Let f : R { 2 , 6 } R be real valued function defined as f ( x ) = x 2 + 2 x + 1 x 2 - 8 x + 12 . Then range of f is [2023]

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Solution

Q.
Let $f : R - {2, 6} \to R$ be real valued function defined as $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12} .$ Then range of $f$ is [2023]

1 $(- \infty, - \frac{21}{4}] \cup [\frac{21}{4}, \infty)$

2 $(- \infty, - \frac{21}{4}] \cup [0, \infty)$

3 $(- \infty, - \frac{21}{4}) \cup (0, \infty)$

4 $(- \infty, - \frac{21}{4}] \cup [1, \infty)$

Ans.
(2)

Let $y = \frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12}$

By cross multiplying, we get

$y x^{2} - 8 x y + 12 y - x^{2} - 2 x - 1 = 0$

$\Rightarrow x^{2} (y - 1) - x (8 y + 2) + (12 y - 1) = 0$

Case I: When $y \neq 1$ ; $D \geq 0$

$\Rightarrow {(8 y + 2)}^{2} - 4 (y - 1) (12 y - 1) \geq 0$

$\Rightarrow y (4 y + 21) > 0$

$\Rightarrow y \in (- \infty, - \frac{21}{4}] \cup [0, \infty) - {1}$

Case II: When $y = 1$

$x^{2} + 2 x + 1 = x^{2} - 8 x + 12$

$\Rightarrow 10 x = 11$

$\Rightarrow x = \frac{11}{10}$

So, $y$ can be 1. Hence, $y \in (- \infty, - \frac{21}{4}] \cup [0, \infty)$

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