# CMI 2015 Objective & Subjective | Problems & Solutions

This post contains Chennai Mathematical Institute, CMI, 2015 Objective, and Subjective Problems and Solutions. Please contribute problems and solutions in the comments.

## Objective

1. For all finite word strings comprising A and B only, A string is arranged by dictionary order. eg. ABAA
2.  For any arbitrary string w, with another string y<w, there cannot always exist a string x, w<x<y
3. There is an infinite set of strings a1,a2… such that ai<a(i+1) for all i.
4. There are fewer than 50 strings less than AABBABBA
• Ten people are seated in a circle. One person contributes five hundred rupees. Every person contributes the average of the money contributed by his two neighbors.
1. What is the sum contributed by all the ten?
1. >5000
2. < 5000
3. .=5000
4. Cannot say.
2. 2. What is maximum contribution by an individual?
1. 500
2. =500
3. none
• There are 4 bins and 4 balls. Let $P(E_i)$ be the probability of first n balls falling into distinct bins.
Find

1. $P(E_4)$
2. $P(E_4|E_3)$
3. $P(E_4|E_2)$
4. $P(E_3|E_4)$
• Let $f(x) = \sin^{-1} (\sin (\pi x))$. Find
1. f(2.7).
2. f'(2.7)
3. integral from 0 to 2.5 of f(x)dx
4. value of x for which f'(x) does not exist
• In some country number plates are formed by 2 digits and 3 vowels. It is called confusing if it has both digit 0 and vowel o.
1. How many such number plates exist?
2. How many are not confusing
• A number is called magical if a and b are not coprime to n, a+b is also not coprime to n. For example, 2 is magical as all even numbers are not coprime to 2. Find whether the following numbers are magical
1. 129
2. 128
3. 127
4. 100
• a) In the expansion of $(1+ \sqrt 2)^10 = \sum_0^10 C_i (\sqrt 2)^i$, the term with maximum value is
b) If $(1+\sqrt 2)^n = p_n+q_n \sqrt 2$ , where $p_n$ and $q_n$ are integers, $\lim_{ n to \infty} \frac {p_n}{q_n} ^{10}$ is

## Subjective

1. In a circle, AB be the diameter.. X is an external point. Using straight edge construct a perpendicular to AB from X
1. If X is inside the circle then how can this be done
Discussion
2. a be a positive integer from set {2, 3, 4, ... 9999}. Show that there are exactly two positive integers in that set such that 10000 divides a*a-1.
1. Put $n^2 - 1$ in place of 9999. How many positive integers a exists such that $n^2$ divides a(a-1)
Discussion
3. P(x) is a polynomial. Show that $\displaystyle { \lim_{t to \infty} \frac{P(t)} {e^t} }$ exists. Also show that the limit does not depend on the polynomial.
4. We define function $\displaystyle { f(x) = \frac {e^{\frac{-1}{x}}}{x}}$ when x< 0; f(x) = 0 if x=0 and $\displaystyle { f(x) = \frac {e^{\frac{-1}{x}}}{x}}$ when x > 0 . Show that the function is continuous and differentiable. Find limit at x =0
5. p,q,r any real number such that $p^2 + q^2 + r^2 = 1$
1. Show that $3*(p^2 q + p^2 r) + 2(r^3 +q^3) \le 2$
2. Suppose $f(p,q,r) = 3(p^2 q + p^2 r ) + 2(r^3 +q^3)$ .  At what values (p,q, r) does f(p,q,r) maximizes and minimizes?
6. Let g(n) is GCD of (2n+9) and $6n^2+11n-2$ then then find greatest value of g(n)

### 62 comments on “CMI 2015 Objective & Subjective | Problems & Solutions”

1. soln of q.no.-1
Let 2 line drawn meeting A to X and B to X in P and Q at the pepheri of circle respectively.Now join A to Q and B to P.As AB is diameter of circle so angleAPB and angleBQA are 90*.So if we extend the linea AQ and BP,Let it meets at point O ie ORTHOCENTRE of triangle ABX.Now join X to O ie automatically perpendicular to AB.
HENCE SOLVED

2. Hint q.no.-3
as exponential function is more increasing then polynomial functin.
See derivative of both thats why any polynomial fucnyion upon exponential like e^x as x->infinity is 0 so it is independent of any particular polynomial.

