Prove that the solution of `(x+1)/(x+3)lt3`" is "(-oo,-4)cup(-3,oo)l.`

Solve (x+1)/(x+3)lt3

The solution set of |3x-2| lt 1 is

Show that range of the function (1)/(2cosx-1)" is "(-oo,-(1)/(3)]cup[1,oo)

For x^2-(a+3)|x|+4=0 to have real solutions, the range of a is (-oo,-7]uu[1,oo) b. (-3,oo) c. (-oo,-7) d. [1,oo)

If the intercepts made by the line y=mx by lines x=2 and x=5 is less than 5, then the range of values of m is a. (-oo,-4/3)uu(4/3,oo) b. (-4/3,4/3) c. (-3/4,4/3) d. none of these

Let f(x) be a quadratic expression such that f(-1)+f(2)=0 . If one root of f(x)=0 is 3 , then the other root of f(x)=0 lies in (A) (-oo,-3) (B) (-3,oo) (C) (0,5) (D) (5,oo)

If (log)_3(x^2-6x+11)lt=1, then the exhaustive range of values of x is: (a) (-oo,2)uu(4,oo) (b) [2,4] (c) (-oo,1)uu(1,3)uu(4,oo) (d) none of these