First slide

Theory of equations

Question

For all real values of $x, | \frac{12 x}{4 x^{2} + 9} |$

Moderate

A

$\leq 1$
B

$\leq 2$
C

>1
D

>2

Solution

Let,
$\begin{array}{r} \frac{12 x}{4 x^{2} + 9} = y \\ 4 {yx}^{2} - 12 x + 9 y = 0 \end{array}$
As x is real,
$\begin{aligned} D = 144 - 4 \cdot 4 y \cdot 9 y \geq 0 \Rightarrow 1 - y^{2} \geq 0 \\ \Rightarrow y^{2} \leq 1 \\ ∴ | y | \leq 1 \\ Hence, | \frac{12 x}{4 x^{2} + 9} | \leq 1 \end{aligned}$

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