If `(|x-2|)/(x-2)geq0,` then `x in [2,oo)` b. `x in (2,oo)` c. `x in (-oo,2)` d. `x in (-oo,2]`

If |x+3|geq10 , then x in (-13 ,\ 7] b. x in (-oo,-13)uu(7,oo) c. x in (-13 ,7) d. x in (-oo,\ -13]uu[7,oo)

If x\ a n d\ a are real numbers such that a >0 and |x|> a , then (a) x in (-a ,oo) b. x in [-oo, a) c. x in (-a , a) d . x in (-oo,-a)U(a ,oo)

Solution set of the inequation, (x^2-5x+6)/(x^2+x+1)<0 is x in (-oo,2) (b) x in (2,3) x in (-oo,2)uu(3,oo) (d) x in (3,oo)

If |x+2|lt=9 then x in (-7, 11) b. x in [-11 ,7] c. x in (-oo,7)uu(11 ,oo) d. x in (-oo,-7)uu[11 ,oo)

If |x+2|lt=9 then x in (-7, 11) b. x in [-11 ,7] c. x in (-oo,7)uu(11 ,oo) d. x in (-oo,-7)uu[11 ,oo)

If x and a are real numbers such that a>0 and |x|>a, then x in(-a,oo) b.x in[-oo,a) c.x in(-a,a) d.x in(-oo,-a)uu(a,oo)

Complete solution set of the inequation (x^2-1)(x^2-6x+8)geq0 is (a) x in [-1,1]uu[2,4] (b) x in (-oo,-1)uu[1,2]uu[4,oo] (c) x in [-1,2]uu[4,oo] (d) none of these

Solution set of the inequation,(x^(2)-5x+6)/(x^(2)+x+1)<0 is x in(-oo,2) (b) x in(2,3)x in(-oo,2)uu(3,oo)( d) x in(3,oo)

If the function f(x)=x^2-k x+5 is increasing on [2,\ 4] , then (a) k in (2,\ oo) (b) k in (-oo,\ 2) (c) k in (4,\ oo) (d) k in (-oo,\ 4)

Let f(x)=sqrt(|x|-{x})(w h e r e{dot} denotes the fractional part of (x)a n dX , Y are its domain and range, respectively). Then (a) X in (-oo,1/2) and Y in (1/2,oo) (b) X in (-oo in ,1/2)uu[0,oo)a n dY in (1/2,oo) (c) X in (-oo,-1/2)uu[0,oo)a n dY in [0,oo) (d) none of these