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National and Regional Contests
India Contests
ISI B.Stat Entrance Exam
2011 ISI B.Stat Entrance Exam
2011 ISI B.Stat Entrance Exam
Part of
ISI B.Stat Entrance Exam
Subcontests
(10)
5
1
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ISI B-Stat 2011, P5: Prove that PQCD is a parallelogram
A
B
C
D
ABCD
A
BC
D
is a trapezium such that
A
B
∥
D
C
AB\parallel DC
A
B
∥
D
C
and
A
B
D
C
=
α
>
1
\frac{AB}{DC}=\alpha >1
D
C
A
B
=
α
>
1
. Suppose
P
P
P
and
Q
Q
Q
are points on
A
C
AC
A
C
and
B
D
BD
B
D
respectively, such that
A
P
A
C
=
B
Q
B
D
=
α
−
1
α
+
1
\frac{AP}{AC}=\frac{BQ}{BD}=\frac{\alpha -1}{\alpha+1}
A
C
A
P
=
B
D
BQ
=
α
+
1
α
−
1
Prove that
P
Q
C
D
PQCD
PQC
D
is a parallelogram.
9
1
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P9: sum of products of the reciprocals
Consider all non-empty subsets of the set
{
1
,
2
⋯
,
n
}
\{1,2\cdots,n\}
{
1
,
2
⋯
,
n
}
. For every such subset, we find the product of the reciprocals of each of its elements. Denote the sum of all these products as
S
n
S_n
S
n
. For example,
S
3
=
1
1
+
1
2
+
1
3
+
1
1
⋅
2
+
1
1
⋅
3
+
1
2
⋅
3
+
1
1
⋅
2
⋅
3
S_3=\frac11+\frac12+\frac13+\frac1{1\cdot 2}+\frac1{1\cdot 3}+\frac1{2\cdot 3} +\frac1{1\cdot 2\cdot 3}
S
3
=
1
1
+
2
1
+
3
1
+
1
⋅
2
1
+
1
⋅
3
1
+
2
⋅
3
1
+
1
⋅
2
⋅
3
1
(i) Show that
S
n
=
1
n
+
(
1
+
1
n
)
S
n
−
1
S_n=\frac1n+\left(1+\frac1n\right)S_{n-1}
S
n
=
n
1
+
(
1
+
n
1
)
S
n
−
1
.(ii) Hence or otherwise, deduce that
S
n
=
n
S_n=n
S
n
=
n
.
7
1
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P7: k primes (each > k) in AP with CD < k+2 is impossible
(i) Show that there cannot exists three peime numbers, each greater than
3
3
3
, which are in arithmetic progression with a common difference less than
5
5
5
.(ii) Let
k
>
3
k > 3
k
>
3
be an integer. Show that it is not possible for
k
k
k
prime numbers, each greater than
k
k
k
, to be in an arithmetic progression with a common difference less than or equal to
k
+
1
k+1
k
+
1
.
2
1
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P3: a^m+b^m=c^m and a^n+b^n=c^n cant hold simultaneously
Consider three positive real numbers
a
,
b
a,b
a
,
b
and
c
c
c
. Show that there cannot exist two distinct positive integers
m
m
m
and
n
n
n
such that both
a
m
+
b
m
=
c
m
a^m+b^m=c^m
a
m
+
b
m
=
c
m
and
a
n
+
b
n
=
c
n
a^n+b^n=c^n
a
n
+
b
n
=
c
n
hold.
6
1
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P6: Complex numbers: 6|n and (alpha)^n=1
Let
α
\alpha
α
be a complex number such that both
α
\alpha
α
and
α
+
1
\alpha+1
α
+
1
have modulus
1
1
1
. If for a positive integer
n
n
n
,
1
+
α
1+\alpha
1
+
α
is an
n
n
n
-th root of unity, then show that
α
\alpha
α
is also an
n
n
n
-th root of unity and
n
n
n
is a multiple of
6
6
6
.
4
1
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f(0)=1, f(x) \ge 0 \ge f'(x), f"(x)\le f'(x) for x\ge 0
Let
f
f
f
be a twice differentiable function on the open interval
(
−
1
,
1
)
(-1,1)
(
−
1
,
1
)
such that
f
(
0
)
=
1
f(0)=1
f
(
0
)
=
1
. Suppose
f
f
f
also satisfies
f
(
x
)
≥
0
,
f
′
(
x
)
≤
0
f(x) \ge 0, f'(x) \le 0
f
(
x
)
≥
0
,
f
′
(
x
)
≤
0
and
f
′
′
(
x
)
≤
f
(
x
)
f''(x) \le f(x)
f
′′
(
x
)
≤
f
(
x
)
, for all
x
≥
0
x\ge 0
x
≥
0
. Show that
f
′
(
0
)
≥
−
2
f'(0) \ge -\sqrt2
f
′
(
0
)
≥
−
2
.
10
1
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Trigo relation in a right angled. ISIBS2011P10
Show that the triangle whose angles satisfy the equality
sin
2
A
+
sin
2
B
+
sin
2
C
cos
2
A
+
cos
2
B
+
cos
2
C
=
2
\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2
cos
2
A
+
cos
2
B
+
cos
2
C
sin
2
A
+
sin
2
B
+
sin
2
C
=
2
is right angled.
3
1
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f(f(f(x)))=x ; ISI BS 2011, P3
Let
R
\mathbb{R}
R
denote the set of real numbers. Suppose a function
f
:
R
→
R
f: \mathbb{R} \to \mathbb{R}
f
:
R
→
R
satisfies
f
(
f
(
f
(
x
)
)
)
=
x
f(f(f(x)))=x
f
(
f
(
f
(
x
)))
=
x
for all
x
∈
R
x\in \mathbb{R}
x
∈
R
. Show that(i)
f
f
f
is one-one,(ii)
f
f
f
cannot be strictly decreasing, and(iii) if
f
f
f
is strictly increasing, then
f
(
x
)
=
x
f(x)=x
f
(
x
)
=
x
for all
x
∈
R
x \in \mathbb{R}
x
∈
R
.
8
1
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I_n = \int (0 to n*pi) sinxdx/(1+x)
Let
I
n
=
∫
0
n
π
sin
x
1
+
x
d
x
,
n
=
1
,
2
,
3
,
4
I_n =\int_{0}^{n\pi} \frac{\sin x}{1+x} \, dx , \ \ \ \ n=1,2,3,4
I
n
=
∫
0
nπ
1
+
x
sin
x
d
x
,
n
=
1
,
2
,
3
,
4
Arrange
I
1
,
I
2
,
I
3
,
I
4
I_1, I_2, I_3, I_4
I
1
,
I
2
,
I
3
,
I
4
in increasing order of magnitude. Justify your answer.
1
1
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An easy ineq; ISI BS 2011, P1
Let
x
1
,
x
2
,
⋯
,
x
n
x_1, x_2, \cdots , x_n
x
1
,
x
2
,
⋯
,
x
n
be positive reals with
x
1
+
x
2
+
⋯
+
x
n
=
1
x_1+x_2+\cdots+x_n=1
x
1
+
x
2
+
⋯
+
x
n
=
1
. Then show that
∑
i
=
1
n
x
i
2
−
x
i
≥
n
2
n
−
1
\sum_{i=1}^n \frac{x_i}{2-x_i} \ge \frac{n}{2n-1}
i
=
1
∑
n
2
−
x
i
x
i
≥
2
n
−
1
n