Monday, May 20, 2013

Q13/045) If x is real show that x/x^2-5x+9 always lie in the interval [-1/11,1]?



let y = x/(x^2-5x+9)
so yx^2 - 5yx + 9y = x
or yx^2 - (5y+ 1) x + 9y = 0
for it to have real root x(as x is real)  discriminate > = 0 or
(5y+1) ^2 - 36y^2 >= 0
=> (5y + 1 + 6y)(5y+1 - 6y) >= 0
or (11y + 1)(1-y) >= 0 or y >= -1/11 and <= 1
so x/x^2-5x+9 always lie in the interval [-1/11,1]

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