Question

Find the range of the following function :-

$f(x) = \dfrac{x^2-3x+2}{x^2+x-6}$​

Answer 1

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

Given function is

$\rm \: f(x) = \dfrac{ {x}^{2} - 3x + 2}{ {x}^{2} + x - 6}$

can be rewritten as

$\rm \: f(x) = \dfrac{ {x}^{2} - 2x - x + 2}{ {x}^{2} + 3x - 2x - 6}$

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

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

$\rm\implies \:x \: \ne \: 2, \: - 3$

Now,

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

Let assume that

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

$\rm \: xy + 3y = x - 1$

$\rm \: xy - x = - 3y - 1$

$\rm \: x(y - 1) = - (3y + 1)$

$\rm \: x(1 - y) = 3y + 1$

$\rm \: x = \dfrac{3y + 1}{1 - y}$

$\rm\implies \:y \: \ne \: 1$

And

$\rm \: When \: x \: = \: 2$

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

$\rm \: 2 - 2y = 3y + 1$

$\rm \: - 2y - 3y =1 - 2$

$\rm \: - 5y = - 1$

$\rm\implies \:y = \dfrac{1}{5}$

So, it means,

$\rm\implies \:y \: \ne \: 1, \: \dfrac{1}{5}$

$\rm\implies \:y \: \in \: R \: - \: \bigg\{1, \: \dfrac{1}{5} \bigg\}$

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ADDITIONAL INFORMATION

If a and b are two real positive numbers such that a < b, then

$\boxed{\sf{ \:(x - a)(x - b) < 0 \: \: \rm\implies \:a < x < b \: }}$

$\boxed{\sf{ \:(x - a)(x - b) \leqslant 0 \: \: \rm\implies \:a \leqslant x \leqslant b \: }}$

$\boxed{\sf{ \:(x - a)(x - b) > 0 \: \: \rm\implies \:x < a \: \: or \: \: x > b \: }}$

$\boxed{\sf{ \:(x - a)(x - b) \geqslant 0 \: \: \rm\implies \:x \leqslant a \: \: or \: \: x \geqslant b \: }}$

$\boxed{\sf{ \: |x| < y \: \: \rm\implies \: \: - y < x < y \: }}$

$\boxed{\sf{ \: |x| \leqslant y \: \: \rm\implies \: \: - y \leqslant x \leqslant y \: }}$

Answer 2

Question:-

Find the range of the following function :-

$\sf \large f(x) = \dfrac{x^2-3x+2}{x^2+x-6}.$

Given:-

$\sf \large f(x) = \dfrac{x^2-3x+2}{x^2+x-6}.$

To Find:-

• The range of the following function.

Solution:-

$\sf \large f(x) = y = \frac{x {}^{2} - 3x + 2}{x {}^{2} + x - 6 }$

$\sf \large \Rightarrow y = \frac{x {}^{2} - 2x - x + 2}{x {}^{2} + 3x - 2x - 6}$

$\sf \large \Rightarrow y = \frac{(x - 2)(x - 1)}{(x + 3)(x - 2)}$

y is not defined when (x + 3) (x – 2) = 0.

i.e., x = – 3 and x = 2.

$\sf \large \therefore D_f = R - \big \{ - 3,2 \big \}$

$\sf \large So, \: y = \frac{x - 1}{x + 3}$

$\sf \large \Rightarrow xy + 3y = x - 1$

$\sf \large \Rightarrow 3y + 1 = x - xy$

$\sf \large \Rightarrow \frac{3y + 1}{1 - y} = x$

$\sf \large \because1 - y ≠0$

$\sf \large \therefore y≠1$

$\sf \large For \: x = 2,y = \frac{1}{5} \\ \: \: \sf \large \big \{but \: x = 2 \: is \: not \: in \: decimal \big \}$

$\sf \large \therefore y = \frac{1}{5} \: cannot \: be \: range$

Answer:-

${ \boxed{ \sf \large So, \: The \: Range \: is =R - \bigg \{ \frac{1}{5} ,1 \bigg \}. }}$

Hope you have satisfied. ⚘