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    Maths-
    General
    Easy

    Question

    Range of the function f left parenthesis x right parenthesis equals fraction numerator x squared plus x plus 2 over denominator x squared plus x plus 1 end fraction semicolon x element of R text  is  end text

    1. left parenthesis 1 comma straight infinity right parenthesis
    2. left parenthesis 1 comma 11 divided by 7 right parenthesis
    3. left square bracket 1 comma 7 divided by 3 right square bracket
    4. left parenthesis 1 comma 7 divided by 5 right parenthesis

    The correct answer is: left square bracket 1 comma 7 divided by 3 right square bracket

      For 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
      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 negative 1 less or equal than log subscript 3 invisible function application x less or equal than 1 end cell row cell not stretchy rightwards double arrow 3 to the power of negative 1 end exponent less or equal than x less or equal than 3 end cell end table
      therefore text  Domain of  end text 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 open square brackets 1 third comma 3 close square brackets text .  end text

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      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

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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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      The value of  tan invisible function application open curly brackets cos to the power of negative 1 end exponent invisible function application open parentheses negative 2 over 7 close parentheses minus pi over 2 close curly brackets is

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      sin invisible function application open parentheses 1 half cos to the power of negative 1 end exponent invisible function application 4 over 5 close parentheses is equal to

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      cos invisible function application open square brackets cos to the power of negative 1 end exponent invisible function application open parentheses negative 1 over 7 close parentheses plus sin to the power of negative 1 end exponent invisible function application open parentheses negative 1 over 7 close parentheses close square brackets text  is equal to  end text

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      tan to the power of negative 1 end exponent invisible function application fraction numerator x over denominator square root of a squared minus x squared end root end fraction is equal to

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      If theta equals sin to the power of negative 1 end exponent invisible function application x plus cos to the power of negative 1 end exponent invisible function application x minus tan to the power of negative 1 end exponent invisible function application x comma 1 less or equal than x less than straight infinity , then the smallest interval in which θ lies is

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