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May 14, 2012

Solutions to I.S.I. 2012 Subjective (B.Stat, B.Math)

Q7. Consider two circles with radii a, and b and centers at (b, 0), (a, 0) respectively with b<a. Let the crescent shaped region M has a third circle which at any position is tangential to both the inner circle and the outer circle. Find the locus of center c of the third circle as it traverses through the region M (remaining tangential to both the circle.


Join AC and BC. AC passes through, (T_1) the point of tangency of the smaller circle with the circle with center at (a, 0) and BC when extended touches (T_2) which is the other point of the tangency.
Assume the radius of the moving (and growing circle) to be r at a particular instance. Then AC = a+r and BC = b-r.
Then AC+BC = a+b which is a constant for any position of C. Hence C is a point whose some of distances from two fixed points at any instant is a constant. This is the locus definition of an ellipse with foci at (a, 0) and (b, 0).

Q8. Let S = {1, 2, ... , n}. Let (f_1 , f_2 , ... )  be functions from S to S (one-one and onto). For any function f, call D, subset of S, to be invariant if for all x in D, f(x)  is also in D. Note that for any function the null set and the entire set are 'invariant' sets. Let \deg(f) be the number of invariant subsets for a function.
 a) Prove that there exists a function with \deg(f)=2 .
 b) For a particular value of k prove that there exist a function with \deg(f) = (2^k)



Consider the function defined piecewise as f(x) = x - 1 is (x \ne 1) and f(x) = n if x = 1

Of course null set and the entire sets are invariant subsets. We prove that there are no other invariant subsets.

Suppose D = {(a_1 , a_2 , ... , a_k )} be an invariant subset with at least one element.

Since we are working with natural numbers only, it is possible to arrange the elements in ascending order (there is a least element by well ordering principle).

Suppose after rearrangement D = {(b_1 , b_2 , ... , b_k )} where (b_1) is the least element of the set

If (b_1 \ne 1) then (f(b_1) = b_1 -1) is not inside D as (b_1) is the smallest element in D. Hence D is no more an invariant subset which is contrary to our initial assumption.

This (b_1) must equal to 1.

As D is invariant subset (f(b_1) = n ) must belong to D. Again f(n) = n-1 is also in D and so on. Thus all the elements from 1 to n are in D making D=S.

Hence we have proved that degree of this function is 2.


For a natural number 'k' to find a function with {\deg(f)} = (2^k) define the function piecewise as

f(x) = x for (1\le x \le k-1)
= n for x=k
= x-1 for the rest of elements in 'n'

To construct an invariant subset the 'k-1' elements which are identically mapped, and the entirety of the 'k to n' elements considered as a unit must be considered. Thus there are total k-1 + 1 elements with which subsets are to be constructed. There are (2^k) subsets possible.

6 comments on “Solutions to I.S.I. 2012 Subjective (B.Stat, B.Math)”

  1. Solution of 3: Let i denote the row number and j denote the column numberThen by inspection,elements in first row are given by formulaa_{1j}=(j(j+1))/2in general any element of array is given by formulaa_{ij}=((i+j-1)(i+j-2))/2+jIf a_{ij}=20096If we put j=196, we get i=5.Therefore 20096 lies in 5th row and 196 column of array.Remark: Solving problem by guessing is difficult. I myself could only figure formula in exam.Regards,Sumit Kumar Jha

  2. Could we solve it this way?
    Let us assume that f is a many-one function that assumes exactly K values out of the set of n Natural numbers for x=1,2,3....n. Thus we have (n-k) values that do not feature in the range of f. Now , we do an interesting operation . We consider all subsets of these (n-k) elements ( 2^(n-k)) and insert the range set of f into each. thus all these subsets are now invariant under f. Therefore, Deg(f) =2^n-k
    Replacing k by (n-1), we have a function whose degree is 2.
    Replacing k by (n-k) ,we have afunction whose degree is 2^k.

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