Welcome to Ctanujit Institute of Statistics & Mathematics...An Initiative by 'ISI'ian

Urgent Notice:

Due to unavoidable reason, Ctanujit Classes will be closed from 1st November to 30th November 2018.
Postal Package will be available from 1st December 2018.



This is a kind of initiative from me to state a trick & solve problems using those tricks from tifr,isi,cmi,nbhm papers. This will be like a series. This is the beginning one.

Trick 1:-
If the sum of two positive quantities is a constant(given),then their product is maximum when two quantities are equal.
Ex. Let a+b=12,then max(ab)=36 , i.e. When a equals b.

Problem1. (CMI-2010 M.Sc entrance) True/False- For x<0 , e^x (1 - e^x) ≤ 1/4 .
Sol:- True. e^x,1-e^x both are positive. Now e^x +(1 - e^x)=1
So their product is max when e^x=1-e^x=1/2,so max[e^x (1 - e^x)] is 1/4.

Problem2. (ISI MMA PAPER 2010) If a,b are positive real variables whose sum is a constant k, then the minimum value of root of{(1+1/a)(1+1/b)} is
A. k-1/k         B. k+2/k        C. 1+2/k         D. none
Sol: (C). The given root is minimum when ab is maximum. Now ab is maximum when a=b=k/2. So value of
 root of{(1+1/a)(1+1/b)}=(1+2/k).