∀x∈R The maximum value of xx2−5x+9 is

Let y=xx2−5x+9⇒yx2−(5y+1)x+9y=0Since x is real⇒b2−4ac≥0(5y+1)2−4(y)(9y)≥0⇒25y2+1+10y−36y2≥0−11y2+10y+1≥0⇒11y2−10y−1≤0⇒11y2−11y+y−1≤0⇒11y(y−1)+1(y−1)≤0⇒(11y+1)(y−1)≤0y∈[−111,1] ∴ Maximum value is 1

∀x∈R The maximum value of xx2−5x+9 is

Let y=xx2−5x+9⇒yx2−(5y+1)x+9y=0Since x is real⇒b2−4ac≥0(5y+1)2−4(y)(9y)≥0⇒25y2+1+10y−36y2≥0−11y2+10y+1≥0⇒11y2−10y−1≤0⇒11y2−11y+y−1≤0⇒11y(y−1)+1(y−1)≤0⇒(11y+1)(y−1)≤0y∈[−111,1] ∴ Maximum value is 1

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$\forall x \in R$ The maximum value of $\frac{x}{x^{2} - 5 x + 9}$ is

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$\frac{- 1}{11}$

$- 1$

$\frac{1}{11}$

answer is D.

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Detailed Solution

Let $y = \frac{x}{x^{2} - 5 x + 9}$

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

Since x is real

$\begin{array}{l} \Rightarrow b^{2} - 4 a c \geq 0 \\ {(5 y + 1)}^{2} - 4 (y) (9 y) \geq 0 \\ \Rightarrow 25 y^{2} + 1 + 10 y - 36 y^{2} \geq 0 \\ - 11 y^{2} + 10 y + 1 \geq 0 \end{array}$

$\begin{array}{l} \Rightarrow 11 y^{2} - 10 y - 1 \leq 0 \\ \Rightarrow 11 y^{2} - 11 y + y - 1 \leq 0 \\ \Rightarrow 11 y (y - 1) + 1 (y - 1) \leq 0 \\ \Rightarrow (11 y + 1) (y - 1) \leq 0 \end{array}$