Indian Statistical Institute , the leading organisation in India in mathematics and research works conducts entrance tests every year. This test are based solely on mathematics on a higher level than grade school.
Here I present in this note the best problems so far available from the Institute's entrance test of Undergraduate admission. It must be noted that ISI never publishes any solution to it's given problems, the solutions presented are thus my own and might have faults which you are welcome to correct.
Comment problems from ISI that you have I would add to the note if possible with the solution. So here we go !
Problem : A class has students. Let , denote the number of friends the i-th student has in the class. For each , let denote the number of students having at least friends. Prove that :
Number of friends of the -th student
Number of students having at least friends
It is harmless to renumber each student so that their corresponding number of friends form an increasing sequence . Since now denotes the number of students having at least friends it follows that there is some for which , and since is an increasing sequence for all , will have at least friends. So is exactly the number of elements of the set where each of these students numbering from on wards have friends and number of these students is value of .
Now we need to put these in a more formal way to help establish a general formula to get our . If we define a function by :
Now note that beginning with we find an such that and for all , and we have .
So we need to count the number of entries for each from some where it has to .Our function perfectly fits that requirement and we have ,
Now to prove the given we apply summation again on both sides to get,
Thus proved !
Problem : For Prove that
Define . So has a critical point at . For , and thus is increasing. So,
Similarly For , and thus is decreasing we have,
Thus combining we get the result !
Problem : Let be a function that is differentiable times for some positive integer . The -th derivative is denoted by . Suppose, . Prove that for some .
By Rolle's Theorem since we have for some . Now if we define then we have by rolle's theorem that for some since . Proceeding in this manner we will have for some since . Since we have have proved the result that for some where .
Problem: Let be a continuous function such that has no real solution. Prove that has no real solution.
It suffices to prove that is solvable whenever is solvable. Let be the unique solution to so that . So it follows that is true and is the unique fixed point of . Since it is given that has no solution so does . Thus proved !
Problem: Let be a real-valued differentiable function on the real line such that exists and is finite. Prove that
Hence for the limit to exist and be finite it follows that and thus proved.
Problem: Let and be polynomials with integer coefficients. Assume that for integers . Show that there exists a negative integer such that
Solution: Define a dummy polynomial so that has roots . Since it's a -th degree polynomial it must have 7 roots. Moreover since is monic we may write for some integer which is the -th root.
Since it follows that the sum of the roots is . Therefore which makes . Thus which implies and we are proved.
Problem: Given the polynomial with real coefficients such that . Show that not all roots of can be real.
Solution: Let the roots of the polynomial be for . Since and it follows that which implies which is a contradiction with and thus our assumption that all of 's are real was false. Thus not all real roots are possible.
Problem:Let and be two non-decreasing twice differentiable functions defined on an interval such that for each , and . Suppose that is linear in on . Show that we must have for all
Solution: Let and by differentiating twice we have . Since are non decreasing we have and for all . By putting and we have . But since and and thus . So implies and if sum of two or more squares are zero they must be individually zero. Thus
Problem: Let be a positive integer such that is the square of a positive integer. Then determine all possible values of .
Solution: Let for some . Since takes positive integral values the discriminant of the quadratic must be a perfect square. Hence for some positive integer . This factorizes to .
Since is a prime so RHS can be factored in exactly two ways as a product of two factors namely and . Since so we must have which implies and from which we get and this gives as the only positive integer solution.
Problem(Subjective 173): Let be polynomials in , each having all integer coefficients, such that . Assume that is not the zero polynomial. Show that and
Solution: Since is not the zero polynomial with integer coefficients it's obvious that . This gives, . But since a perfect square is always non-negative we have . This together implies that and thus each of 's are zero that is . Substituting this we have which makes since . Thus proved.
Problem: Let where is an odd integer. Let be a function defined on taking values in such that : & for all . Show that
Solution: Since and also for any we have and thus it follows that is just a permutation of . Say for some and any we have . Now it is safe to assume or rather substitute to have . In a similar manner for some we have for any , and assuming we have . This process repeated times will give for some where it must be noted that . So this proves that .
Problem: Consider the equation . Prove that it has only one real root which lies between and and further the root must be irrational.
Solution: Let and applying Descarte's rule of signs to we have only sign change so it may have at most one positive real root. Applying same to produces no change so there are no negative real roots. Since an equation of odd degree always has a real root(SInce complex roots occur in pairs) so it has 1 real root.
Next it is easy to show by Intermediate Value Theorem that the root lies between and since and . Now had the root been rational it has to be an integer by the Rational Root Theorem. Clearly isn't a root and for all integers . So there must be an irrational root.
Problem: Let . Consider two circles with radii and and centers and respectively. Let be the center of any circle in between the two circles and tangent to both. Determine the locus of the center of any such circle.
We have assumed to be the center of , to be the center of and to be the center of . Joining AC and BC we find that if be the radius of the variable circle at any instant then and respectively. Thus is constant for any such circle. The sum of distances of the center of the smaller circle from two points (the center of the other two circles respectively) is constant which is a property of the ellipse. Thus the locus of is ellipse with it's foci at and respectively.
To be continued....