QJEEM23Ma6b00: JEE Main 2023 Mathematics - Class 11 JEEM23Ma6b00 Solution: Let be real valued function defined as . Then range of is

Question

Verified Solution • 2025–26 Session moderate

Let

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

Accepted Answer

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

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

Let

y \neq 1

, then

D \geq 0

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

\Rightarrow 16 y^{2}+1+8 y-\left(12 y^{2}-13 y+1\right) \geq 0

\Rightarrow 4 y^{2}+21 y \geq 0

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

\text { for } y=1 \text {, }

\begin{aligned}&amp; -8 x+12=2 x+1 \\\\&amp; x=\frac{11}{10} \quad \therefore I \in R\end{aligned}

\therefore

Range

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

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