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Question

f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left left columnspacing 1em end attributes row cell left square bracket x right square bracket      text  if  end text minus 3 less than x less or equal than negative 1 end cell row cell vertical line x vertical line      text  if  end text minus 1 less than x less than 1 end cell row cell vertical line left square bracket negative x right square bracket vertical line      text  if  end text 1 less or equal than x less or equal than 3 end cell end table text  then  end text left curly bracket x colon f left parenthesis x right parenthesis greater or equal than 0 right curly bracket equals close

  1. left parenthesis negative 1 comma 3 right parenthesis
  2. left square bracket negative 1 comma 3 right parenthesis
  3. left parenthesis negative 1 comma 3 right square bracket
  4. left square bracket negative 1 comma 3 right square bracket

Hint:

For the given function we have to find the value of x when f(x) will be greater than or equals to 0. The given function has three cases we will put the value of x from each case and then determine when would f(x) will be 0 or positive.

The correct answer is: left parenthesis negative 1 comma 3 right square bracket


    For, f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left left columnspacing 1em end attributes row cell left square bracket x right square bracket      text  if  end text minus 3 less than x less or equal than negative 1 end cell row cell vertical line x vertical line      text  if  end text minus 1 less than x less than 1 end cell row cell vertical line left square bracket negative x right square bracket vertical line      text  if  end text 1 less or equal than x less or equal than 3 end cell end table close, we have to find x when f(x) will be greater than or equals to 0.
    As per the given function,
    f(x) will be negative whennegative 3 less than x less or equal than negative 1
    f(x) will be positive or 0 when negative 1 less than x less than 1  (as f(x)= open vertical bar x close vertical bar)
    f(x) will be positive when 1 less or equal than x less or equal than 3    (as f(x)= open vertical bar open square brackets negative x close square brackets close vertical bar)
    So, for f left parenthesis x right parenthesis greater or equal than 0 comma space x equalsleft parenthesis negative 1 comma 3 right square bracket

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