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if the range of parameter $t$ in the interval  $\left(0, 2\pi\right)$ satisfying

$\displaystyle \frac{(-2x^2+5x-10)}{(\sin t) x^2 + 2(1+ \sin t )x + 9\sin t +4} > 0$  for all real values of  $x$

is  $(a,b)$ and  $a+ b=k\pi$ . then finding  $k$

above $\displaystyle \displaystyle \frac{(2x^2-5x+10)}{(\sin t) x^2 + 2(1+ \sin t )x + 9\sin t +4} <0$

and $\displaystyle 2x^2-5x+10>0$ for all real values of $x$

so $(\sin t)x^2+2(1+\sin t)x+9\sin t+4<0$

wan,t be able to go further, could some help me with this

1 Answers 1

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$ax^2+bx+c<0$ for all real values of $x$ iff $a<0$ and $b^2<4ac$.

In our case, $a = \sin t, b=2(1+\sin t), c = 9\sin t +4$. So, $\sin t$ is negative.

Moreover, $b^2<4ac$ gives $$4(1 + 2\sin t +\sin^2t)<4(9\sin^2 t + 4\sin t)$$ $$1 + 2\sin t +\sin^2t<9\sin^2 t + 4\sin t$$ $$0<8\sin^2 t + 2\sin t - 1$$ $$0<(4\sin t -1)(2\sin t +1)$$ As $\sin t$ is negative, the first term is always negative. So, we want the second term to be negative, which means $\sin t < -\frac12$, i.e. $t\in (\frac{4\pi}3, \frac{5\pi}3)$. So, $a=\frac43$ and $b=\frac53$. Thus, $k = 3$.

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    i did not understand $(4\sin t-1)<0$ for all real $x$, explain me, thanks – 2017-01-16
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    Read the second line. It says $\sin t$ is negative. Then, $4\sin t$ is negative. So, $4\sin t-1$ is also negative. – 2017-01-16
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    thanks Emre, but using walframalpha https://www.wolframalpha.com/input/?i=all+real+t+in+(%5Csin+t)x%5E2%2B2(1%2B%5Csin+t)x%2B9%5Csin+t%2B4%3C0,+t+%5Cin+(0,2%5Cpi) – 2017-01-17
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    What you have written on Wolframalpha has no meaning. I'll be happy to help you out if you did not understand a part of my solution. – 2017-01-17