Answer

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**Hint:**Firstly, change the given expression completely in the form of a quadratic equation. Then, proceeding further we compare the coefficients of the obtained equation with the general quadratic equation. As we know that the discriminant value for a real variable is always greater than or equal to zero. Substituting the coefficients in the condition, we get the interval for the given expression.

**Formula Used:**For a given real variable, we have the discriminant condition as $D \geqslant 0$ .

\[ \Rightarrow {b^2} - 4ac \geqslant 0\]

**Complete step-by-step solution:**

Suppose the given expression is $y = \dfrac{{x + 2}}{{2{x^2} + 3x + 6}}$ .

On cross-multiplication, we get

$2y{x^2} + 3yx + 6y = x + 2$

Subtracting both the sides $(x + 2)$

$2y{x^2} + 3yx + 6y - x - 2 = 0$

Further, we can write it as

$2y{x^2} + (3y - 1)x + 6y - 2 = 0$

As we can see, the above equation has become a quadratic equation of the form $a{x^2} + bx + c = 0$ .

So, we have $a = 2y$ , $b = (3y - 1)$ and $c = (6y - 2)$

Given, $x$ is real so, $D \geqslant 0$ i.e., \[{b^2} - 4ac \geqslant 0\] .

On substituting values, we have

${(3y - 1)^2} - 4(2y)(6y - 2) \geqslant 0$

which on further solving gives

$ \Rightarrow 9{y^2} + 1 - 6y - 48{y^2} + 16y \geqslant 0$

$ \Rightarrow - 39{y^2} + 10y + 1 \geqslant 0$

On multiplying both the sides by negative sign, the "greater than or equal to" changes to "less than or equal to". Hence, the inequality becomes

$ \Rightarrow 39{y^2} - 10y - 1 \leqslant 0$

Split the middle term and make the quadratic inequality in the form of multiplication of two terms as below,

$39{y^2} - 13y + 3y - 1 \leqslant 0$

$(3y - 1)(13y + 1) \leqslant 0$

Which gives

$\therefore $ $y = \left[ { - \dfrac{1}{{13}},\dfrac{1}{3}} \right]$

**Hence, the correct option is B.**

**Note:**For sets that have a "less than or equal to" or "greater than or equal to" member, the rectangular bracket symbols $[,]$ are used respectively. They are equivalent to the $ \geqslant $ and $ \leqslant $ symbols. Sets with a lower bound or upper bound are denoted by the parenthesis symbols $(,)$, respectively. They are equivalent to the symbols $ > $ and $ < $ .

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