Let f:R−{2,6}→R be real valued function defined as f(x)=x2+2x+1x2−8x+12.Then range offx is.

Let y=x2+2x+1x2−8x+12By cross multiplying yx2−8xy+12y−x2−2x−1=0x2(y−1)−x(8y+2)+(12y−1)=0Case 1,y≠1D≥0 ⇒ 8y-22−4(y−1)(12y−1)≥0⇒y(4y+21)≥0 y∈(−∞,−214]∪[0,∞)−{1}Case2,y=1x2+2x+1=x2−8x+1210x=11x=1110 So,y can be 1

Let f:R−{2,6}→R be real valued function defined as f(x)=x2+2x+1x2−8x+12.Then range offx is.

Let y=x2+2x+1x2−8x+12By cross multiplying yx2−8xy+12y−x2−2x−1=0x2(y−1)−x(8y+2)+(12y−1)=0Case 1,y≠1D≥0 ⇒ 8y-22−4(y−1)(12y−1)≥0⇒y(4y+21)≥0 y∈(−∞,−214]∪[0,∞)−{1}Case2,y=1x2+2x+1=x2−8x+1210x=11x=1110 So,y can be 1

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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 $f (x)$ is.

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

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

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

answer is A.

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Detailed Solution

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

By cross multiplying

$\begin{array}{l} y x^{2} - 8 x y + 12 y - x^{2} - 2 x - 1 = 0 \\ x^{2} (y - 1) - x (8 y + 2) + (12 y - 1) = 0 \\ C a s e 1, y \neq 1 \\ D \geq 0 \end{array} \begin{array}{l} \Rightarrow {(8 y - 2)}^{2} - 4 (y - 1) (12 y - 1) \geq 0 \\ \Rightarrow y (4 y + 21) \geq 0 \end{array}$

$\begin{array}{l} y \in (- \infty, \frac{- 21}{4}] \cup [0, \infty) - {1} \\ C a s e 2, y = 1 \\ x^{2} + 2 x + 1 = x^{2} - 8 x + 12 \\ 10 x = 11 \\ x = \frac{11}{10} S o, y c a n b e 1 \end{array}$