3. q.no.-4
it done ab-initio method or basics principle of limit,continuty and derivability

4. one of the today cmi prob-
let g(n) is GCD of (2n+9) and 6n^2+11n-2 then then find largest +ve integer of g(n)
something like that

1. sahil says:

We need to find greatest value of g(n).

5. Harisha says:

Part A
Ten people are seated in a circle. One person contributes five hundred rupees. Every person contributes the average of the money contributed by his two neighbors.
1. What is the sum contributed by all the ten?
A.5000
C.=5000
D. Cannot say.
2. What is maximum contribution by an individual?
A.500
C.=500
D.none

1. Debjit Mandal says:

=5000....everyone is giving the same amount=500

6. Debjit Mandal says:

Anyone has solved question 5???.....part 1 is easy. I wanna know about part 2. Maximum value is '2'....is the minimum value '-2'???....actually, I have proved it, though I have a confusion!. Please reply!

1. soham says:

how did you solve part 1

7. Krishna Moorthi says:

A11. There are 4 bins and 4 balls. Let P(Ei) be the probability of first n balls falling into distinct bins.
Find a) P(E4) b)P(E4|E3) c)P(E4|E2) d)P(E3|E4)

1. Krishna Moorthi says:

b) 1/4
c)1/8
d)1

2. Sr says:

a is 1/256 b is 1/4 c is 1/16 d is 1

1. Krishna Moorthi says:

How? a) 4/4*3/4*2/4/1/4
c)2/4*1/4
Or so I think. The first ball can go into any bin. THe next can go into any bin but that particular one. The next in the remaining two and last one in the specific bin.

8. Krishna Moorthi says:

A9. Let f(x) = arcsin (sin (pi*x)).
Find a) f(2.7). b) f'(2.7) c)integral from 0 to 2.5 of f(x)dx
d)value of x for which f'(x) does not exist

1. Krishna Moorthi says:

b)pi
c)17*pi/8
d)Does not exist

1. You see, range of arcsin(x) is from pi/2 to -pi/2 . Your answer (.7pi > .5pi) so it is obviously wrong. Now since .7pi is range pi/2 to pi, (for example, if have 2pi/3, then it will become , pi-2pi/3. Similarly pi-7pi/10=3pi/10

2. I think answer to d) will be n+1/2 where n is any integer.

3. Krishna Moorthi says:

Ah yes guys. Sorry. My mistake.

2. Sr says:

again a is 3pi/10 c is pi/8 and d is 1/pi

1. Krishna Moorthi says:

Sr, can you tell me how? I am still getting the same answers

9. Krishna Moorthi says:

A8. In some country number plates are formed by 2 digits and 3 wovels. It is called confusing if it has both digit 0 and vowel o. a) How many such number plates exist? b) How many are not confusing

1. Krishna Moorthi says:

b) Not sure - 11000 something (Sorry!)

1. Sr says:

The qn was how many are confusing

2. Krishna Moorthi says:

The question was how many are NOT confusing. WIth the NOT being in bold letters. And the answer is 125*81+64*19. As there can be 125*81 combinations with no 0. And there can be 64*19 combinations with no o. Both cannot be there together. Any one of them can exist

10. Krishna Moorthi says:

A3. A number is called magical if a and b are not coprime to n, a+b is also not coprime to n. For example, 2 is magical as all even numbers are not coprime to 2. Find whether the following numbers are magical a)129 b)128 c)127 d)100

1. Krishna Moorthi says:

b)Yes
c)Yes
d)No

1. tiyacmi says:

why is 127 magical ? 127 = 1*127. 1 and 127 are both not coprime to 127, but 1+127=128 is coprime to 127. so 127 is not magical, right ?

2. tiyacmi says:

why is 127 magical ?

3. Krishna Moorthi says:

Tiyacmi, by that logic neither is 2 a prime as neither 1 nor 2 are coprime to 2 but 3 is coprime to 2. But they give in the question itself that 2 is magical. I do not think 1 is counted as it is neither a prime nor a composite. Take all other numbers coprime to 127. You will find that it is magical. (In fact, all prime numbers are magical)

11. Debjit Mandal says:

Subjective 2)
we know that there exist a number 'a' in that set, such that a^2=a(mod n^2). Now let a+b=n^2 [There must be an integer 'b' in the set]. then a^2=(n^2-b)^2=b^2(mod n^2) and a^2=a=n^2-b=-b(mod n^2). That means, b^2=-b(mod n^2), so b(b+1)=0(mod n^2) That means, our second integer is (b+1)= (n^2+1-a). Now, if 'a' and 'b+1' are distinct, then we can find two distinct integers. if not, then a=b+1 => a-b=1 and we know that a+b=n^2, Hence a=(n^2+1)/2 and b=(n^2-1)/2. Now, if n^2 is even, then 'a' and 'b' are not integers. So, we can find only one integer 'a' if n^2 is odd and two distinct integers 'a' and (n^2+1-a) when n^2 is even.
[Is there any problem in this solution? I am not sure!]

