Question

find the domain and range of 2x-3/x-5
pls tell​

Answer 1

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

Given that,

$\rm :\longmapsto\:f(x) = \dfrac{2x - 3}{x - 5}$

Domain :-

For f(x) to be defined,

$\rm :\longmapsto\:x - 5 \ne \: 0$

$\rm :\longmapsto\:x \ne \: 5$

$\bf\implies \:x \: \in \: R - \{5 \}$

Hence,

Domain of

$\rm :\longmapsto\:f(x) = \dfrac{2x - 3}{x - 5}$

is

$\bf\implies \:x \: \in \: R - \{5 \}$

Now, we have to find the range of f(x)

Given that

$\rm :\longmapsto\:f(x) = \dfrac{2x - 3}{x - 5}$

Let us assume that

$\rm :\longmapsto\:y = \dfrac{2x - 3}{x - 5}$

$\rm :\longmapsto\:xy - 5y = 2x - 3$

$\rm :\longmapsto\:xy - 2x = 5y - 3$

$\rm :\longmapsto\:x(y - 2) = 5y - 3$

$\rm :\longmapsto\:x = \dfrac{5y - 3}{y - 2}$

For x to be defined,

$\rm :\longmapsto\:y - 2 \ne \: 0$

$\rm :\longmapsto\:y \ne \: 2$

$\bf\implies \:y \: \in \: R - \{2 \}$

Hence,

Range of

$\rm :\longmapsto\:f(x) = \dfrac{2x - 3}{x - 5}$

is

$\bf\implies \:f(x) \: \in \: R - \{2 \}$

Additional Information :-

Let us consider few more examples :-

1. Find the domain of the following functions :-

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

Solution :-

Now, f(x) is defined when

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

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

$\bf\implies \:x \: \in \: [1, \: \infty )$

$\bf :\longmapsto\:(b) \: f(x) = \sqrt{3 - x}$

Solution :-

Now, f(x) is defined when

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

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

$\rm :\longmapsto\:x \leqslant 3$

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

$\bf :\longmapsto\:(c) \: f(x) = \sqrt{9 - {x}^{2} }$

Now, f(x) to be defined when

$\rm :\longmapsto\:9 - {x}^{2} \geqslant 0$

$\rm :\longmapsto\: - ({x}^{2} - 9) \geqslant 0$

$\rm :\longmapsto\: ({x}^{2} - 9) \leqslant 0$

$\rm :\longmapsto\:(x - 3)(x + 3) \leqslant 0$

$\bf\implies \:x \: \in \: [ - 3, \: 3]$