Home
Class 11
MATHS
The relation 'f' is defined by: f(x) = {...

The relation 'f' is defined by: `f(x) = {(x^(2)","0 le x le 3),(3x" ,"3 le x le 10):}`. The relation g is defined by `g(x) = {(x^(2)"," 0 le x le 2),(3x"," 2 le x le 10):}`
Show that f is a function and g is not a function.

Text Solution

AI Generated Solution

Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • RELATIONS AND FUNCTIONS

    MODERN PUBLICATION|Exercise Exercise|10 Videos

  • RELATIONS AND FUNCTIONS

    MODERN PUBLICATION|Exercise Revision Exercise|18 Videos

  • RELATIONS AND FUNCTIONS

    MODERN PUBLICATION|Exercise Ncert File Question for Ncert Book (Exercise2.3)|5 Videos

  • PROBABILITY

    MODERN PUBLICATION|Exercise CHAPTER TEST|12 Videos

  • SEQUENCES AND SERIES

    MODERN PUBLICATION|Exercise CHAPTER TEST|12 Videos

Similar Questions

Explore conceptually related problems

The functin 'f' defined by: f(x) = {(x^(2)" ," 0 le x le 3),(3x ", " 3 le x le 10):} The relation 'g' is defined by g(x) = {(x^(2)"," 0 le x le 2),(3x " ," 2 le x le 10):} Show that 'f' is a function and 'g' is not a function

The relation f is defined by f(x) ={(3x+2", "0le x le2),(x^(3)", "2 le x le 5):}. The relation g is defined by g(x) ={(3x+2", "0le x le1),(x^(3)", "1 le x le 5):} Show that f is a function and g is not a function

Knowledge Check

  • If f(x)={{:(,x^(2)+1,0 le x lt 1),(,-3x+5, 1 le x le 2):}

    A
    It has a relative minimum at x = 1
    B
    It has a relative maximum at x =1
    C
    It is not continuous at x = 1
    D
    None of these

  • If f(x) = {{:(3x^(2) + 12 x -1, -1 le x le 2),(37-x, 2 lt x le 3):} , then

    A
    f(x) is increasing on `[-1,2]`
    B
    f (x) is continuous in `[-1,3]`
    C
    f'(2) does not exist
    D
    f(x) has the maximum value at x = 2

  • Let f(x) ={:{(x, "for", 0 le x lt1),( 3-x,"for", 1 le x le2):} Then f(x) is

    A
    continuous at x =1
    B
    Right continuous at x =1
    C
    Left continuous at x =1
    D
    Limit exists at x =1

    • Similar Questions

    Explore conceptually related problems

    Let f(x)={: (x^2 ,"when" 0 le x le 2),(2x , "when" 2 le x le 5) :} g(x)={: (x^2 ,"when" 0 le x le 3),(2x , "when" 3 le x le 5) :} Show that f is a function while g is not a function .

    Let f(x)={{:(1+x",", 0 le x le 2),(3-x"," ,2 lt x le 3):} find (fof) (x).

    If f(x)={:{(x-1", for " 1 le x lt 2),(2x+3", for " 2 le x le 3):} , then at x=2

    If f(x) = {{:( 3x ^(2) + 12 x - 1",", - 1 le x le 2), (37- x",", 2 lt x le 3):}, then

    If f (x)= {{:(1+x, 0 le x le 2),( 3x-2, 2 lt x le 3):}, then f (f(x)) is not differentiable at:

    Home
    Profile