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

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

(B) $\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty)$

(C) $\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)$

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

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Sets Relations and Functions

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

Answer: (A)

Writing

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

and clearing denominators,

(y - 1) x^{2} - (8 y + 2) x + (12 y - 1) = 0.

For real

x

with

y \neq = 1

the discriminant must be non-negative:

(8 y + 2)^{2} - 4 (y - 1) (12 y - 1) = 16 y^{2} + 84 y = 4 y (4 y + 21) \geq 0

, giving

y \leq - \frac{21}{4}

y \geq 0.

The value

y = 1

is attained at

x = \frac{11}{10}

(in

[0, \infty)

) and

y = 0

x = - 1

, both admissible. The range is

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

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Single correct option