Question

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

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{1}{ {x}^{2} - 3x + 2 }$

Can be rewritten as

$\rm :\longmapsto\:f(x) = \dfrac{1}{ {x}^{2} - 2x - x + 2}$

$\rm :\longmapsto\:f(x) = \dfrac{1}{x(x - 2) - 1(x - 2)}$

$\rm :\longmapsto\:f(x) = \dfrac{1}{(x - 1)(x - 2)}$

For, f(x) to be defined,

$\rm :\longmapsto\:x - 1 \: \ne \: 0 \: \: \: and \: \: \: x - 2 \: \ne \: 0$

$\rm :\longmapsto\:x \: \ne \: 1 \: \: \: and \: \: \: x \: \ne \: 2$

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

Hence,

Domain of

$\bf :\longmapsto\:f(x) = \dfrac{1}{ {x}^{2} - 3x + 2 }$

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

Additional Information :-

Let us take some more examples.

Question :- Find the domain of the following functions :-

$\bf :\longmapsto\:(a) \: \: f(x) = \sqrt{x + 2}$

For, f(x) to be defined,

$\rm :\longmapsto\:x + 2 \geqslant 0$

$\rm :\longmapsto\:x \geqslant - \: 2$

$\bf\implies \:x \: \in \: [ - \: 2, \: \infty )$

$\bf :\longmapsto\:(b) \: \: f(x) = \sqrt{ - x + 1}$

For, f(x) to be defined,

$\rm :\longmapsto\: - x + 1 \geqslant 0$

$\rm :\longmapsto\: - x \geqslant - 1$

$\rm :\longmapsto\: x \leqslant 1$

$\bf\implies \:x \: \in \: ( - \: \infty , \: 1]$

$\bf :\longmapsto\:(c) \: \: f(x) = \sqrt{ 1 - {x}^{2} }$

For, f(x) to be defined,

$\rm :\longmapsto\:1 - {x}^{2} \geqslant 0$

$\rm :\longmapsto\: - ({x}^{2} - 1) \geqslant 0$

$\rm :\longmapsto\: {x}^{2} - 1 \leqslant 0$

$\rm :\longmapsto\:(x - 1)(x + 1) \leqslant 0$

$\rm :\longmapsto\: - 1 \leqslant x \leqslant 1$

$\bf\implies \:x \: \in \: [ - 1, \: 1]$

$\bf :\longmapsto\:(d) \: \: f(x) = \dfrac{1}{ \sqrt{3 - x} }$

For, f(x) to be defined,

$\rm :\longmapsto\:3 - x > 0$

$\rm :\longmapsto\: - x > - 3$

$\rm :\longmapsto\: x < 3$

$\bf\implies \:x \: \in \: ( - \: \infty , \: 3)$

Answer 2

SOLUTION

TO DETERMINE

The domain of the function

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

EVALUATION

Here the given function is

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

Now we factorise the denominator

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

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

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

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

The function is well defined for all values of x except where the denominator vanishes

Now the denominator vanish

When ( x - 1 ) ( x - 2 ) = 0

⇒ x = 1 , 2

Hence the required Domain of the function

$\sf = \mathbb{R} - \{1, 2 \}$

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