If x is real, prove that xx2-5x+9lies between -111and 1

y=xx2-5x+9; yx2-5x+9=x yx2-5yx+9y-x=0 yx2+-5y-1x+9y=0 Compare with ax2+bx+c=0 then we get a=y, b=-5y-1, c=9y Since x is real, ∆≥0 ⇒b2-4ac≥0 -5y-12-4y9y≥0 25y2+1+10y-36y2≥0 -11y2+10y+1≥0; 11y2 -10y-1≤0 11y2-11y+y-1≤0; 11y+1y-1≤0 y+111y-1≤0;-111≤y≤1 y lies between -111and 1.

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If x is real, prove that $\frac{x}{x^{2} - 5 x + 9}$ lies between $\frac{- 1}{11} a n d 1$

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Detailed Solution

$y = \frac{x}{x^{2} - 5 x + 9}$ ; $y (x^{2} - 5 x + 9) = x$

$y x^{2} - 5 y x + 9 y - x = 0 y x^{2} + (- 5 y - 1) x + 9 y = 0 C o m p a r e w i t h a x^{2} + b x + c = 0 t h e n w e g e t a = y, b = (- 5 y - 1), c = 9 y S i n c e x i s r e a l, ∆ \geq 0 \Rightarrow b^{2} - 4 a c \geq 0 {(- 5 y - 1)}^{2} - 4 (y) (9 y) \geq 0 25 y^{2} + 1 + 10 y - 36 y^{2} \geq 0 - 11 y^{2} + 10 y + 1 \geq 0; 11 y^{2^{}} - 10 y - 1 \leq 0 11 y^{2} - 11 y + y - 1 \leq 0; (11 y + 1) (y - 1) \leq 0 (y + \frac{1}{11}) (y - 1) \leq 0; \frac{- 1}{11} \leq y \leq 1 y l i e s b e t w e e n \frac{- 1}{11} a n d 1 .$