1plz solve these .................these may be easy for u all .............bt these ques puzzled me..................
8 Answers
1
14 is already discussed
http://targetiit.com/iit-jee-forum/posts/functions-10481.html
13) this means that 2 sin $ -1 can lie between 0 and 1
ie sin $ can lie between 1/2 and 1
so $ can lie between [pi/6+n pi, npi+pi-pi/6]
12.
f(r/2n) = kr/2nkr/2n + k1/2
f((2n-r)/2n) = k(2n-r)/2nk(2n-r)/2n + k1/2 k-1/2 + r/2nk-1/2 + r/2n = k1/2kr/2n + k1/2
f(r/2n) + f(2n-r/2n) = 1
so, f(1/2n) + f(2n-1/2n) = f(2/2n) + f(2n-2/n) =.....= f(n-1/2n) + f(n+1/2n)... (n-1 pairs)
so summation = n-1 + f(n/2n) = n-1 + k1/2k1/2 + k1/2 = n - 12