Question

Complete solution set of the equation |x² - 1 + cosx| = |x²− 11 + |cosx| belonging to (-2π, π), is

• A. $\left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right] \sim ( - 1,1 )$
• B. $\left[ - \frac { 3 \pi } { 2 } , - \frac { \pi } { 2 } \right] \cup [ - 1,1 ] \cup \left[ \frac { \pi } { 2 } , \pi \right)$
• C. $\left[ - \frac { 3 \pi } { 2 } , - \frac { \pi } { 2 } \right] \cup \left[ \frac { \pi } { 2 } , \pi \right)$
• D. $\left( - 2\pi, - \frac{3\pi}{2} \right\rbrack \cup \left\lbrack - \frac{\pi}{2}, - 1 \right\rbrack \cup \left\lbrack 1,\frac{\pi}{2} \right\rbrack$

Answer 1

Answer: (D)

$\left( - 2\pi, - \frac{3\pi}{2} \right\rbrack \cup \left\lbrack - \frac{\pi}{2}, - 1 \right\rbrack \cup \left\lbrack 1,\frac{\pi}{2} \right\rbrack$

Answer 2

|x² - 1 + cosx| = |x² -1| + |cosx|. It implies that (x² -1).

cos ≥ 0 because |x + y| = |x| + |y| if x y ≥ 0. Sign scheme of

(x² - 1) cosx is

Thus solution is

$\left[ - \frac { \pi } { 2 } , - 1 \right] \cup \left[ 1 , \frac { \pi } { 2 } \right] \cup \left( - 2 \pi , \frac { 3 \pi } { 2 } \right]$