Functions Question 587
Question: Let $ 2{{\sin }^{2}}x+3\sin x-2>0 $ and $ x^{2}-x-2<0 $ (x is measured in radians). Then x lies in the interval
[IIT 1994]
Options:
A) $ ( \frac{\pi }{6},\ \frac{5\pi }{6} ) $
B) $ ( -1,\ \frac{5\pi }{6} ) $
C) $ (-1,\ 2) $
D) $ ( \frac{\pi }{6},\ 2 ) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ 2,{{\sin }^{2}}x+3,\sin x-2>0 $
$ 2,{{\sin }^{2}}x+4,\sin x-\sin x-2>0 $
$ 2,\sin x,(\sin x+2)-1,(\sin x+2)>0 $
$ ,(\sin x+2),(2\sin x-1)>0 $
$ 2,\sin x-1>0\Rightarrow \sin x>1/2 $
$ x>\pi /6\Rightarrow x\in (\pi /6,\infty ) $ …..(i)
and $ x^{2}-x-2<0,\Rightarrow ,x^{2}-2x+x-2<0 $
$ x,(x-2)+1,(x-2)<0 $
$ (x+1),(x-2)<0,\Rightarrow x\in (-1,2) $ …..(ii)
From (i) and (ii), $ x\in (\pi /6,2) $ .
BETA
