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May 10, 2015

ISI B.Stat, B.Math Paper 2015 Subjective| Problems & Solutions

Here, you will find all the questions of ISI Entrance Paper 2014 from Indian Statistical Institute's B.Stat Entrance. You will also get the solutions soon of all the previous year problems.

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

Problem 1:

Let $ y = x^2 + ax + b $ be a parabola that cuts the coordinate axes at three distinct points. Show that the circle passing through these three points also passes through $(0,1)$. Discussion

Problem 2:

Find all such Natural number $n$ such that $7$ divides $5^n + 1$.


Problem 3:

Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R},$ satisfying
for all $x, y \in \mathbb{R} .$ Justify your answer.


Problem 4:

Say $0 < a_1 < a_2 < ... < a_n $ be $n$ real numbers. Show that the equation $\frac{a_1}{a_1 - x } + \frac{a_2}{a_2 - x} + \cdot + \frac{a_n}{a_n - x} = 2015 $ has $n$ real solutions.

Problem 5:

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 $(j-1)2^j + 1 $

Problem 6:

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.

Problem 7:

Let $m_1< m_2 < \cdots <m_{k-1}< m_k$ be $k$ distinct positive integers such that their reciprocals are in arithmetic progression.

$1$. Show that $k< m_1 + 2$.

$2$. For any integer $k>0$, give an example of a sequence of $k$ positive integers whose reciprocals are in arithmetic progression.

Problem 8:

Let $P(x) = x^7 + x^6 + b_5 x^5 + b_4 x^4 + \cdots + b_0$ and $Q(x) = x^5 + c_4 x^4 + c_3 x^3 + \cdots + c_0$ be polynomial with integer coefficients , Assume that $P(i) = Q(i)$,Β for integers $i= 1,2,3 \cdot 6$Β  . Show that there exists a negative integer $r$ such that $P(r) = Q(r)$ .

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84 comments on “ISI B.Stat, B.Math Paper 2015 Subjective| Problems & Solutions”

  1. 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

    1. 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)

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

      1. 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 πŸ˜‰

    1. 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.

  3. 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

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

      1. 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.

  5. 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.

    1. 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

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

    2. 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).

  6. 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.

  7. 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'.

  8. 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.

    1. 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...

    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

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

    1. 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. πŸ˜‰

      1. 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)

    1. Oh sure, I do understand...
      Anyway, I am pretty glad to be selected. Cheenta helped a lot, thank you!

  10. 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

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