24. $\quad \int _ { 0 } ^ { \pi / 2 } \sin x d x$ is equal to\\
(A) $\frac { \pi } { 8 } ( 1 + \sqrt { 2 } )$\\
(B) $\frac { \pi } { 4 } ( 1 + \sqrt { 2 } )$\\
(C) $\frac { \pi } { 8 \sqrt { 2 } }$\\
(D) $\frac { \pi } { 4 \sqrt { 2 } }$

Sol. (A)

$$\begin{aligned}
& \int _ { 0 } ^ { \pi / 2 } \sin x d x = \frac { \frac { \pi } { 2 } + 0 } { 4 } \left( \sin ( 0 ) + \sin \left( \frac { \pi } { 2 } \right) + 2 \sin \left( \frac { 0 + \frac { \pi } { 2 } } { 2 } \right) \right) \\
& = \frac { \pi } { 8 } ( 1 + \sqrt { 2 } )
\end{aligned}$$

\begin{enumerate}
  \setcounter{enumi}{24}
  \item Data could not be retrieved.
  \item If $\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) < 0 \forall \mathrm { x } \in ( \mathrm { a } , \mathrm { b } )$ and c is a point such that $\mathrm { a } < \mathrm { c } < \mathrm { b }$, and (c, $\left. \mathrm { f } ( \mathrm { c } ) \right)$ is the point lying on the curve for which $\mathrm { F } ( \mathrm { c } )$ is maximum, then $\mathrm { f } ^ { \prime } ( \mathrm { c } )$ is equal to\\
(A) $\frac { f ( b ) - f ( a ) } { b - a }$\\
(B) $\frac { 2 ( f ( b ) - f ( a ) ) } { b - a }$\\
(C) $\frac { 2 f ( b ) - f ( a ) } { 2 b - a }$\\
(D) 0
\end{enumerate}

Sol. (A)

$$\begin{aligned}
& \left( F ^ { \prime } ( c ) = ( b - a ) f ^ { \prime } ( c ) + f ( a ) - f ( b ) \right. \\
& F ^ { \prime \prime } ( c ) = f ^ { \prime \prime } ( c ) ( b - a ) < 0 \\
& \Rightarrow F ^ { \prime } ( c ) = 0 \Rightarrow f ^ { \prime } ( c ) = \frac { f ( b ) - f ( a ) } { b - a }
\end{aligned}$$

\section*{Comprehension III}
Let ABCD be a square of side length 2 units. $\mathrm { C } _ { 2 }$ is the circle through vertices $\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D }$ and $\mathrm { C } _ { 1 }$ is the circle touching all the sides of the square ABCD . L is a line through A .\\