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# If $x$ is real, the expression $\dfrac{{x + 2}}{{2{x^2} + 3x + 6}}$ takes all value in the interval A. $\left( {\dfrac{1}{{13}},\dfrac{1}{3}} \right)$ B. $\left[ { - \dfrac{1}{{13}},\dfrac{1}{3}} \right]$ C. $\left( { - \dfrac{1}{3},\dfrac{1}{{13}}} \right)$ D. None of these

Last updated date: 22nd Jul 2024
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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 $<$ .