If for any real x,y = frac{11 x^2+12 x+6}{x^2 | vedclass

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(B) Given $y = \frac{11 x^2+12 x+6}{x^2+4 x+2}$.Rearranging the terms,we get:$y(x^2+4x+2) = 11x^2+12x+6$$(y-11)x^2 + (4y-12)x + (2y-6) = 0$Since $x$ i

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$-5, 3$

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(B) Given $y = \frac{11 x^2+12 x+6}{x^2+4 x+2}$.Rearranging the terms,we get:$y(x^2+4x+2) = 11x^2+12x+6$$(y-11)x^2 + (4y-12)x + (2y-6) = 0$Since $x$ is a real number,the discriminant $D$ must be greater than or equal to $0$:$D = (4y-12)^2 - 4(y-11)(2y-6) \geq 0$$16(y-3)^2 - 8(y-11)(y-3) \geq 0$$8(y-3) [2(y-3) - (y-11)] \geq 0$$8(y-3)(2y-6-y+11) \geq 0$$8(y-3)(y+5) \geq 0$This inequality holds when $y \leq -5$ or $y \geq 3$.Comparing this with the given condition $y < a$ or $y \geq b$,we identify $a = -5$ and $b = 3$.

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

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