this is a work in progress. post problems, solutions and correction in the comment section

Let be a parabola that cuts the coordinate axes at three distint points. Show that the circle passing through these three points also passes through (0,1). Discussion

Find all such Natural number n such that 7 divides Discussion

Find all functions f, such that |f(x) – f(y)| = 2|x-y|. Discussion

Say be n real numbers. Show that the equation has n real solutions.

Consider the set S = {1, 2, 3, …, j}. In a subset P of S, Max P be the maximum element of that subset. Show that the sum of all Max P (over all subsets of the set) is

There are three unit circles each of which is tangential to the other two. A triangle is drawn such that each side of the triangle is tangential to exactly two of the circle. Find the length of sides of this triangle.

Let \(m_1< m_2 < \ldots m_{k-1}< m_k\) be \(k\) distinct positive integers such that their reciprocals are in arithmetic progression.
1.Show that \(k< m_1 + 2\).
2. Give an example of such a sequence of length \(k\) for any positive integer \(k\).

Let and , P(i) = Q(i), i= 1,2,3 …, 6 . Show that there exists a negative integer r such that P(r) = Q(r) .

Dear reader, correct me if I am wrong-
1. Basic concepts.
2. Answer is all integers of the form 3,9,15,21…
3. f(x)= 2x
4. Needs concept of complex numbers.
5. Can be shown by induction.
6. Answer is 2+2sqrt.(3)
7. m1 is (k-1)! Rest is by induction.
8. r= -22

Okey, I got the 3rd one wrong, a silly mistake really, given that my approach was correct. Let’s hope the rest is right. (Yes, I was lucky, I solved ALL the sums)

Well, even by considering that its a special case of malffati circles, its as clear as day that they are equilateral! (Because the circles are equiradial)
For a detailed proof, one can use coordinate…
Btw, the diagram too showed the triangle to be equilateral lol 😉

A symmetric argument may be cancelled as, symmetry is only a observation. A better method- Join the centers of circles, expand it 3/2 times with same center, proof that the expanded sides are tangents. This proof is better and is of less than a half page.

The four points are A(0,b), B(0,1), C(x_1,0) and D(x_2,0) where X_1 and x_2 are the two real roots of the equations and suppose, x_2>x_1. Now, case-1 when both the roots are positive(Similar for both negative). Consider a point M on the X axis toward +infinity [just for the figure], Now, if ABCD is cyclic, the angle(ADM)=angle(ABC)=angle(270-BCD)
Use this relation and find the slopes respectively and equate.
case-2 when one root is positive and the other is negative. Use same concept of checking equal angles and slopes respectively.

You can go for it in a different way. f can be written as (x-e)*(x-f). Clearly cut points are (e,0) ,(f,0) ,(0,ef). Take (e,0) and (f,0) as extremities of diameter, satisfy with the third point. Get the equation of the circle. Probably ef will come out to be 1. The equation of circle will be satisfied by (0,1).

5) For an element ‘i’ in S, there will be (1+1)^(i-1) [actually, the expansion of this term in binomial coefficients] subsets where ‘i’ is the MAX. Hence sum of the MAX of those sets is i*2^(i-1).
So, required sum will be A= (sum over i=1 to j) i*2^(i-1).
Now, by A=2*A-A and simple algebra, we are done.

3) |{f(x)-f(y)}/(x-y)|=2 =>{f(x)-f(y)}/(x-y)=+ or – 2. for, y=0, {f(x)-f(0)}/x=+ or -2, that means f(x) is linear. So, f(x)=+ or -2x +f(0). [f(0) is the constant term of f(x)]
so, f(x)=2x+c and -2x+c for all real numbers ‘c’.

4) We know that the equation will have n roots, we have to prove that all are real. So, rewrite the equation as x/(a1-x) + x/(a2-x)… + x/(an-x) = 2015-n.
Now, assume a complex root z. We know that complex roots always occur in pairs with their conjugate, so let another root be z* (conjugate of z). Thus, z/(a1-z) + z/(a2-z) + z/(a3-z) …= z*/(a1-z*) + z*/(a2-z*) + z*/(a3-z*) …
Under the given conditions, it is impossible unless z=z*, thus the complex root degenerates to a real.

