Question

The value of $m$ for which expression $\frac{2 x^2-5 x+3}{4 x-m}$ can take all real values for $x \in R-\left\{\frac{m}{4}\right\}$ is

• A. $(4,6)$
• B. $[4,6]$
• C. $R$
• D. $R-\{4,6\}$

Answer 1

Answer: (A)

$y=\frac{2 x^2-5 x+3}{4 x-m} \Rightarrow 4 x y-m y=2 x^2-5 x+3$
$ \Rightarrow 2 x^2-x(5-4 y)+m y+3=0$
$D \geq 0 $
$(5-4 y)^2-4 \cdot 2(m y+3) \geq 0 $
$\Rightarrow 16 y^2+y(40-8 m)+1 \geq 0$
$D \leq 0 $
$(40-8 m)^2-4 \cdot 16 \leq 0$
$m \in[4,6]$
$\text { But } x \neq m / 4$
$\text { so, } m \in(4,6)$