Question

If the range of the parameter $t$ in the interval $(0,2\pi )$, satisfying $\frac{-2x^2+5x-10}{\left(\sin t\right)x^2+2(1+\sin t)x+(9\sin t+4)}>0$ for all real values of $x$ is $(a,b)$, and $(a+b)=k\pi$, then the value of $k$ is

Answer 1

$-2x^2+5x-10<0,\forall x\in R$
Let $\sin t=a$

$a{x}^2+2(1+a)x+(9a+4)<0,\forall x\in R$
It is only possible if $a<0$ and $4(1+a{)}^2-4a(9a+4)<0\Rightarrow 8a^2+2a-1>0\Rightarrow (4a-1\left)\right(2a+1)>0a>\frac{1}{4},\text{which is not possible}\Rightarrow a<-\frac{1}{2}\Rightarrow \sin t<-\frac{1}{2}\Rightarrow t\in (\frac{7\pi }{6},\frac{11\pi }{6})\Rightarrow a+b=3\pi \Rightarrow k=3$