The real solution of `[x] + 5x +[10x]+[20x] = 36k + 35`,`k in I`, if the fractional part of x lies in `[1/10,1/5)`

The complete solution set of the equation sgn ((x^(2)-5x+4)/(x))=-1, where {*} is fractional part function is

If f'(x)=|x|-{x} where {x} denotes the fractional part of x, then f(x) is decreasing in (-1/2, k) , find k.

Assertion (A) : The number of positive integral solutions of x_(1)+x_(2)+x_(3)=10 is 36. Reason (R) : The number positive integral solutions of the equation x_(1)+x_(2)+x_(3)+.......+x_(k)=n is ""^(n-1)C_(k-1) , The correct answer is

If 2x-1 is a factor of 4x^(3) - 16x^(2) + 10x + k , find the value of k.

If (x+5) is a factor of p(x)=x^3-20x+5k then k=?

If domain of the function f(x)=sqrt(10x-x^(2)) is A then possible value(s) of x in A satisfying (6{x}^(2)-5{x}+1)=0 is/are (where {-} denotes fractional part function)