Question

f(x) = $\int _ { 0 } ^ { x } \left( e ^ { t } - 1 \right) ( t - 1 ) ( \sin t - \cos t ) \sin t d t$, ∀ x ∈ $\left( - \frac { \pi } { 2 } , 2 \pi \right)$ then f(x) is decreasing in

• A. $\left( - \frac { \pi } { 2 } , 0 \right) \cup \left( \frac { \pi } { 4 } , 1 \right) \cup \left( \pi , \frac { \pi } { 4 } \right)$
• B. $\left( - \frac { \pi } { 2 } , \frac { \pi } { 4 } \right) \cup ( 1 , \pi ) \cup \left( \frac { 5 \pi } { 4 } , 2 \pi \right)$
• C. $\left( \frac { \pi } { 4 } , 1 \right) \cup \left( \pi , \frac { 5 \pi } { 4 } \right)$
• D. $\left( 0 , \frac { \pi } { 4 } \right) \cup ( 1 , \pi ) \cup \left( \frac { 5 \pi } { 4 } , 2 \pi \right)$

Answer 1

Answer: (C)

$\left( \frac { \pi } { 4 } , 1 \right) \cup \left( \pi , \frac { 5 \pi } { 4 } \right)$

Answer 2

f '(x) = (e^x − 1) (x − 1) (sinx − cosx) sinx. Sign scheme of f '(x) is

clearly f(x) is increasing in and decreasing in $\left( \frac { \pi } { 4 } , 1 \right) \cup \left( \pi , \frac { 5 \pi } { 4 } \right)$