BC
is maximized when
∠
B
C
A
=
∠
C
A
B
\angle BCA=\angle CAB
∠
BC
A
=
∠
C
A
B
(c) Hence or otherwise, show that the area of the triangle
A
B
C
ABC
A
BC
is maximum when the triangle is equilateral.
2
1
Hide problems
Problem on irrationals...... [ISI(BS) 06#2]
Suppose that
a
a
a
is an irrational number.(a) If there is a real number
b
b
b
such that both
(
a
+
b
)
(a+b)
(
a
+
b
)
and
a
b
ab
ab
are rational numbers, show that
a
a
a
is a quadratic surd. (
a
a
a
is a quadratic surd if it is of the form
r
+
s
r+\sqrt{s}
r
+
s
or
r
−
s
r-\sqrt{s}
r
−
s
for some rationals
r
r
r
and
s
s
s
, where
s
s
s
is not the square of a rational number).(b) Show that there are two real numbers
b
1
b_1
b
1
and
b
2
b_2
b
2
such thati)
a
+
b
1
a+b_1
a
+
b
1
is rational but
a
b
1
ab_1
a
b
1
is irrational.ii)
a
+
b
2
a+b_2
a
+
b
2
is irrational but
a
b
2
ab_2
a
b
2
is rational. (Hint: Consider the two cases, where
a
a
a
is a quadratic surd and
a
a
a
is not a quadratic surd, separately).
8
1
Hide problems
Solve log[base 2] x=cx
Show that there exists a positive real number
x
≠
2
x\neq 2
x
=
2
such that
log
2
x
=
x
2
\log_2x=\frac{x}{2}
lo
g
2
x
=
2
x
. Hence obtain the set of real numbers
c
c
c
such that
log
2
x
x
=
c
\frac{\log_2x}{x}=c
x
lo
g
2
x
=
c
has only one real solution.
9
1
Hide problems
N=4*M, M is the reverse of N
Find a four digit number
M
M
M
such that the number
N
=
4
×
M
N=4\times M
N
=
4
×
M
has the following properties.(a)
N
N
N
is also a four digit number(b)
N
N
N
has the same digits as in
M
M
M
but in reverse order.
6
1
Hide problems
f(x)=x-xe^(-1/x)
(a) Let
f
(
x
)
=
x
−
x
e
−
1
x
,
x
>
0
f(x)=x-xe^{-\frac1x}, \ \ x>0
f
(
x
)
=
x
−
x
e
−
x
1
,
x
>
0
. Show that
f
(
x
)
f(x)
f
(
x
)
is an increasing function on
(
0
,
∞
)
(0,\infty)
(
0
,
∞
)
, and
lim
x
→
∞
f
(
x
)
=
1
\lim_{x\to\infty} f(x)=1
lim
x
→
∞
f
(
x
)
=
1
.(b) Using part (a) or otherwise, draw graphs of
y
=
x
−
1
,
y
=
x
,
y
=
x
+
1
y=x-1, y=x, y=x+1
y
=
x
−
1
,
y
=
x
,
y
=
x
+
1
, and
y
=
x
e
−
1
∣
x
∣
y=xe^{-\frac{1}{|x|}}
y
=
x
e
−
∣
x
∣
1
for
−
∞
<
x
<
∞
-\infty<x<\infty
−
∞
<
x
<
∞
using the same
X
X
X
and
Y
Y
Y
axes.
1
1
Hide problems
Equation of the normal
If the normal to the curve
x
2
3
+
y
2
3
=
a
2
3
x^{\frac{2}{3}}+y^{\frac23}=a^{\frac23}
x
3
2
+
y
3
2
=
a
3
2
at some point makes an angle
θ
\theta
θ
with the
X
X
X
-axis, show that the equation of the normal is
y
cos
θ
−
x
sin
θ
=
a
cos
2
θ
y\cos\theta-x\sin\theta=a\cos 2\theta
y
cos
θ
−
x
sin
θ
=
a
cos
2
θ
7
1
Hide problems
a combinatorial inequality
for any positive integer
n
n
n
greater than
1
1
1
, show that
2
n
<
(
2
n
n
)
<
2
n
∏
i
=
0
n
−
1
(
1
−
i
n
)
2^n<\binom{2n}{n}<\frac{2^n}{\prod\limits_{i=0}^{n-1} \left(1-\frac{i}{n}\right)}
2
n
<
(
n
2
n
)
<
i
=
0
∏
n
−
1
(
1
−
n
i
)
2
n
3
1
Hide problems
A verrrrry well known one!
Prove that
n
4
+
4
n
n^4 + 4^{n}
n
4
+
4
n
is composite for all values of
n
n
n
greater than
1
1
1
.