Question

The inequality sin^–1 (sin 5) > x² – 4x holds if-

• A. x = 2 –$\sqrt { 9 - 2 \pi }$
• B. x = 2 + $\sqrt { 9 - 2 \pi }$
• C. xĪ (2 –$\sqrt { 9 - 2 \pi }$, 2 +$\sqrt { 9 - 2 \pi }$)
• D. x > 2 +$\sqrt { 9 - 2 \pi }$

Answer 1

Answer: (C)

xĪ (2 –$\sqrt { 9 - 2 \pi }$, 2 +$\sqrt { 9 - 2 \pi }$)

Answer 2

Since $\frac { 3 \pi } { 2 } < 5 < 2$,

We have sin 5 < 0, so sin^–1 (sin 5) = 2p – 5

Thus the given inequality can be written as

2p – 5 > x² – 4x or x² – 4x – (2p – 5) < 0

Ž $\left[ x - \frac { 4 - \sqrt { 16 - 4 ( 2 \pi - 5 ) } } { 2 } \right]$

$\left[ x - \frac { 4 + \sqrt { 16 - 4 ( 2 \pi - 5 ) } } { 2 } \right]$< 0

Ž [x – 2 –$\sqrt { 9 - 2 \pi }$] [x – (2 +$\sqrt { 9 - 2 \pi }$)] < 0

x Ī (2 –$\sqrt { 9 - 2 \pi }$), (2 +$\sqrt { 9 - 2 \pi }$).