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The maximum value of (3x*x + 9x + 17)/(3x*x + 9x + 7) is.x is real

The maximum value of (3x*x + 9x + 17)/(3x*x + 9x + 7) is.x is real

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1 Answers

Arun
25750 Points
6 years ago
Dear Harsh
 
  1. First of all, distribute the fraction.
    1. (3x^2+9x+17)/(3x^2+9x+7) = 1 + 10/(3x^2+9x+7)
  2. Now, for this sum to be Maximum, the fraction should be at its maximum value. For this to happen, the denominator of the fraction should be at its minimum. Consider the polynomial 3x^2+9x+7
    • Try to make the polynomial a Perfect Square. This way, it'll be easier to calculate its Minimum.
      • 3x^2+9x+7 = [{(√3)x}^2 + 2*(√3)*{3(√3)/2}*x + 27/4] +(7 - 27/4)
      • This is nothing but, [(√3)x + 3*(√3)/2]^2 + 1/4.
      • This polynomial can have its minimum value as 1/4, because the Perfect Square can have its minimum value as 0, since x is Real.
    • Thus min value of polynomial is 1/4.
    • Hence total value(maximum) = 1+ 10/(1/4)
    • = 1+ 40 = 41
 
Regards
Arun (askIITians forum expert)

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