MathDB
Problems
Contests
National and Regional Contests
India Contests
ISI Entrance Examination
2012 ISI Entrance Examination
2012 ISI Entrance Examination
Part of
ISI Entrance Examination
Subcontests
(8)
8
1
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degree of f=2^k
Let
S
=
{
1
,
2
,
3
,
…
,
n
}
S = \{1,2,3,\ldots,n\}
S
=
{
1
,
2
,
3
,
…
,
n
}
. Consider a function
f
:
S
→
S
f\colon S\to S
f
:
S
→
S
. A subset
D
D
D
of
S
S
S
is said to be invariant if for all
x
∈
D
x\in D
x
∈
D
we have
f
(
x
)
∈
D
f(x)\in D
f
(
x
)
∈
D
. The empty set and
S
S
S
are also considered as invariant subsets. By
deg
(
f
)
\deg (f)
de
g
(
f
)
we define the number of invariant subsets
D
D
D
of
S
S
S
for the function
f
f
f
.i) Show that there exists a function
f
:
S
→
S
f\colon S\to S
f
:
S
→
S
such that
deg
(
f
)
=
2
\deg (f)=2
de
g
(
f
)
=
2
.ii) Show that for every
1
≤
k
≤
n
1\leq k\leq n
1
≤
k
≤
n
there exists a function
f
:
S
→
S
f\colon S\to S
f
:
S
→
S
such that
deg
(
f
)
=
2
k
\deg (f)=2^{k}
de
g
(
f
)
=
2
k
.
7
1
Hide problems
Locus of the centre of Gamma
Let
Γ
1
,
Γ
2
\Gamma_1,\Gamma_2
Γ
1
,
Γ
2
be two circles centred at the points
(
a
,
0
)
,
(
b
,
0
)
;
0
<
a
<
b
(a,0),(b,0);0<a<b
(
a
,
0
)
,
(
b
,
0
)
;
0
<
a
<
b
and having radii
a
,
b
a,b
a
,
b
respectively.Let
Γ
\Gamma
Γ
be the circle touching
Γ
1
\Gamma_1
Γ
1
externally and
Γ
2
\Gamma_2
Γ
2
internally. Find the locus of the centre of of
Γ
\Gamma
Γ
6
1
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square has maximum area... isi 2012 #6
i) Let
0
<
a
<
b
0<a<b
0
<
a
<
b
.Prove that amongst all triangles having base
a
a
a
and perimeter
a
+
b
a+b
a
+
b
the triangle having two sides(other than the base) equal to
b
2
\frac {b}{2}
2
b
has the maximum area.ii)Using
i
)
i)
i
)
or otherwise, prove that amongst all quadrilateral having give perimeter the square has the maximum area.
5
1
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number with 6 and 0 cant be square.... isi 2012 #5
Let
m
m
m
be a number containing only
0
0
0
and
6
6
6
as its digits.Show that
m
m
m
can't be a perfect square.
4
1
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x^8-x^7+x^2-x+15
Prove that the polynomial equation
x
8
−
x
7
+
x
2
−
x
+
15
=
0
x^{8}-x^{7}+x^{2}-x+15=0
x
8
−
x
7
+
x
2
−
x
+
15
=
0
has no real solution.
3
1
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Array of numbers... isi 2012 #3
Consider the numbers arranged in the following way:
1
3
6
10
15
21
⋯
2
5
9
14
20
⋯
⋯
4
8
13
19
⋯
⋯
⋯
7
12
18
⋯
⋯
⋯
⋯
11
17
⋯
⋯
⋯
⋯
⋯
16
⋯
⋯
⋯
⋯
⋯
⋯
⋮
⋮
⋮
⋮
⋮
⋮
⋱
\begin{array}{ccccccc} 1 & 3 & 6 & 10 & 15 & 21 & \cdots \\ 2 & 5 & 9 & 14 & 20 & \cdots & \cdots \\ 4 & 8 & 13 & 19 & \cdots & \cdots & \cdots \\ 7 & 12 & 18 & \cdots & \cdots & \cdots & \cdots \\ 11 & 17 & \cdots & \cdots & \cdots & \cdots & \cdots \\ 16 & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array}
1
2
4
7
11
16
⋮
3
5
8
12
17
⋯
⋮
6
9
13
18
⋯
⋯
⋮
10
14
19
⋯
⋯
⋯
⋮
15
20
⋯
⋯
⋯
⋯
⋮
21
⋯
⋯
⋯
⋯
⋯
⋮
⋯
⋯
⋯
⋯
⋯
⋯
⋱
Find the row number and the column number in which the the number
20096
20096
20096
occurs.
2
1
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g(x)=(\alpha+|x|)^{2}e^{(5-|x|)^{2}}
Consider the following function
g
(
x
)
=
(
α
+
∣
x
∣
)
2
e
(
5
−
∣
x
∣
)
2
g(x)=(\alpha+|x|)^{2}e^{(5-|x|)^{2}}
g
(
x
)
=
(
α
+
∣
x
∣
)
2
e
(
5
−
∣
x
∣
)
2
i) Find all the values of
α
\alpha
α
for which
g
(
x
)
g(x)
g
(
x
)
is continuous for all
x
∈
R
x\in\mathbb{R}
x
∈
R
ii)Find all the values of
α
\alpha
α
for which
g
(
x
)
g(x)
g
(
x
)
is differentiable for all
x
∈
R
x\in\mathbb{R}
x
∈
R
.
1
1
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trigonometric identity..... isi 2012 #1
i)If
X
,
Y
,
Z
X,Y,Z
X
,
Y
,
Z
be the angles of a triangle then show that
tan
X
2
tan
Y
2
+
tan
Y
2
tan
Z
2
+
tan
Z
2
tan
X
2
=
1
\tan {\frac{X}{2}}\tan {\frac{Y}{2}}+\tan {\frac{Y}{2}}\tan {\frac{Z}{2}}+\tan {\frac{Z}{2}}\tan {\frac{X}{2}}=1
tan
2
X
tan
2
Y
+
tan
2
Y
tan
2
Z
+
tan
2
Z
tan
2
X
=
1
ii) Prove using (i) or otherwise that
tan
X
2
tan
Y
2
tan
Z
2
≤
1
3
3
\tan {\frac{X}{2}}\tan {\frac{Y}{2}}\tan {\frac{Z}{2}}\leq\frac {1}{3\sqrt{3}}
tan
2
X
tan
2
Y
tan
2
Z
≤
3
3
1