If for any real x,11x2+12x+6x2+4x+2=y is such that y<a or y≥b, then a,b are

⇒yx2+4xy+2y=11x2+12x+6 ⇒(y=11)x2+4x(y−3)+2(y−3)=0⇒x=−4(y−3)±16(y−3)2−8(y−11)(y−3)2(y−11) Since, x∈ real number D≥0⇒16−8(y−11)(y−3)≥0⇒8(y−3)(2y−6−y+11)≥0⇒8(y−3)(y+5)≥0⇒y≤−5 or y≥3∴a=−5 and b=3

If for any real x,11x2+12x+6x2+4x+2=y is such that y<a or y≥b, then a,b are

⇒yx2+4xy+2y=11x2+12x+6 ⇒(y=11)x2+4x(y−3)+2(y−3)=0⇒x=−4(y−3)±16(y−3)2−8(y−11)(y−3)2(y−11) Since, x∈ real number D≥0⇒16−8(y−11)(y−3)≥0⇒8(y−3)(2y−6−y+11)≥0⇒8(y−3)(y+5)≥0⇒y≤−5 or y≥3∴a=−5 and b=3

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If for any real $x, \frac{11 x^{2} + 12 x + 6}{x^{2} + 4 x + 2} = y$ is such that $y < a o r y \geq b,$ then a,b are

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3, 5

-5, 3

-4, 5

-6,4

answer is B.

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Detailed Solution

$\Rightarrow y x^{2} + 4 x y + 2 y = 11 x^{2} + 12 x + 6 \begin{array}{l} \Rightarrow (y = 11) x^{2} + 4 x (y - 3) + 2 (y - 3) = 0 \\ \Rightarrow x = \frac{- 4 (y - 3) \pm \sqrt{\frac{16 {(y - 3)}^{2} - 8}{(y - 11) (y - 3)}}}{2 (y - 11)} \end{array} \begin{array}{l} S i n c e, x \in r e a l n u m b e r D \geq 0 \\ \Rightarrow 16 - 8 (y - 11) (y - 3) \geq 0 \\ \Rightarrow 8 (y - 3) (2 y - 6 - y + 11) \geq 0 \\ \Rightarrow 8 (y - 3) (y + 5) \geq 0 \\ \Rightarrow y \leq - 5 o r y \geq 3 \\ ∴ a = - 5 a n d b = 3 \end{array}$