If [a, b] is the range of the function f(x) = | vedclass

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(B) Let $y = \frac{x+2}{2x^2+3x+6}$.Since $x \in \mathbb{R}$,we have $2yx^2 + 3xy + 6y = x + 2$,which simplifies to $2yx^2 + (3y-1)x + (6y-2) = 0$.For

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$a  0$

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(B) Let $y = \frac{x+2}{2x^2+3x+6}$.Since $x \in \mathbb{R}$,we have $2yx^2 + 3xy + 6y = x + 2$,which simplifies to $2yx^2 + (3y-1)x + (6y-2) = 0$.For $x$ to be a real number,the discriminant $D$ must be greater than or equal to $0$.$D = (3y-1)^2 - 4(2y)(6y-2) \geq 0$.$9y^2 - 6y + 1 - 8y(6y-2) \geq 0$.$9y^2 - 6y + 1 - 48y^2 + 16y \geq 0$.$-39y^2 + 10y + 1 \geq 0$.$39y^2 - 10y - 1 \leq 0$.Factoring the quadratic: $(13y+1)(3y-1) \leq 0$.This gives the range $y \in [-\frac{1}{13}, \frac{1}{3}]$.Thus,$a = -\frac{1}{13}$ and $b = \frac{1}{3}$.Since $a  0$,the correct option is $B$.

If $[a, b]$ is the range of the function $f(x) = \frac{x+2}{2x^2+3x+6}$ for $x \in \mathbb{R}$,then:

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