Question

The domain of the function f(x) = 1/(x² – 3x + 2) is

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\sf \: f(x) = \frac{1}{ {x}^{2} - 3x + 2 }$

$\sf \: f(x) = \frac{1}{ {x}^{2} - 2x - x + 2 }$

$\sf \: f(x) = \frac{1}{ x(x - 2) - 1(x - 2) }$

$\sf \: f(x) = \frac{1}{ (x - 2) \: (x - 1) }$

Now, f(x) is defined if

$\sf \: (x - 2)(x - 1) \: \ne \: 0$

$\sf \:  \implies \: x \: \ne \: 1, \: 2$

$\sf \:  \implies \: x \: \in \: R - \{1, \: 2 \}$

Hence,

$\sf \implies Domain \: of \: \frac{1}{ {x}^{2} - 3x + 2 } \: is \: x \: \in \: R - \{1, \: 2 \}$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = sinx & \sf   - 1 \leqslant y \leqslant 1\\ \\ \sf y = cosx & \sf  - 1 \leqslant y \leqslant 1 \\ \\ \sf y = tanx & \sf y \:  \in \: ( -  \infty , \infty )\\ \\ \sf y = cosec & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = secx & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = cotx & \sf y \:  \in \: ( -  \infty , \infty ) \end{array}} \\ \end{gathered}$

Answer 2

Sᴏʟᴜᴛɪᴏɴ :-

Given function :-

$\begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: f(x) = \frac{1}{x {}^{2} - 3x + 2} } \\ \end{gathered}$

So,

$\begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: \frac{1}{x {}^{2} - 3x + 2} } \\ \end{gathered}$

$\begin{gathered}\\ \large\mathsf { \: \: x {}^{2} - 3x + 2 } \\ \end{gathered}$

Dᴏᴍᴀɪɴ :-

$\begin{gathered}\\ \large\dashrightarrow \boxed{\mathsf \red { \: \: x \: \in \: \mathbb{R} \: \: - \bigg \{1 \: \: and \: \: 2 \: \bigg \}}} \\ \end{gathered}$

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