Home
Class 11
MATHS
If 1/(3x-4) lt 0 , then (i) x in (4/3 ,...

If `1/(3x-4) lt 0` , then (i) `x in (4/3 , oo)` (ii) `x in (-oo , 4/3)` (iii) `x in (0,oo)` (iv) ` x in (0,4/3)`

A

`x in (4/3 , oo)`

B

`x in (-oo , 4/3)`

C

`x in (0,oo)`

D

` x in (0,4/3)`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • LINEAR INEQUALITIES

    ICSE|Exercise MULTIPLE CHOICE QUESTION|16 Videos

  • LIMITS AND DERIVATIVES

    ICSE|Exercise Multiple Choice Questions |31 Videos

  • MATHEMATICAL INDUCTION

    ICSE|Exercise Exercise (a)|1 Videos

Similar Questions

Explore conceptually related problems

The solution of system of linear inequalities in x represented on line as (i) x in ( - oo,-4) cup [3,oo) (ii) x in (- oo, -4] cup (5,oo) (iii) x in (- oo , -3) cup (5,oo) (iv) x in (-oo , -4) cup [ 5,oo)

If |x-1|>5, then a. x in (-4,6) b. x in [-4,6] c. x in (-oo,-4)uu(6,oo) d. x in (-oo,-4)uu[6,oo)

If (2x+1)/(x-2) le 1 , x in R , then the solution set is (i) ( -oo,1/2) (ii) (- oo,-1/2] (iii) [-3,2] (iv) [-3,2)

Let f(x) lt 0 AA x in (-oo, 0) and f (x) gt 0 ,AA x in (0,oo) also f (0)=0, Again f'(x) lt 0 ,AA x in (-oo, -1) and f '(x) gt 0, AA x in (-1,oo) also f '(-1)=0 given lim _(x to -oo) f (x)=0 and lim _(x to oo) f (x)=oo and function is twice differentiable. If f'(x) lt 0 AA x in (0,oo)and f'(0)=1 then number of solutions of equation f (x)=x ^(2) is : (a) 1 (b) 2 (c) 3 (d) 4

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 difference between the roots of the equation x^(2)+ax+1=0 is less than sqrt(5) , then the set of possible values of a is (i) (3,oo) (ii) (-oo,-3) (iii) (-3,3) (iv) (-3,oo)

A function f(x) having the following properties, (i) f(x) is continuous except at x=3 (ii) f(x) is differentiable except at x=-2 and x=3 (iii) f(0) =0 lim_(x to 3) f(x) to - oo lim_(x to oo) f(x) =3 , lim_(x to oo) f(x)=0 (iv) f'(x) gt 0 AA in (-oo, -2) uu (3,oo) " and " f'(x) le 0 AA x in (-2,3) (v) f''(x) gt 0 AA x in (-oo,-2) uu (-2,0)" and "f''(x) lt 0 AA x in (0,3) uu(3,oo) Then answer the following questions Find the Maximum possible number of solutions of f(x)=|x|

Let f(x) lt 0 AA x in (-oo, 0) and f (x) gt 0 AA x in (0,oo) also f (0)=0, Again f'(x) lt 0 AA x in (-oo, -1) and f '(x) gt 0 AA x in (-1,oo) also f '(-1)=0 given lim _(x to -oo) f (x)=0 and lim _(x to oo) f (x)=oo and function is twice differentiable. The minimum number of points where f'(x) is zero is: (a) 1 (b) 2 (c) 3 (d) 4

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

If 1+2x +3x^2+4x^3 +…..oo ge 4 then