1. Sr says:

By Prime factorisation easy solution 625 and 9376

2. Krishna Moorthi says:

Not exactly sure. But I think it depends on the number of prime factors of n as well. Example: n=125

3. There exist (2) elements in set when the number can be expressed in form ab, where a,b neq 1 and gcd(a,b)=1. You have to first prove that , if n is prime or prime power , then there will be no solutions.

4. I am not sure . . . . but i think you just proved that if n is even
then there are even number of solutions. Isn't it?
How can u say only 2 solutions?

12. Krishna Moorthi says:

A2 for all finite word strings comprising A and B only, A string is arranged by dictionary order. eg. ABAA<ABB Also, A string with same components as a lesser string but with ehere xtra digits is greater than it eg. AB<ABAA. Write true or false
A) For any arbitrary string w, with another string y<w, there cannot always exist a string x, w<x<y
B) There is an infinite set of strings a1,a2... such that ai<a(i+1) for all i.
C) There are fewer than 50 strings less than AABBABBA

1. Krishna Moorthi says:

B) True???
C) False

1. I think second statement is false. Because there exist a small string that is A. No string is smaller than A. Hence Statement (2) is False.

2. Krishna Moorthi says:

Yes but they are not saying that every set of string is like that are they. There is ANY one set. Take the set of 500 billion B's as the first set. Every B can be replaced by as many number of A's as we want and we will get an infinite set

13. Krishna Moorthi says:

A7. a) In the expansion of (1+√2)^10 = sigma(0-10) Ci*(√2)^i, the term with maximum value is
b)If (1+√2)^n = pn+qn√2, where pn and qn are integers, limit (n --> infinity) (pn/qn)^10 is

1. Krishna Moorthi says:

b)1???

1. Krishna Moorthi says:

Ah yes. pn/qn = √2. I missed it in the heat of the moment I Guess

2. Krishna Moorthi says:

Debjani, the maximum term is 10C6*√2^6. It is i=6. Just calculate

3. thats what. how do u get p(n) and q(n) in the first place??

2. chavan3825 says:

can i get an explanatn, plz?

1. Krishna Moorthi says:

a) Self explanatory - Just expand the binomial expansion.
b) The ratio converges to √2. √2^10 = 32

2. Krishna Moorthi says:

Chavan, take the case of (1+√2)^2 = 1+2+2√2 = 3+2√2. Here, p(2)=3 and q(2)=2. In case of (1+√2)^3 = 1+3√2+6+2√2 = 7+5√2. Here, p(3) = 7 and q(3) = 5. This is how you get p(n) and q(n)

14. Krishna Moorthi says:

Sir, what is the expected cutoff to be selected in CMI 2015?

15. tiyacmi says:

For A.8 part b) how many are not confusing ?
first if it does not contain o : 100*64
second if it does not contain 0 : 81*125
now, from this we have to subtract the case where it contains neither o nor 0 since we have counted it twice ... so the answer will be :
100*64 + 81*125 - 81*64 = 11341

1. Krishna Moorthi says:

Yes. It is what I got too

16. […] CMI 2015 […]

17. Soham Ghosh says:

is the answer of the 6th problem for the G.C.D 70? please confirm

18. Soham Ghosh says:

yes I mistook indeed...its 35

19. Soham Ghosh says:

I could not avoid a non calculus solution to 5:
Part1:
We have after simplifying and substituting p^2 by 1-q^2-r^2,
f(q,r)=(q+r)(3-(q+r)^2)
Let q+r=t
f(t)=t(3-t^2)
f'(t)=3-3t^2=0 implies t=+1 or t=-1
It is easy to notice that the function attains maxima when t=+1
Putting t=1 in the ineq. it is indeed true
For 2nd part,
Plug t=-1 and get -2 as minima

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