Domain and Range!

I have solved this problem without using calculus.

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Let $\sin x = y$ and equate the expression to a value $z$ . Hence, we have

$\dfrac{y^2+y-1}{y^2-y+2} = z \implies y^2(1-z) + y(1+z) - (1+2z) = 0$

We know that $y$ is real and so the discriminant of the quadratic in terms of $y$ must be greater than or equal to zero. This gives

$\Delta \ge 0 \\ \implies {(1+z)}^2 - 4 \left[ - (1+2z) \right] (1-z) \ge 0 \\ \implies -7z^2 + 6z + 5 \ge 0 \\ \implies 7z^2 - 6z - 5 \le 0$

From here, we get

$z \in \left[ \dfrac{3-2\sqrt{11}}{7} , \dfrac{3+2\sqrt{11}}{7} \right]$

But since $\dfrac{3+2\sqrt{11}}{7} \approx 1$ , and the expression assumes its maximum value of $\dfrac 12$ at $y = \sin x = 1$ , therefore the upper bound is $\dfrac 12$ .

Which gives the range as

$x \in \left[ \dfrac{3-2\sqrt{11}}{7} , \dfrac 12 \right]$

Thus, $A+B+C+D+E+F = 3+2+11+7+1+2 = \boxed{26}$ .

[(3-2√11)/7,1/2]. By differentiating and finding roots of the equation differentiated by equating it to zero..