First slide

Theory of expressions

Question

For the real values of x, maximum value of $\frac{x}{x^{2} - 5 x + 9}$ is

Easy

A

1
B

2
C

3
D

0

Solution

Let $y = \frac{x}{x^{2} - 5 x + 9}$
$\Rightarrow y x^{2} - (5 y + 1) x + 9 y = 0$

Here $a = y, b = - (5 y + 1), c = 9 y$

$Δ \geq 0 \Rightarrow {(5 y + 1)}^{2} - 4. y .9 y \geq 0$       $(∵ b^{2} - 4 a c \geq 0)$

                $\Rightarrow 11 y^{2} - 10 y - 1 \leq 0$

              $\Rightarrow (11 y + 1) (y - 1) \leq 0$

              $\Rightarrow y \in [\frac{- 1}{11}, 1]$

                $y = 1, \frac{- 1}{11}$

$∴$ Maximum value =1

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