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Question

Given an in equation open vertical bar 1 plus fraction numerator 2 over denominator x end fraction close vertical bar greater than 3 whose solution set is given by left parenthesis a comma 0 right parenthesis union left parenthesis 0 comma b right parenthesis then answer the following questions If solution set for left parenthesis x plus 1 right parenthesis to the power of 2 end exponent less than left parenthesis 7 x minus 3 right parenthesis blankis (c,d), then a+b+c+d=

  1. 5    
  2. 8    
  3. fraction numerator 9 over denominator 2 end fraction    
  4. fraction numerator 11 over denominator 2 end fraction    

The correct answer is: fraction numerator 11 over denominator 2 end fraction


    left parenthesis x plus 1 right parenthesis to the power of 2 end exponent less than left parenthesis 7 x minus 3 right parenthesis blank rightwards double arrow x to the power of 2 end exponent plus 2 x plus 1 less than left parenthesis 7 x minus 3 right parenthesis
    rightwards double arrow x to the power of 2 end exponent minus 5 x plus 4 less than 0 rightwards double arrow left parenthesis x minus 1 right parenthesis blank left parenthesis x minus 2 right parenthesis blank less than 0
    rightwards double arrow x element of left parenthesis 12 right parenthesis blank therefore c equals 1 comma d equals 4 comma a equals fraction numerator negative 1 over denominator 2 end fraction comma b equals 1
    a plus b plus c plus d equals negative fraction numerator 1 over denominator 2 end fraction plus 1 plus 1 plus 4 equals fraction numerator 11 over denominator 2 end fraction

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