Question

Range of f(x) = $\frac{x - 1}{x^{2} - 2x + 3}$, is

• A. [1 −$\sqrt{3}$, 1+ $\sqrt{3}$]
• B. [−1/4, 1/4]
• C. R
• D. $\left\lbrack \frac{1}{4}(1 - \sqrt{3}),\frac{1}{4}(1 + \sqrt{3}) \right\rbrack$

Answer 1

Answer: (D)

$\left\lbrack \frac{1}{4}(1 - \sqrt{3}),\frac{1}{4}(1 + \sqrt{3}) \right\rbrack$

Answer 2

Domain of f(x) is all real numbers.

f(x) = $\frac { - ( x - \sqrt { 3 } ) ( x + \sqrt { 3 } ) } { \left( x ^ { 2 } - 2 x + 3 \right) ^ { 2 } }$

$f ( \sqrt { 3 } ) = \frac { 1 } { 4 } ( 1 + \sqrt { 3 } ) , f [ - \sqrt { 3 } ) = \frac { 1 } { 4 } [ 1 - \sqrt { 3 } )$

also $\operatorname { Lim } _ { x \rightarrow \infty } f ( x ) = 0$ , $\operatorname { Lim } _ { x \rightarrow - \infty }$f(x) = 0

⇒ Range is $\left[ \frac { 1 } { 4 } ( 1 - \sqrt { 3 } ) , \frac { 1 } { 4 } ( 1 + \sqrt { 3 } ) \right]$