9Let f(x)=1cos{x}where [y] and {y} denote greatest integer and fractional part functions respectivelyand g(x)=2x2 – 3x(k+1)+(3k + 1). If g(f(x)]lt0 Vxe Rthen find the number of integral valuesof k.

Let f(x)=(-1)^([x]) where [.] denotes the greatest integer function),then

Let f(x)=sec^(-1)[1+cos^(2)x], where [.] denotes the greatest integer function. Then the

Let f(x)=sec^(-1)[1+cos^(2)x], where [.] denotes the greatest integer function. Then the

f(x)=1+[cos x]x, in 0<=x<=(x)/(2) (where [.] denotes greatest integer function)

Let f(x) = [x]^(2) + [x+1] - 3 , where [.] denotes the greatest integer function. Then

Let f(x) = [x]^(2) + [x+1] - 3 , where [.] denotes the greatest integer function. Then

If f(x)=([{x}]tan^(-1)((x^(2)-3x-1)/(x^(2)-3x-5))+3-x^(7))^(1/7) . Where [k] and {k} denotes greatest integer and fractional part functions of k respectively, then the value of f^(-1)(50)-f(50)+f(f(100)) , is

Let f(x)=1+|x|,x =-1, where [* denotes the greatest integer function.Then f{f(-2.3)} is equal to