*Please contribute problems and solutions in the comments.*

## Objective

- For all finite word strings comprising A and B only, A string is arranged by dictionary order. eg. ABAA
- For any arbitrary string w, with another string y<w, there cannot always exist a string x, w<x<y
- There is an infinite set of strings a1,a2… such that ai<a(i+1) for all i.
- 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.
- What is the sum contributed by all the ten?
- >5000
- < 5000
- .=5000
- Cannot say.

- 2. What is maximum contribution by an individual?
- 500
- =500
- none

- What is the sum contributed by all the ten?
- There are 4 bins and 4 balls. Let \(P(E_i)\) be the probability of first n balls falling into distinct bins.

Find- \(P(E_4) \)
- \(P(E_4|E_3) \)
- \(P(E_4|E_2) \)
- \(P(E_3|E_4) \)

- Let \(f(x) = \sin^{-1} (\sin (\pi x)) \). Find
- f(2.7).
- f'(2.7)
- integral from 0 to 2.5 of f(x)dx
- 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.
- How many such number plates exist?
- 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
- 129
- 128
- 127
- 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

- In a circle, AB be the diameter.. X is an external point. Using straight edge construct a perpendicular to AB from X
- If X is inside the circle then how can this be done

Discussion

- If X is inside the circle then how can this be done
**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.- Put \(n^2 – 1 \) in place of 9999. How many positive integers
**a**exists such that \(n^2 \) divides a(a-1)

Discussion

- Put \(n^2 – 1 \) in place of 9999. How many positive integers
- 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.
- 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
- p,q,r any real number such that \(p^2 + q^2 + r^2 = 1 \)
- Show that \(3*(p^2 q + p^2 r) + 2(r^3 +q^3) \le 2 \)
- 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?

- Let g(n) is GCD of (2n+9) and \(6n^2+11n-2 \) then then find greatest value of g(n)

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

Hint of q.no.-2

use Chinese remainder theorem.

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.

q.no.-4

it done ab-initio method or basics principle of limit,continuty and derivability

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

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

good

SAHIL

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

A. >5000

B.<5000

C.=5000

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

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!

how did you solve part 1

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)

Answers – a) 3/32

b) 1/4

c)1/8

d)1

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

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.

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

Answers – a)0.7

b)pi

c)17*pi/8

d)Does not exist

ans-0.3*pi

a)0.7. How? please explain.I am getting the answer as 0.3*phi=0.94

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

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

Please anybody confirm.

Ah yes guys. Sorry. My mistake.

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

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

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

Answers – a)12500

b) Not sure – 11000 something (Sorry!)

The qn was how many are confusing

answer is 250

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

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

Answers – a)No

b)Yes

c)Yes

d)No

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 ?

why is 127 magical ?

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)

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!]

By Prime factorisation easy solution 625 and 9376

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

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.

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?

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

Answers – A) True

B) True???

C) False

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.

yes.i think so.

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

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

Answers – a) i=6

b)1???

b)its 32 man

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

a)i=5. how u got 6? please post your solution.

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

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

can i get an explanatn, plz?

a) Self explanatory – Just expand the binomial expansion.

b) The ratio converges to √2. √2^10 = 32

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)

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

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

Yes. It is what I got too

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

I think it is 35

yes I mistook indeed…its 35

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