function doubts...............

plz solve these .................these may be easy for u all .............bt these ques puzzled me..................

3) 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]

5 and 6 try graphs..

for 7, write the question better.

2.

f(r/2n) = k^r/2nk^r/2n + k^1/2

f((2n-r)/2n) = k^(2n-r)/2nk^(2n-r)/2n + k^1/2 k^{-1/2 + r/2n}k^{-1/2 + r/2n} = k^1/2k^r/2n + k^1/2

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 + k^1/2k^1/2 + k^1/2 = n - 12