Let `alpha` be the smallest integral value of x, x>0 such that `tan 19x=(cos96^0+sin96^0)/(cos96^0-sin96^0)`. The last digit of `alpha` is

The least positive integer n for which tan(107n)^(0)=(cos96^(@)+sin96^(@))/(cos96^(@)-sin96^(@))

The smallest value of the polynomial x^(3)-18x^(2)+96x is [0,9] is

The smallest value of th polynomial x^(3) - 18 x^(2) + 96 x in [0,9] is

Let the smallest positive value of x for which the function f(x)=sin""(x)/(3)+sin""(x)/(11), ( x in R ) achieves its maximum value be x_(0) . Express x_(0) in degree i.e. x_(0)=alpha^(0) . Then , the sum of the digits in alpha is

Let the smallest positive value of x for which the function f(x)=sin""(x)/(3)+sin""(x)/(11), ( x in R ) achieves its maximum value be x_(0) . Express x_(0) in degree i.e. x_(0)=alpha^(0) . Then , the sum of the digits in alpha is

Let the smallest positive value of x for which the function f(x)=sin""(x)/(3)+sin""(x)/(11), ( x in R ) achieves its maximum value be x_(0) . Express x_(0) in degree i.e. x_(0)=alpha^(0) . Then , the sum of the digits in alpha is

Let the smallest positive value of x for which the function f(x)=sin""(x)/(3)+sin""(x)/(11), ( x in R ) achieves its maximum value be x_(0) . Express x_(0) in degree i.e. x_(0)=alpha^(0) . Then , the sum of the digits in alpha is

The smallest value of the polynomial x^(3)-18x^(2)+96x in [0, 9] is …………..

For any trarisister alpha = 0.96 , what is the value of its beta ?—