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If $f(x)=\frac{2 x-3}{(x-2)(x-3)}$ is a real valued function then the value that $f(x)$ does not take is
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Verified Answer
The correct answer is:
-2
$f(x)=\frac{2 x-3}{(x-2)(x-3)}$
Let $y=\frac{2 x-3}{x^2-5 x+6}$
$$
\begin{aligned}
& \Rightarrow \mathrm{y}\left(\mathrm{x}^2-5 \mathrm{x}+6\right)-(2 \mathrm{x}-3)=0 \\
& \Rightarrow \mathrm{yx}^2-(5 \mathrm{y}+2) \mathrm{x}+(6 \mathrm{y}+3)=0
\end{aligned}
$$
for real $\mathrm{x}, \mathrm{D} \geq 0$
$$
\Rightarrow(5 \mathrm{y}+2)^2-4 \mathrm{y}(6 \mathrm{y}+3) \geq 0
$$
On solving we get $\mathrm{y} \in(-\infty,-2-2 \sqrt{3}] \cup[-2+2 \sqrt{3}, \infty)$
$$
\Rightarrow \mathrm{x} \neq-2
$$
Let $y=\frac{2 x-3}{x^2-5 x+6}$
$$
\begin{aligned}
& \Rightarrow \mathrm{y}\left(\mathrm{x}^2-5 \mathrm{x}+6\right)-(2 \mathrm{x}-3)=0 \\
& \Rightarrow \mathrm{yx}^2-(5 \mathrm{y}+2) \mathrm{x}+(6 \mathrm{y}+3)=0
\end{aligned}
$$
for real $\mathrm{x}, \mathrm{D} \geq 0$
$$
\Rightarrow(5 \mathrm{y}+2)^2-4 \mathrm{y}(6 \mathrm{y}+3) \geq 0
$$
On solving we get $\mathrm{y} \in(-\infty,-2-2 \sqrt{3}] \cup[-2+2 \sqrt{3}, \infty)$
$$
\Rightarrow \mathrm{x} \neq-2
$$
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