ISI Entrance Paper 2008 – from Indian Statistical Institute’s B.Stat Entrance

Also see: ISI and CMI Entrance Course at Cheenta

- Of all triangles with given perimeter, find the triangle with the maximum area. Justify your answer
- A 40 feet high screen is put on a vertical wall 10 feet above your eye-level. How far should you stand to maximize the angle subtended by the screen (from top to bottom) at your eye?
- Study the derivatives of the function

and sketch its graph on the real line. - Suppose P and Q are the centres of two disjoint circles and respectively, such that P lies outside and Q lies outside . Two tangents are drawn from the point P to the circle , which intersect the circle at point A and B. Similarly, two tangents are drawn from the point Q to the circle , which intersect the circle at points M and N. Show that AB=MN
- Suppose ABC is a triangle with inradius r. The incircle touches the sides BC, CA, and AB at D,E and F respectively. If BD=x, CE=y and AF=z, then show that
- Evaluate:
- Consider the equation . Show that

(a) the equation has only one real root;

(b) this root lies between 1 and 2;

(c) this root must be irrational. - In how many ways can you divide the set of eight numbers into 4 pairs such that no pair of numbers has equal to 2?
- Suppose S is the set of all positive integers. For , define

For example 8*12=6.

Show that exactly two of the following three properties are satisfied:

(i) If , then .

(ii) for all .

(iii) There exists an element such that for all - .Two subsets A and B of the (x,y)-plane are said to be equivalent if there exists a function f: Ato B which is both one-to-one and onto.

(i) Show that any two line segments in the plane are equivalent.

(ii) Show that any two circles in the plane are equivalent.

## 3 replies on “ISI Entrance Paper 2008 – B.Stat Subjective”

How to solve problem number 5 and 6?

Solution to 6: Let y be the given limit. Then e^y = (2nCn)^(1/2n). Now 2nCn = (2n)!/(n!)^2 . We divide both Nr and Dr by (2n)^2n. Hence e^y = [ {(2n)!/(2n)^2n}^(1/2n)]/[{n!/n^n}^2]^(1/2n)*2. Now limit n tends to infinity (n!/n^n)^(1/2n) is e^-1. Hence on simplification limit n tends to infinity e^y =2 and finally result is log 2. Do it yourself to understand the solution.

you can expand the factorial and use definite integral as a sum of infinite terms. the things multiplied in the log can be split and then take as a summation which can be used as an integral by putting r/n as x and 1/n as dx

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