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Chennai Mathematical Institute B.Sc. Entrance Exam
2018 CMI B.Sc. Entrance Exam
2018 CMI B.Sc. Entrance Exam
Part of
Chennai Mathematical Institute B.Sc. Entrance Exam
Subcontests
(6)
3
1
Hide problems
Value of $f(0)$
Let
f
f
f
be a function on non-negative integers defined as follows
f
(
2
n
)
=
f
(
f
(
n
)
)
and
f
(
2
n
+
1
)
=
f
(
2
n
)
+
1
f(2n)=f(f(n))~~~\text{and}~~~f(2n+1)=f(2n)+1
f
(
2
n
)
=
f
(
f
(
n
))
and
f
(
2
n
+
1
)
=
f
(
2
n
)
+
1
(a) If
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
, find
f
(
n
)
f(n)
f
(
n
)
for every
n
n
n
. (b) Show that
f
(
0
)
f(0)
f
(
0
)
cannot equal
1
1
1
. (c) For what non-negative integers
k
k
k
(if any) can
f
(
0
)
f(0)
f
(
0
)
equal
2
k
2^k
2
k
?
6
1
Hide problems
CMI 2018 #6
Imagine the unit square in the plane to be a carrom board. Assume the striker is just a point, moving with no friction (so it goes forever), and that when it hits an edge, the angle of reflection is equal to the angle of incidence, as in real life. If the striker ever hits a corner it falls into the pocket and disappears. The trajectory of the striker is completely determined by its starting point
(
x
,
y
)
(x,y)
(
x
,
y
)
and its initial velocity
(
p
,
q
)
→
\overrightarrow{(p,q)}
(
p
,
q
)
. If the striker eventually returns to its initial state (i.e. initial position and initial velocity), we define its bounce number to be the number of edges it hits before returning to its initial state for the
1
st
1^{\text{st}}
1
st
time. For example, the trajectory with initial state
[
(
.
5
,
.
5
)
;
(
1
,
0
)
→
]
[(.5,.5);\overrightarrow{(1,0)}]
[(
.5
,
.5
)
;
(
1
,
0
)
]
has bounce number
2
2
2
and it returns to its initial state for the
1
st
1^{\text{st}}
1
st
time in
2
2
2
time units. And the trajectory with initial state
[
(
.
25
,
.
75
)
;
(
1
,
1
)
→
]
[(.25,.75);\overrightarrow{(1,1)}]
[(
.25
,
.75
)
;
(
1
,
1
)
]
has bounce number
4
4
4
.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
a
)
<
/
s
p
a
n
>
<span class='latex-bold'>(a)</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
a
)
<
/
s
p
an
>
Suppose the striker has initial state
[
(
.
5
,
.
5
)
;
(
p
,
q
)
→
]
[(.5,.5);\overrightarrow{(p,q)}]
[(
.5
,
.5
)
;
(
p
,
q
)
]
. If
p
>
q
⩾
0
p>q\geqslant 0
p
>
q
⩾
0
then what is its velocity after it hits an edge for the
1
st
1^{\text{st}}
1
st
time ? What if
q
>
p
⩾
0
q>p\geqslant 0
q
>
p
⩾
0
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
b
)
<
/
s
p
a
n
>
<span class='latex-bold'>(b)</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
b
)
<
/
s
p
an
>
Draw a trajectory with bounce number
5
5
5
or justify why it is impossible.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
c
)
<
/
s
p
a
n
>
<span class='latex-bold'>(c)</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
c
)
<
/
s
p
an
>
Consider the trajectory with initial state
[
(
x
,
y
)
;
(
p
,
0
)
→
]
[(x,y);\overrightarrow{(p,0)}]
[(
x
,
y
)
;
(
p
,
0
)
]
where
p
p
p
is a positive integer. In how much time will the striker
1
st
1^{\text{st}}
1
st
return to its initial state ?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
d
)
<
/
s
p
a
n
>
<span class='latex-bold'>(d)</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
d
)
<
/
s
p
an
>
What is the bounce number for the initial state
[
(
x
,
y
)
;
(
p
,
q
)
→
]
[(x,y);\overrightarrow{(p,q)}]
[(
x
,
y
)
;
(
p
,
q
)
]
where
p
,
q
p,q
p
,
q
are relatively prime positive integers, assuming the striker never hits a corner ?
