Question

Let $f: R -\\{2,6\\} \rightarrow R$ 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. $\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)$
• B. $\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$
• C. $\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty)$
• D. $\left(-\infty,-\frac{21}{4}\right] \cup[1, \infty)$

Answer 1

Answer: (B)

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

Answer 2

Let y=x2−8x+12x2+2x+1​
By cross multiplying
yx2−8xy+12y−x2−2x−1=0
x2(y−1)−x(8y+2)+(12y−1)=0
Case 1,y=1
D≥0
⇒(8y+2)2−4(y−1)(12y−1)≥0
⇒y(4y+21)≥0

y∈(−∞,4−21​]∪[0,∞)−{1}
Case 2,y=1
x2+2x+1=x2−8x+12
10x=11
x=1011​ So, y can be 1
Hence y∈(−∞,4−21​]∪[0,∞)
So, the correct option is (B) : $\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$