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    Maths-
    General
    Easy

    Question

    The domain of  sin to the power of negative 1 end exponent invisible function application open parentheses log subscript 3 invisible function application x close parentheses text  is  end text

    1. left square bracket negative 1 comma 1 right square bracket
    2. left square bracket 0 comma 1 right square bracket
    3. left square bracket 0 comma straight infinity right square bracket
    4. open square brackets 1 third comma 3 close square brackets

    The correct answer is: left square bracket 0 comma straight infinity right square bracket

      The given function is defined when  x squared minus 1 semicolon 3 plus x greater than 0 text  and  end text 3 plus x not equal to 1
      table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell not stretchy rightwards double arrow x squared greater than 1 semicolon 3 plus x greater than 0 text  and  end text x not equal to negative 2 end cell row cell not stretchy rightwards double arrow negative 1 greater than x greater than 1 semicolon x greater than negative 3 comma x not equal to negative 2 end cell end table
      thereforeDomain of the function is
      D subscript f equals left parenthesis negative 3 comma negative 2 right parenthesis union left parenthesis negative 2 comma negative 1 right parenthesis union left parenthesis 1 comma straight infinity right parenthesis

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      text  If  end text f left parenthesis x right parenthesis equals fraction numerator 1 over denominator square root of vertical line x vertical line minus x end root end fraction text  then, domain of  end text f left parenthesis x right parenthesis text  is  end text

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