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Easy

Question

How many real solutions does the equation x to the power of 7 end exponent plus 14 x to the power of 5 end exponent plus 16 x to the power of 3 end exponent plus 30 x minus 560 equals 0 have?

  1. 5  
  2. 7  
  3. 1  
  4. 3  

The correct answer is: 1


    Let f open parentheses x close parentheses equals x to the power of 7 end exponent plus 14 x to the power of 5 end exponent plus 16 x to the power of 3 end exponent plus 30 x minus 560
    f to the power of ´ end exponent open parentheses x close parentheses equals 7 x to the power of 6 end exponent plus 70 x to the power of 4 end exponent plus 48 x to the power of 2 end exponent plus 30 greater than 0 comma blank for all x element of R

    therefore f left parenthesis x right parenthesis is increasing

    therefore blank f open parentheses x close parentheses equals 0 has only one solution

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