The symmetry of the figure suggests that all the angles are equal (60 deg). Now, by the same symmetry (read: congruence), the two angles that the line joining the centre of a circle and the vertex of the triangle nearest to it will be equal (30 deg). Rest is trigo…

PERHEPS THE BEST SOLUTION OF Q.No.-4 SUBJECTIVE PART
If we take 2015 in L.H.S. and consider a1/(a1-x)+………+an/(an-x)-2015 as f(x) and draw the graph of f(x),there are n-virtical line in graph as 0<a1<a2<……<an at each of these numbers denominator of one of the terms of function becomes ZERO.Now if you see a1 from R.H.S. the value of function is minus infinity and from L.H.S. of a2 is plus infinity and obviously in between a1 and a2 graph is continuous so the the graph cuts the X-axis and in this way graph repeats except behind a1 and forward an.But using this rest u will prove the rest.
Thank u

They asked me in interview in starting 1 prob prove square of integer is of the form either 4k or 8k+1,with litle bit hint i just manage to solve and i didn’t get answer.Another prob was find all soln of
sinx^5+cosx^3=1.I felt even hint felt like curse in interview for me.It was really bad in my entire life i never felt so inconfidant.
BUT u guys can definetly do well its about me not about u
THANK U
GOOD BYE

Q8. Q(x)=x^5+c4x^4+c3x^3+c2x^2+c1x^1+c0

and is ans. fr Q.6 sq root(7)+sq root(3)

No, its 2+2sqrt.(3)

Dear reader, correct me if I am wrong-

1. Basic concepts.

2. Answer is all integers of the form 3,9,15,21…

3. f(x)= 2x

4. Needs concept of complex numbers.

5. Can be shown by induction.

6. Answer is 2+2sqrt.(3)

7. m1 is (k-1)! Rest is by induction.

8. r= -22

3rd one is 2x+c

Oh yeah, my bad

Also -2x+c

Okey, I got the 3rd one wrong, a silly mistake really, given that my approach was correct. Let’s hope the rest is right. (Yes, I was lucky, I solved ALL the sums)

problem no.7 solution please…

6) The total figure is symmetric because all the three circles are unit circles. Hence the triangle will be also symmetric, hence equilateral.

are you sure it will be equilateral because it is a special case of malffati circles

Well, even by considering that its a special case of malffati circles, its as clear as day that they are equilateral! (Because the circles are equiradial)

For a detailed proof, one can use coordinate…

Btw, the diagram too showed the triangle to be equilateral lol 😉

A symmetric argument may be cancelled as, symmetry is only a observation. A better method- Join the centers of circles, expand it 3/2 times with same center, proof that the expanded sides are tangents. This proof is better and is of less than a half page.

5^3 = -1 (mod 7)

=> 5^(3k)=(-1)^k (mod 7)

so, k should be odd.

Hence, n=3k where, k is odd natural numbers

Sorry, I forgot to mention the question number. It is question 2

nah, answer is n is integer of form 6k+3.

1) Just try to compare the angles of the quadrilateral i.e. the slopes and the other angles.

I don’t exactly get you, I used a different method, can you please elaborate?

The four points are A(0,b), B(0,1), C(x_1,0) and D(x_2,0) where X_1 and x_2 are the two real roots of the equations and suppose, x_2>x_1. Now, case-1 when both the roots are positive(Similar for both negative). Consider a point M on the X axis toward +infinity [just for the figure], Now, if ABCD is cyclic, the angle(ADM)=angle(ABC)=angle(270-BCD)

Use this relation and find the slopes respectively and equate.

case-2 when one root is positive and the other is negative. Use same concept of checking equal angles and slopes respectively.

8) Not sure, but I think, there should be some other conditions upon the coefficients of the two polynomials, like b_0 and c_0 are not equal etc.

I would suggest that you put p(I)-q(I) = m(I).

{1,2…6} are 6 roots of m(I)=0… sum of roots should be -1

Oh yeah!!!….I’m so sorry…..That’s the actual argument. We need not think about the other coefficients….Nice one

You can go for it in a different way. f can be written as (x-e)*(x-f). Clearly cut points are (e,0) ,(f,0) ,(0,ef). Take (e,0) and (f,0) as extremities of diameter, satisfy with the third point. Get the equation of the circle. Probably ef will come out to be 1. The equation of circle will be satisfied by (0,1).

5) For an element ‘i’ in S, there will be (1+1)^(i-1) [actually, the expansion of this term in binomial coefficients] subsets where ‘i’ is the MAX. Hence sum of the MAX of those sets is i*2^(i-1).

So, required sum will be A= (sum over i=1 to j) i*2^(i-1).

Now, by A=2*A-A and simple algebra, we are done.

I see, nice method, I used induction (it helps that j+1 will be greater than all 1 to j!)

Induction is really cool in this kind of counting problems.

3) |{f(x)-f(y)}/(x-y)|=2 =>{f(x)-f(y)}/(x-y)=+ or – 2. for, y=0, {f(x)-f(0)}/x=+ or -2, that means f(x) is linear. So, f(x)=+ or -2x +f(0). [f(0) is the constant term of f(x)]

so, f(x)=2x+c and -2x+c for all real numbers ‘c’.

