Question

Let 2sin² x + 3 sin x –2 > 0 and x² – x – 2 < 0, (x is measured in radians). Then x lies in the interval:

• A. (π/6, 5π/6)
• B. (–1, 5π/6)
• C. (–1, 2)
• D. (π/6,2)

Answer 1

Answer: (D)

(π/6,2)

Answer 2

x² – x – 2 < 0 ⇒ (x + 1) (x –2) < 0 ⇒ –1 < x < 2 ….…(i)

2sin²x + 3sin x –2 > 0

⇒ (2 sinx –1) (sinx + 2) > 0 ⇒ sinx < –2 or sinx >1/2

⇒ sin x > $\frac{1}{2}$ ⇒ $\frac{\pi}{6}$< x < $\frac{5\pi}{6}$ …..(ii)

From (i) and (ii), $\frac{\pi}{6}$ < x < 2