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Join AC and BC. AC passes through, the point of tangency of the smaller circle with the circle with center at (a, 0) and BC when extended touches 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 . Let 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 be the number of invariant subsets for a function.**

** a) Prove that there exists a function with .**

** b) For a particular value of k prove that there exist a function with = **

**Discussion:**

**(a) **

Consider the function defined piecewise as f(x) = x - 1 is 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 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 where is the least element of the set

If then is not inside D as is the smallest element in D. Hence D is no more an invariant subset which is contrary to our initial assumption.

This must equal to 1.

As D is invariant subset 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.

**(b) **

For a natural number 'k' to find a function with = define the function piecewise as

f(x) = x for

= 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 subsets possible.

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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

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.