Maths-
General
Easy

Question

The domain of function, satisfying f left parenthesis x right parenthesis plus f open parentheses x to the power of negative 1 end exponent close parentheses equals fraction numerator x to the power of 3 end exponent plus 1 over denominator x end fraction

  1. An empty set    
  2. A singleton    
  3. A finite set    
  4. An infinite set    

The correct answer is: A finite set


    f left parenthesis x right parenthesis plus open parentheses fraction numerator 1 over denominator x end fraction close parentheses equals x to the power of 2 end exponent plus fraction numerator 1 over denominator x end fraction
    text replacing  end text rightwards arrow fraction numerator 1 over denominator x end fraction semicolon open parentheses fraction numerator 1 over denominator x end fraction close parentheses minus f left parenthesis x right parenthesis equals fraction numerator 1 over denominator x to the power of 2 end exponent end fraction plus x
    fraction numerator 1 over denominator x to the power of 2 end exponent end fraction plus x equals x to the power of 2 end exponent plus fraction numerator 1 over denominator x end fraction rightwards double arrow x minus fraction numerator 1 over denominator x end fraction equals x to the power of 2 end exponent minus fraction numerator 1 over denominator x to the power of 2 end exponent end fraction rightwards double arrow open square brackets x minus fraction numerator 1 over denominator x end fraction close square brackets equals open square brackets x minus fraction numerator 1 over denominator x end fraction close square brackets open square brackets x plus fraction numerator 1 over denominator x end fraction close square brackets
    open square brackets x minus fraction numerator 1 over denominator x end fraction close square brackets open square brackets x plus fraction numerator 1 over denominator x end fraction minus 1 close square brackets equals 0
    x equals fraction numerator 1 over denominator x end fraction semicolon
    x plus fraction numerator 1 over denominator x end fraction equals 1 left parenthesis text  rejected  end text right parenthesis
    text hence x  end text equals 1 text  or  end text minus 1 rightwards double arrow
    left curly bracket 1 comma negative 1 rightwards double arrow 3 right curly bracket

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