Functions Question 19
Question 19 - 31 January - Shift 2
Let $f: \mathbb{R}-{2,6} \to \mathbb{R}$ be real valued function defined as $f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$. Then range of $f$ is
(1) $(-\infty,-\frac{21}{4}] \cup[0, \infty)$
(2) $(-\infty,-\frac{21}{4}) \cup(0, \infty)$
(3) $(-\infty,-\frac{21}{4}] \cup[\frac{21}{4}, \infty)$
(4) $(-\infty,-\frac{21}{4}] \cup[1, \infty)$
Show Answer
Answer: (1)
Solution:
Formula: Operations on functions, Range of function, Nature of roots of quadratic equation
Let $y=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$
By cross multiplying
$yx^{2}-8 xy+12 y-x^{2}-2 x-1=0$
$x^{2}(y-1)-x(8 y+2)+(12 y-1)=0$
Case 1: $ y \neq 1$
$D \geq 0$
$\Rightarrow(8 y+2)^{2}-4(y-1)(12 y-1) \geq 0$
$\Rightarrow y(4 y+21) \geq 0$
$y \in(-\infty, \frac{-21}{4}] \cup[0, \infty)-{1}$
Case 2: $y=1$
$ \begin{aligned} & x^{2}+2 x+1=x^{2}-8 x+12 \\ & 10 x=11 \\ & x=\frac{11}{10} \quad \text{ So, } y \text{ can be } 1 \end{aligned} $
Therefore, range of f is $(-\infty,-\frac{21}{4}] \cup[0, \infty)$