4) We know that the equation will have n roots, we have to prove that all are real. So, rewrite the equation as x/(a1-x) + x/(a2-x)… + x/(an-x) = 2015-n.

Now, assume a complex root z. We know that complex roots always occur in pairs with their conjugate, so let another root be z* (conjugate of z). Thus, z/(a1-z) + z/(a2-z) + z/(a3-z) …= z*/(a1-z*) + z*/(a2-z*) + z*/(a3-z*) …

Under the given conditions, it is impossible unless z=z*, thus the complex root degenerates to a real.

can I know that how many questions would u have to do correct in order to get admission

How is the triangle question done?

Please could u upload the solution of 6th question? Triangle one

The symmetry of the figure suggests that all the angles are equal (60 deg). Now, by the same symmetry (read: congruence), the two angles that the line joining the centre of a circle and the vertex of the triangle nearest to it will be equal (30 deg). Rest is trigo…

Thank you so much

Glad to help!

Btw, you are a knockout tbh.

You should go through the previous comments also! 🙂

Indeed.

C’mon guys, be a sport, 😛

PERHEPS THE BEST SOLUTION OF Q.No.-4 SUBJECTIVE PART

If we take 2015 in L.H.S. and consider a1/(a1-x)+………+an/(an-x)-2015 as f(x) and draw the graph of f(x),there are n-virtical line in graph as 0<a1<a2<……<an at each of these numbers denominator of one of the terms of function becomes ZERO.Now if you see a1 from R.H.S. the value of function is minus infinity and from L.H.S. of a2 is plus infinity and obviously in between a1 and a2 graph is continuous so the the graph cuts the X-axis and in this way graph repeats except behind a1 and forward an.But using this rest u will prove the rest.

Thank u

What do u mean exactly mister?

Was just complimenting you… though I’m pretty sure it has been done a jillion times be4. U mind?

sphadan what’ your mean

what’s your mean Anannnnya

what’s your mean

I have typeset the answer to Question 5: http://www.texpaste.com/n/lw8wf29m

Well, I guess everybody @ this page already have it, thanks anyway!

I have typeset the answer to Q.5 at http://www.texpaste.com/n/84gc43bw

Whats urs bro

Thnx fr the compliments but i have got many 🙂

Well, I was smart enough to figure that out. Anyway, I hope we both end up @ ISI B’lore, it would be ‘interesting’ to have you @ the campus. 😉

according to bstat results 2015 are you selected?????

Yes, but I had changed my application to B.Math btw, so I’m gonna meet the dean of ISI Kolkata to sort that out. (Am in Kolkata now, btw)

Hmm i wish the same

Btw where are u from

I am from Mumbai. What about you?

right above

how did you solve the last question of this year’s isi subjective paper

Btw, this is a public forum, you might reach me @ [email protected]

When will the results of ISI B.math be announced?

Really eager to know…

see on isi official site left hand down named RESULT FRAMEWORK DOCUMENT

But that doesn’t declare the result.

MAN AND WOMAN I DON’T WANT TO LISTEN ANTTHING

THE RESULT IS DECLARED RIGHT NOW AND I AM SELECTED

BSTK-CC-0060

Congratulations

thank u sir for that u accomodated through blog

thank a lot

‘grats.

CONGRATULATION! Spandan Bhattacharjee

Why do you congratulate me?

Oh I see, thanks!

because u are selected in isi written

Only B.Stat results are announced. B.Math is not yet announced.

Rahul, Spandan and Ananya I am deleting comments regarding facebook etc. I hope you understand why

Oh sure, I do understand…

Anyway, I am pretty glad to be selected. Cheenta helped a lot, thank you!

All the best Spandan

Thanks again!

Ys we do respect ur move

I will keep this comment for 24 hour (so that the intended viewers may see it). Then I will delete this. I hope you understand why.

I hope that wasn’t intended for me!!!

and sir it is good if i share about interview coz my interview is on 12/06/2015 sharp 10:00am

1st day 1st show

Yes please do that

sorry for repatation

and sir it is good if i share about interview coz my interview is on 12/06/2015 sharp 10:00am

1st day 1st show

They asked me in interview in starting 1 prob prove square of integer is of the form either 4k or 8k+1,with litle bit hint i just manage to solve and i didn’t get answer.Another prob was find all soln of

sinx^5+cosx^3=1.I felt even hint felt like curse in interview for me.It was really bad in my entire life i never felt so inconfidant.

BUT u guys can definetly do well its about me not about u

THANK U

GOOD BYE

‘Grats m8! U R SELECTED!

Overall paper was easy and answer to all questions was possible in the two hours.