Question

If $\sqrt{2} \sin ^{2} x+(3 \sqrt{2}+1) \sin x+3>0$ and
$x^{2}-7 x+10<0$, then $x$ lies in the interval

• A. $\left(\frac{-\pi}{4}, \frac{3 \pi}{4}\right)$
• B. $\left(2, \frac{5 \pi}{4}\right)$
• C. $\left(0, \frac{3 \pi}{2}\right)$
• D. $\left(\frac{5 \pi}{4}, 5\right)$

Answer 1

Answer: (B)

We have,
$ \sqrt{2} \sin ^{2} x+(3 \sqrt{2}+1) \sin x+3>0 $
$ \Rightarrow (\sqrt{2} \sin x+1)(\sin x+3)>0 $
$ \Rightarrow \sin x>\frac{-1}{\sqrt{2}} x \in\left(\pi, \frac{5 \pi}{4}\right)\,....(i) $
and $x^{2}-7 x+10 < 0 $
$(x-5)(x-2)<0$
$x \in(2,5)\,....(ii)$
From Eqs. (i) and (ii), we get
$x \in\left(2, \frac{5 \pi}{4}\right)$