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if x is real then find the upper and lower limits of (9x^2-3x+1)/(9x^2+3x+1)

if x is real then find the upper and lower limits of (9x^2-3x+1)/(9x^2+3x+1)

Grade:11

1 Answers

Ajay
209 Points
7 years ago
Basically you have to find the range of the function as follows.
 let y =  (9x^2-3x+1)/(9x^2+3x+1)
y*(9x^2+3x+1) =  (9x^2-3x+1)
Rearranging the terms so that it forms a quatratic equation for x.
9(y-1)x2 + 3x(y+1) + (y-1) = 0.
Since x is real D for this equation should be greater than zero hence
9(1+y)2 – 36(y-1)2 is greater or equal to zero
simplifying we get
3y2 – 10y -3 less than or equal to zero
(y-1)(3y-1) less than zero.
 
y lies in to [1/3, 3].
Hence upper limit is 3 and lower  limit is 1/3.
 

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