If x is real, the expression frac{{x + 2}}{{2 | vedclass

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Question

(b) If the given expression be $y$, then $y = 2{x^2}y + (3y - 1)x + (6y - 2) = 0$

If $y 
e 0$then $\Delta \ge 0$ for real $x$ i.e. ${B^2} - 4AC \g

Vedclass · Accepted Answer

$\left[ { - \frac{1}{{13}},\frac{1}{3}} \right]$

Vedclass · Answer

(b) If the given expression be $y$, then $y = 2{x^2}y + (3y - 1)x + (6y - 2) = 0$

If $y 
e 0$then $\Delta \ge 0$ for real $x$ i.e. ${B^2} - 4AC \ge 0$

or -$39{y^2} + 10y + 1 \ge 0$ or $(13y + 1)(3y - 1) \le 0$

==> $ - 1/13 \le y \le 1/3$

If $y = 0$ then $x = - 2$ which is real and this value of $y$ is included in the above range.

If $x$ is real, the expression $\frac{{x + 2}}{{2{x^2} + 3x + 6}}$ takes all value in the interval

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