5
1
Hide problems
CMI 2018 #5
An
alien
\textrm{alien}
alien
script has
n
n
n
letters
b
1
,
b
2
,
…
,
b
n
b_1,b_2,\dots,b_n
b
1
,
b
2
,
…
,
b
n
. For some
k
<
n
/
2
k<n/2
k
<
n
/2
assume that all words formed by any of the
k
k
k
letters (written left to right) are meaningful. These words are called
k
k
k
-words. Such a
k
k
k
-word is considered
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
s
a
c
r
e
d
<
/
s
p
a
n
>
<span class='latex-bold'>sacred</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
s
a
cre
d
<
/
s
p
an
>
if:i. no letter appears twice and, ii. if a letter
b
i
b_i
b
i
appears in the word then the letters
b
i
−
1
b_{i-1}
b
i
−
1
and
b
i
+
1
b_{i+1}
b
i
+
1
do not appear. (Here
b
n
+
1
=
b
1
b_{n+1} = b_1
b
n
+
1
=
b
1
and
b
0
=
b
n
b_0 = b_n
b
0
=
b
n
).For example, if
n
=
7
n = 7
n
=
7
and
k
=
3
k = 3
k
=
3
then
b
1
b
3
b
6
,
b
3
b
1
b
6
,
b
2
b
4
b
6
b_1b_3b_6, b_3b_1b_6, b_2b_4b_6
b
1
b
3
b
6
,
b
3
b
1
b
6
,
b
2
b
4
b
6
are sacred
3
3
3
-words. On the other hand
b
1
b
7
b
4
,
b
2
b
2
b
6
b_1b_7b_4, b_2b_2b_6
b
1
b
7
b
4
,
b
2
b
2
b
6
are not sacred. What is the total number of sacred
k
k
k
-words? Use your formula to find the answer for
n
=
10
n = 10
n
=
10
and
k
=
4
k = 4
k
=
4
.
2
1
Hide problems
CMI 2018 #2
Answer the following questions :
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
a
)
<
/
s
p
a
n
>
<span class='latex-bold'>(a)</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
a
)
<
/
s
p
an
>
Find all real solutions of the equation
(
x
2
−
2
x
)
x
2
+
x
−
6
=
1
\Big(x^2-2x\Big)^{x^2+x-6}=1
(
x
2
−
2
x
)
x
2
+
x
−
6
=
1
Explain why your solutions are the only solutions.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
b
)
<
/
s
p
a
n
>
<span class='latex-bold'>(b)</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
b
)
<
/
s
p
an
>
The following expression is a rational number. Find its value.
6
3
+
10
3
−
6
3
−
10
3
\sqrt[3]{6\sqrt{3}+10} -\sqrt[3]{6\sqrt{3}-10}
3
6
3
+
10
−
3
6
3
−
10
1
1
Hide problems
CMI 2018 #1
Answer the following questions :
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
a
)
<
/
s
p
a
n
>
<span class='latex-bold'>(a)</span>~
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
a
)
<
/
s
p
an
>
A natural number
k
k
k
is called stable if there exist
k
k
k
distinct natural numbers
a
1
,
a
2
,
⋯
,
a
k
a_1, a_2,\cdots, a_k
a
1
,
a
2
,
⋯
,
a
k
, each
a
i
>
1
a_i>1
a
i
>
1
, such that
1
a
1
+
1
a
2
+
⋯
+
1
a
k
=
1
\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}=1
a
1
1
+
a
2
1
+
⋯
+
a
k
1
=
1
Show that if
k
k
k
is stable, then
(
k
+
1
)
(k+1)
(
k
+
1
)
is also stable. Using this or otherwise, find all stable numbers.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
b
)
<
/
s
p
a
n
>
<span class='latex-bold'>(b)</span>
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
b
)
<
/
s
p
an
>
Let
f
f
f
be a differentiable function defined on a subset
A
A
A
of the real numbers. Define
f
∗
(
y
)
:
=
max
x
∈
A
{
y
x
−
f
(
x
)
}
f^*(y):=\max_{x\in A} \left\{yx-f(x)\right\}
f
∗
(
y
)
:=
x
∈
A
max
{
y
x
−
f
(
x
)
}
whenever the above maximum is finite.For the function
f
(
x
)
=
ln
x
f(x)=\ln x
f
(
x
)
=
ln
x
, determine the set of points for which
f
∗
f^*
f
∗
is defined and find an expression for
f
∗
(
y
)
f^*(y)
f
∗
(
y
)
involving only
y
y
y
and constants.
4
1
Hide problems
Interesting geometry
Let
A
B
C
ABC
A
BC
be an equilateral triangle of side length
2
2
2
. Point
A
′
A'
A
′
is chosen on side
B
C
BC
BC
such that the length of
A
′
B
A'B
A
′
B
is
k
<
1
k<1
k
<
1
. Likewise points
B
′
B'
B
′
and
C
′
C'
C
′
are chosen on sides
C
A
CA
C
A
and
A
B
AB
A
B
. with
C
B
′
=
A
C
′
=
k
CB'=AC'=k
C
B
′
=
A
C
′
=
k
. Line segments are drawn from points
A
′
,
B
′
,
C
′
A',B',C'
A
′
,
B
′
,
C
′
to their corresponding opposite vertices. The intersections of these line segments form a triangle, labeled
P
Q
R
PQR
PQR
. Prove that
Δ
P
Q
R
\Delta PQR
Δ
PQR
is an equilateral triangle with side length
4
(
1
−
k
)
k
2
−
2
k
+
4
{4(1-k) \over \sqrt{k^2-2k+4}}
k
2
−
2
k
+
4
4
(
1
−
k
)
.